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\begin{document}

\title[Algebraic groups with few subgroups]{Algebraic groups with 
few subgroups}

\author{Skip Garibaldi}
\address{Department of Mathematics \& Computer Science, Emory University, Atlanta, GA 30322, USA}
\email{skip@member.ams.org}
\urladdr{http://www.mathcs.emory.edu/{\textasciitilde}skip/}

\author{Philippe Gille}
\address{UMR 8552 du CNRS, DMA, \'Ecole normale
sup\'erieure, F-75005 Paris, France}
\email{Philippe.Gille@ens.fr}

\subjclass[2000]{20G15}
\thanks{{\tt Version of \today.}}

\begin{abstract}
Every semisimple linear algebraic group over a field $F$ contains nontrivial connected subgroups, namely maximal tori.  In the early 1990s, J.~Tits proved that some groups of type $E_8$ have no others.  We give a simpler proof of his result, prove that some groups of type $\iiiD$ and $\viD$ have no nontrivial connected subgroups, and give partial results for types $\oE$ and $E_7$.  Our result for $\iiiD$ uses a general theorem on the indexes of Tits algebras which is of independent interest.
\end{abstract}

\maketitle

\tableofcontents

\section{Introduction}

In \cite{Ti:deg}, Jacques Tits proved that there exist algebraic groups of type $E_8$ that have very few subgroups.
Specifically, he proved that they are \emph{almost abelian},\footnote{This term was suggested by A. Premet.} i.e., they 
have no proper, nonzero closed connected subgroups other than maximal tori.
Tits also gave a partial result towards the existence of almost abelian groups of type $E_6$ in \cite{Ti9192}.  

Since his work, new tools have appeared, making the problem of construction almost abelian groups easier.  We give a simpler proof of his result for groups of type $E_8$ (\S\ref{E8.sec}), extend his result for groups of type $E_6$ (Prop.~\ref{oE}), and prove a similar partial result for groups of type $E_7$ (Prop.~\ref{E7}).  Roughly speaking, for groups of types $\oE$ and $E_7$, we settle the question of existence of reductive subgroups up to prime-to-$p$ extensions of the base field for each prime $p$.

For groups of type $D_4$, we can do more.  We prove:
\begin{thm}\label{main}
If $G$ is a 
superversal group of type $\iiiD$ or $\viD$ 
over a field of characteristic $\ne 2, 3$,
then $G$ is almost abelian.
\end{thm}

This answers a question by A.~Premet and is proved in \S\ref{D4.sec}.  The proof uses a result about maximal indexes of Tits algebras for groups of outer type (Th.~\ref{sversal.tits}), which is interesting by itself.

A superversal group is roughly speaking the most general group of the given type, see Definition \ref{sversal.def} below. For global fields, such examples do
not occur, see section \ref{global.sec}.

We remark that groups of type $G_2$ or $F_4$ are never almost abelian, because the long roots generate a subgroup of type $A_2$ or $D_4$ respectively.   Thus we have treated the question of existence of reductive subgroups for all exceptional groups, except that for types $E_6$ and $E_7$ we have only done so up to prime-to-$p$ extensions of the base field for each prime $p$ and for inner forms.
%\footnote{And for type $\dE$ we have not proved a theorem over co-2-closed fields.}

\medskip

The idea behind Tits's proofs and ours is to find groups that require a field extension of large degree to split them.  We combine his methods with the recent computations of torsion indexes by Totaro \cite{Tot:E8tor} and the Rost invariant defined in \cite{MG}.

\subsection*{Notation and conventions}
The groups we discuss are linear algebraic groups in the sense of \cite{Borel}.  We write $\mu_n$ for the group whose $F$-points are the $n$-th roots of unity in $F$.  We define ``reductive group" to include the hypothesis that the group is connected.

For a field $F$, we write $\Gal(F)$ for its absolute Galois group, i.e., the Galois group of a separable closure $\Fsep$ of $F$ over $F$.  If $\Gal(F)$ is a $p$-group (i.e., if every finite separable extension of $F$ is of dimension a power of $p$), then we say that $F$ is \emph{$p$-special}.  The subfield $F_p$ of $\Fsep$ consisting of elements fixed by a $p$-Sylow subgroup of $\Gal(F)$ is $p$-special.  We call $F_p$ a \emph{co-$p$-closure} of $F$.

For a central simple $F$-algebra $D$, we write $\SL_1(D)$ for the algebraic groups whose $F$-points are the norm 1 elements of $D$.

We use extensively the notion of Tits index of a semisimple group as defined in \cite{Ti:Cl}.  The list of possible indexes is given in that paper and also in \cite[pp.~320, 321]{Sp:LAG}.  (There is a typo in index \#14 in Springer's table; it needs to have an additional circle darkened.  His Prop.~17.8.2 is correct.)

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{First steps}

Let $G$ be a semisimple algebraic group over a field $F$ of characteristic zero.  By the \emph{type} of $G$, we mean its Killing-Cartan type over an algebraic closure of $F$.

\begin{borel*}\label{isogeny}
The property of being almost abelian is an invariant of isogeny classes. 
That is, 
for every central isogeny $G \ra G'$, the group $G$ is almost abelian if and only if $G'$ is.
\end{borel*}

\begin{borel*}\label{basic.aniso}
\emph{If $G$ is isotropic, then $G$ is not almost abelian}.  Indeed, if $G$ is isotropic then it contains a rank 1 split torus over $F$, and when $G$ has absolute rank 2 this suffices to show that $G$ is not almost abelian.
If $G$ is isotropic and of type $A_1$, then it is isogenous to $\SL_2$ and so has a Borel subgroup defined over $F$.
\end{borel*}



\begin{borel*}\label{outline}
Our general plan is the following.  For a specific semisimple group $G$ over a field $F$, 
we prove three statements:
\begin{enumerate}
\item \label{o.aniso} $G$ is $F$-anisotropic.
\item \label{o.max} $G$ does not contain any $F$-defined proper semisimple subgroups of the same absolute rank as $G$.
\item \label{o.tori} Every nonzero $F$-torus in $G$ is maximal.
\end{enumerate}
This will show that every nonzero proper (connected) reductive subgroup of $G$ is a maximal torus.
Indeed, let $H$ be a nonzero proper reductive subgroup of $G$.  %By \eqref{o.aniso} and \ref{basic.red}, $H$ is reductive.  
By \eqref{o.tori}, $H$ has the same absolute rank as $G$.  Again by \eqref{o.tori}, the central torus of $H$ is either trivial or all of $H$, i.e., $H$ is semisimple (excluded by \eqref{o.max}) or a maximal torus, as claimed.
\end{borel*}


In order to prove \eqref{o.max}, we note that an $F$-defined semisimple subgroup $H$ of $G$ of the same absolute rank contains---and so is normalized by---a maximal torus of $G$, hence $H$ is generated over an algebraic closure of $F$ by root subgroups of $G$ \cite[13.20]{Borel}.  That is, the roots of $H$ are a subsystem of the roots of $G$, and $H$ is of one of the the types described in \cite{BoSieb} or \cite[Th.~5.3]{Dynk:ssub}, even though we do not assume that $F$ has characteristic zero.



In practice, item \eqref{o.tori} is the difficult one.  For this, our work is cut in half by the following elementary observation.

\begin{symm} \label{symm}
Let $G$ be a reductive algebraic group of absolute rank $n$.  If $G$ contains an $F$-torus of rank $m$, then it contains an $F$-torus of rank $n - m$.
\end{symm}

\begin{proof}
Let $S$ be an $F$-torus in $G$ of rank $m$.  It is contained 
in a maximal $F$-torus $T$ of rank $n$.  We switch to the dual category of finitely generated abelian groups with a continuous action by $\Gal(F)$.  The natural projection $T^* \ra S^*$ splits over $\Q$, i.e., there is a map $f \!: S^* \ot \Q \ra T^* \ot \Q$ that is compatible with the Galois action and such that $f(\beta)\vert_S = \beta$ for all $\beta \in S^*$.  Set $S'_0$ to be the intersection of $f(S^* \ot \Q)$ with $T^*$ and 
\[
S' := \left( \cap_{\alpha \in S'_0} \ker \alpha \right)^0.
\]
Then $S'$ is an $F$-subtorus of $T$
such that $T=S.S'$ and $S\cap S'$ is finite; it is the desired torus.
\end{proof}

\begin{borel*}\label{basic.red}
Suppose now that $F$ is perfect or the characteristic of $F$ is not a torsion prime for $G$ as listed in Table \ref{torsion.table}.  Then every closed, connected unipotent subgroup of $G$ is contained in the radical of a (proper) parabolic subgroup by \cite[3.7]{BoTi:uni1} and \cite[2.6]{Ti:uni2}, hence $G$ is isotropic.  That is, \emph{if $G$ is anisotropic, then every closed, connected subgroup is reductive}.  Arguing as in \ref{outline} allows us to conclude that $G$ is almost abelian.
\begin{table}[t]
\begin{tabular}{|c|c|} \hline
type&torsion primes\\ \hline
$A_n$ ($n \ge 1$), $C_n$ ($n \ge 2$)&none \\
$B_n$ ($n \ge 3$), $D_n$ ($n \ge 4$), $G_2$&2 \\
$F_4, E_6, E_7$&2, 3\\
$E_8$&2, 3, 5 \\ \hline
\end{tabular}
\caption{Torsion primes for simple root systems, cf.~\cite[1.13]{St:tor}} \label{torsion.table}
\end{table}
\end{borel*}

\begin{borel*}\label{3A1}
We illustrate the material above by classifying the almost abelian groups of type $A_1 \times A_1 \times A_1$.  Every group of this type is isogenous to a transfer $R_{L/F}(\SL_1(Q))$, where $L$ is a cubic \'etale $F$-algebra and $Q$ is a quaternion $L$-algebra.  If $L$ is not a field, then such a group is obviously not almost abelian, because it contains an $F$-defined subgroup of type $A_1$.  In case $L$ is a field, we have:
\end{borel*}

\begin{prop}[``type $A_1 \times A_1 \times A_1$"]
With the notation of \ref{3A1}, if $L$ is a field, then the following are equivalent:
\begin{enumerate}
\item $R_{L/F}(\SL_1(Q))$ is almost abelian.
\item $Q$ is a division algebra and does not contain a quadratic field extension of $F$.
\item $Q$ is a division algebra and the corestriction $\cores_{L/F}(Q)$ is not split.
\end{enumerate}
\end{prop}

\begin{proof}
Suppose (2) holds.  As $Q$ is division, $R_{L/F}(\SL_1(Q))$ is anisotropic.  The group has no proper semisimple subgroups of rank 3 over an algebraic closure, so it cannot have one over $F$.  Finally, if $R_{L/F}(\SL_1(Q))$ contains a rank 1 $F$-torus, then there is a quadratic extension of $F$ splitting $Q$, which is impossible by (2).  By the Symmetry Lemma, $R_{L/F}(\SL_1(Q))$ also has no rank 2 tori.  Applying \ref{outline}, we conclude that $R_{L/F}(\SL(Q))$ is almost abelian.

If (2) fails, i.e., if $Q$ is not division or does contain a quadratic extension $K$ of $F$, then $R_{L/F}(\SL_1(Q))$ is isotropic or contains a rank 1 $F$-torus respectively, hence is not almost abelian.  That is, (1) implies (2).

The equivalence of (2) and (3) is \cite[43.9]{KMRT}.
\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Splitting fields of tori}

We recall the well-known divisibility bounds on finite subgroups of $\GL_n(\Z)$ and $\GL_n(\Q)$ given by Table \ref{finite.table}.  (We remark that every finite subgroup of $\GL_n(\Q)$ is conjugate to a finite subgroup in $\GL_n(\Z)$ by \cite[p.~124]{Se:Lie}, so the bounds are the same whether one takes $\Z$ or $\Q$.)
\begin{table}[ht]
\begin{tabular}{|c|cc|}  \hline
$n$&\upstrut{3.5}\parbox{2in}{finite subgroups of $\GL_n(\Z)$ or $\GL_n(\Q)$ have order dividing}&reference\\[1.5ex] \hline
1&2& \\
2&$2^3$ or $2^2 \cdot 3$&e.g.~\cite[4.1]{KuzPav}\\
3&$2^4 \cdot 3$&\cite{Tahara}\\
4&$2^7 \cdot 3^2$ or $2^3 \cdot 5$ or $2^4 \cdot 3 \cdot 5$&\cite{Dade}, \cite[p.~1137]{Ti:deg} \\ \hline
\end{tabular}
\caption{Divisibility bound on finite subgroups of $\GL_n(\Z)$ and $\GL_n(\Q)$} \label{finite.table}
\end{table}

The $F$-isomorphism class of an $n$-dimensional torus $T$ can be specified by a continuous homomorphism $\phi \!: \Gal(F) \ra \GL_n(\Z)$ (representing the natural Galois action on $T^*$), and $T$ is split over $F$ if and only if $\phi$ is trivial.  It follows that every torus of rank $n$ can be split by a field extension of degree $|\im \phi|$, and this order is bounded as in Table \ref{finite.table}.

Let us start with the following elementary fact.

\begin{prop}[``type $^1\!A_{p-1}$''] \label{oAp}
Let $D$ be a division $F$-algebra of prime degree $p$ over a field $F$.  Then the group $\SL_1(D)$  is almost abelian.
\end{prop}

\begin{proof}
We follow \ref{outline}.  The group $\SL_1(D)$ is clearly anisotropic, and \ref{outline}\eqref{o.max} holds because there are no proper semisimple subgroups of $\SL_1(D)$ of rank $p-1$ even over an algebraic closure of $F$.

It remains to show that every non-maximal torus in $\SL_1(D)$ is zero. First, we may replace $F$ by a co-$p$-closure $F_p$,
because $D$ remains division over $F_p$.
The torus $T$ is then a norm one torus $R^1_{K/F}(\Gm)$
for a cyclic Galois  $F$--algebra $K$ of dimension $p$. 
Since $T$ is anisotropic, $K$ is a cyclic field extension of $F$.
The corresponding representation $\Gal(K/F) \to \GL(\Z^{p-1}) \to
\GL(\Q^{p-1})$
is $\Q$-irreducible. Thus $T$ does not admit nonzero proper subtori.
\end{proof}

For contrast, we note that a group of type $^1\!A_{n - 1}$ where $n$ has at least two distinct prime factors is never almost abelian 
because a central simple  algebra of degree $n$
is  decomposable, see e.g.\ \cite[4.5.16]{GilleSz}.

\begin{cor}[``type $^2\!A_{p-1}$"] \label{dAp}
Let $p$ be an odd prime, and let $D$ be a central division algebra of degree $p$ over a quadratic extension $L/F$. 
If $D$ has a unitary involution $\tau$, then the group $\SU(D, \tau)$  is almost abelian.
\end{cor}

\begin{proof}
Over $L$, the group $\SU(D, \tau)$ becomes isomorphic to $\SL_1(D)$, which is almost abelian by Prop.~\ref{oAp}.
\end{proof}

Before we go, we note the following:

\begin{lem} \label{tahara}
Every $3$-dimensional torus over a $3$-special field is isotropic.
\end{lem}

\begin{proof}
Over a 3-special field $F$, a torus $T$ corresponds to a homomorphism 
$\phi \!: \Gal(F) \ra \GL_3(\Z)$ that necessarily factors through $\Zm3$ by Table \ref{finite.table}.  
But any representation $\Zm3 \to \GL_3(\Q)$ admits the trivial representation
as direct summand, hence $T$ is isotropic.  
\end{proof}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Global fields}\label{global.sec}

We assume in this section that $F$ is  a \emph{global} field $F$.

\begin{prop}\label{global} 
If an absolutely almost simple group $G$ over a global field $F$ is almost abelian, then $G$ has type 
type $A_{p-1}$ for some prime $p$.
\end{prop}

Conversely, there exist almost abelian groups of type $^1\!A_{n-1}$ 
and $^2\!A_{n-1}$ over $F$ with $n$ prime by Prop.~\ref{oAp} and Cor.~\ref{dAp}.


For the proof of Prop.~\ref{global}, the only tricky case 
is that of outer type $A$.
From Prasad-Rapinchuk's study of subalgebras of algebras with involutions \cite[App.~A]{PrasadR86},
\cite{PrasadR}, we can derive the following fact. 


\begin{prop}\label{hasse-odd} Let $L$ be a quadratic field extension of $F$.
Let $A$ be a central simple $L$-algebra of  degree $n$ equipped with an involution $\tau$ of the second kind such that $L^\tau = F$.
If $\sqrt{-1} \in F$  or if $n$ is odd, 
there exists a cyclic  field extension $E/F$ of degree $n$ 
such that $E \otimes_F L$ is a field and such that
$(E \otimes_F L, \id \otimes \tau)$ embeds in  $(A, \tau)$. 
\end{prop} 


\begin{proof} 
Let $S$ be a finite set of places such that the group $\SU(A, \tau)$ is quasi-split for places away from $S$.  We assume that $S$ contains a finite place $v_0$ such that $L \ot_F F_v$ is not a field; such a place exists by Cebotarev density and if none are in $S$, then we add one.

For each place $v \in S$, we pick a cyclic \'etale $F_v$-algebra $E_v$ of degree $n$ such that $(E_v, \id) \ot_{F_v} (L \ot F_v, \tau)$ embeds in $(A, \tau) \ot_F F_v$ as an $(L \ot_F F_v)$-algebra with involution.

%If $E$ is a field extension of $K$ such that 
%$E \otimes_F L$ is a field extension, \cite[Th.~3.1]{PrasadR} states that 
%$(E \otimes_F L, \id \otimes \sigma)$ embeds in $(A, \tau)$ if and only
%if $(E \otimes_F F_v) \otimes_F L, 
%\id \otimes \sigma)$ embeds
%in $(A \otimes_L (L \otimes_F F_v),\tau)$ for all places $v$ of $F$. 

%Let $S$ be the  finite subset of places of $F$ such that 
%the unitary group $U(A, \tau)$ is quasi-split  at all places outside $S$.

%

%The idea is to approximate local characters  $(\chi_v)_{v \in S}$
%of degree $n$
%by a global character $\chi \in H^1(F, \Z/n\Z)$ 
%such that the associate cyclic \'etale algebra $E$  is such that
%$E \otimes_F L$ is a field and  
%$( E \otimes_{F} F_v  (L \otimes_F F_v), \id \otimes \sigma)$ embeds
%in \break $(A \otimes_L (L \otimes_F F_v),\tau)$ for all $v \in S$. 
%Indeed the local conditions are automatically satisfied outside 
%$S$ \cite[p. 340]{PlatRap}. 



\smallskip



\noindent{\underline{\emph{Case 1}:}}  Suppose first that $L \ot_F F_v$ is not a field.
Then $A \otimes_L (L \otimes_F F_v) \cong A_v \times A_v$ where $A_v$ is a central $F_v$--algebra.

If $v$ is finite, then by local class field theory,  $A_v$ is split by the unramified extension of $F_v$
of degree $n$.  We take it to be $E_v$.

If $v$ is infinite, then $F_v= \C$ or  $n$ is odd.  In either case,   $A \otimes_L (L \otimes_F F_v)$
is split. Hence $\SU(A, \tau) \times_F F_v$ is split; in this case we delete $v$ from $S$.

\smallskip

\noindent{\underline{\emph{Case 2}:}}
Suppose now that $L \ot_F F_v$ is a field $L_w$.
We claim that the algebra $A \otimes_L L_w$ is split. It is obvious in the 
archimedean case;
in the finite case, it follows from
the bijectivity of  the norm map $\Br(L_w) \to \Br(F_v)$.
Then $(A, \tau) \otimes_F F_v$ is adjoint 
to a hermitian form for $L_w/F_v$.
By diagonalizing it, it follows that  $L_w^n$ embeds in $(A \otimes_F F_v, \tau)$.  We take $E_v$ to be $F_v^n$.


\smallskip
%By Chebotarev density theorem, there exists a finite  place $v_0 \not \in S$
%such that $L \otimes_F F_{v_0} = F_{v_0} \times F_{v_0}$.
%Let $\chi_{v_0} \in H^1(F_{v_0}, \Z/n\Z)$ be a character of order $n$

The map
$$
H^1(F, \Z/n\Z) \to  \prod\limits_{v \in S} \,  H^1(F_v, \Z/n\Z) 
$$
is onto by \cite[Cor.~9.2.3]{NSW} (see also Ex.~1 on p.~459 of that book).
So  there exists a cyclic \'etale $F$-algebra $E$ such that $E \ot F_v$ is isomorphic to $E_v$ for all $v \in S$.  Note that $E$ is a field because $E_{v_0}$ is.  Further, $L \ot_F F_{v_0}$ is not a field, but $E \ot_F F_{v_0}$ is, so $E$ does not contain a copy of $L$.  As $E$ is Galois over $F$, it follows that $E \ot L$ is a field.

By construction, for $v \in S$, the algebra with involution $(E_v, \id) \ot_{F_v} (L \ot F_v, \tau)$ embeds in $(A, \tau) \ot_F F_v$.  But the same is true for $v$ not in $S$, because the  group $\SU(A,\tau) \times F_v$ is quasi-split, cf.~\cite[p.~340]{PlatRap}.
By \cite[Th.~3.1]{PrasadR}\footnote{The January 2008 version of \cite{PrasadR} includes the hypothesis that $\chr F \ne 2$, but it is not used in the proof of Theorem 3.1.} (using that $E \ot L$ is a field), $(E, \id) \ot (L, \tau)$ embeds in $(A, \tau)$ as $F$-algebras with involution.
\end{proof}



%We can now proceed to the proof of Proposition \ref{global}.




\begin{proof}[Proof of Prop.~\ref{global}]
Let $G$ be an absolutely almost simple group over $F$, not of type $A_{n-1}$ for $n$ prime.  We prove case-by-case that $G$ is not almost abelian.

\smallskip
\noindent\underline{\emph{Type $^1\!A$}:} 
If $G$ has type $^1\!A_{n-1}$, then it is isogenous to ${\rm PGL}_1(A)$ where
$A$ is a central simple algebra of degree $n$. By 
the Brauer-Hasse-Noether theorem, $A$ is a cyclic algebra, so contains a 
central simple proper subalgebra $A'$ if $n$ is not prime. Then ${\rm PGL}_1(A')$ is a proper semisimple
subgroup of  ${\rm PGL}_1(A)$, hence ${\rm PGL}_1(A)$ is not almost abelian.

\smallskip
\noindent\underline{\emph{Type $^2\!A$}:} 
If $G$ has type $^2\!A_{n-1}$ with $n$ not prime, it is isogenous to ${\rm SU}(B, \tau)$ where
$B$ is a central simple algebra of degree $n \geq 3$ defined over a quadratic separable extension
$L/F$ and equipped with an involution of the second kind. 
If $n$ is even, we know that $B$ contains a quadratic subalgebra $L'/L$ 
stable under $\tau$ \cite[\S5.7, p.~109]{Kn:GC}.
So $\SU(L',\tau)$ is a rank one $k$-torus of ${\rm SU}(B, \tau)$, hence
$\SU(B, \tau)$ is not almost abelian.
If $n$ is odd, Proposition \ref{hasse-odd} shows that
$(B, \tau)$ admits a subalgebra $(E \otimes_F L, \id \otimes \tau)$
where $E$ is a cyclic Galois  $F$-algebra  of degree $n$.
Because $n$ is not prime, there is a field $E_0$ such that $F \subsetneq E_0 \subsetneq E$, and the special unitary group of $(E_0 \ot L, \id \otimes \tau)$ is a non-maximal, non-zero torus in $\SU(B,\tau)$.
Again ${\rm SU}(B,\tau)$ is not almost abelian.

\smallskip
\noindent\underline{\emph{Other types}:} 
Now suppose that $G$ is not of type $A$.
If $F$ has no real places, we know that 
$G/F$ is isotropic  (see \cite[p.~352, Theorem 6.25]{PlatRap}
for the number field case). It is then not almost abelian.  Further, inspecting the list of possible Tits indexes shows that in this case $G$ has a proper parabolic subgroup $P$ defined over $F$ that is reflexive but not a Borel.

If $F$ has real places, then we write $P$ for the parabolic subgroup of $G$ defined over $F(\sqrt{-1})$ from the previous sentence.  It follows that $G$ has an $F$-subgroup $L$ that is conjugate over $F(\sqrt{-1})$ to a Levi subgroup of $P$ \cite[p.~383, Lemma 6.17$^\prime$]{PlatRap}, hence $G$ is not almost abelian.
\end{proof}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Versal groups} \label{versal.sec}

Here we define versal groups, which are the ``most general" groups of a given quasi-split type.

\begin{defn} \label{versal.def}
Fix a field $F_0$, a connected Dynkin diagram $\D$, and an action of the absolute Galois group $\Gal(F_0)$ by graph automorphisms on $\D$, i.e., a continuous homomorphism $\phi \!: \Gal(F_0) \ra \aut(\D)$.  There is a unique quasi-split simple adjoint group $G^q$ whose Tits index has underlying diagram $\D$ and Galois action $\phi$.  Fix a versal $G^q$-torsor $V$ in the sense of \cite[\S5]{Se:ci}; it is a class $V \in H^1(F, G^q)$ for some finitely-generated, regular extension $F$ of $F_0$.  Twisting $G^q$ by $V$ gives an inner form $G$ of $G^q$ defined over $F$.

A \emph{versal group} is a group $G$ (over $F$) obtained in this manner 
from some $F_0, \D, \phi$. 
As an abbreviation, we say that $G$ is a \emph{versal group of type $^a\!T_n$}, for $T_n$ the 
Killing-Cartan type of $\D$ and $a = {|\!\im \phi|}$.
\end{defn}

There is a further notion of versal groups when we allow $\phi$ to vary
inside the morphisms $\phi : \Gal(F) \to \aut(\D)$ with image in the conjugate
in a subgroup $\Gamma \subset \aut(\D)$.

\begin{defn} \label{sversal.def}
Let $F_0$ and $\D$ be as above, and let $G^d$ be the split
adjoint split group of type $\D$.  Fix a 
subgroup $\Gamma \subset \Aut(\D)$. Write $\aut_\Gamma(G^d)$ for the pull back by 
the inclusion $\Gamma \subset \aut(\D)$ of 
the exact sequence 
\[
1 \to G^d \to \aut(G^d) \to \aut(\D) \to 1.
\]
Fix a versal $\aut_\Gamma(G^d)$-torsor $V$ in the sense of \cite[\S5]{Se:ci}; it
is a class $V \in H^1(F, \aut_\Gamma(G^d))$ for some finitely-generated, regular extension $F$ of $F_0$.  Twisting $G^d$ by $V$ gives a form $G$ of $G^d$ defined over $F$.


A \emph{superversal group} is a group $G$ (over $F$) obtained in this manner 
from some $F_0, \D, \Gamma$. 
As an abbreviation, we say that $G$ is a \emph{superversal group of type $^a\!T_n$}, 
for $T_n$ the  Killing-Cartan type of $\D$ and $a = {|\Gamma|}$.

The case $\Gamma=1$, is nothing but the versal 
inner form.
\end{defn}

%In the case $\Gamma=\aut(\D)$, the superversal group  is
% the versal group of  type $\D$.

\begin{borel*}\label{superversal}
Maintain the notation of Definition \ref{sversal.def}.  Suppose that $M$ is an adjoint group with the same Dynkin diagram $\D$, defined over an extension $E$ of $F_0$, and the $*$-action on the Dynkin diagram of $M$ belongs to the image of $H^1(E, \G) \ra H^1(E, \aut(\D))$.  We claim that $M$ is in the image of the map $H^1(E, \aut_\G(G^d)) \ra H^1(E, \aut(G^d))$.

We have to check that this kind of groups
occur as twisted forms of $G^d$ by cocycles with value in $\aut_\Gamma(G^d)$.  
There exists $[z] \in H^1(E, \aut(G^d))$ such that
$M \cong  {_zG^d}$. 
We  consider the commutative diagram
$$
\begin{CD}
H^1(E, \aut(G^d)) @>{\pi}>> H^1(E, \aut(\D)) \\
@AAA @AAA \\
H^1(E,  \aut_\Gamma(G^d)) @>>> H^1(E, \Gamma) . \\
\end{CD}
$$
Our hypothesis is that $\pi_*[z] \in {\rm Im}\Bigl( H^1(E, \Gamma) \to H^1(E, \aut(\D))  \Bigr)$.
According to \cite[\S I.5, prop.~37]{SeCG}, this is equivalent to the fact that 
$_{\pi_*(z)}( \aut(\D)/ \Gamma)(E) \not = \emptyset$.
This finite set is nothing but  $_{\pi_*(z)}\bigl( \aut(G^d)/ \aut_\Gamma(G^d)\bigr)(E)$ hence the claim.
\end{borel*}

%\begin{lem}\label{superversal}
%The superversal group $G/F$ is a versal object 
%among the adjoint groups $M/E$  of Cartan-Killing type $\D$ such that
%the $*$-action $\varphi_M \in H^1(E, \aut(\D))$
%comes from  $H^1(E, \Gamma)$.
%\end{lem}

%\begin{proof} 
%\end{proof}

We conclude this section by the following observation.


\begin{lem}\label{rigid}
Let $\Psi$ be a root datum in the sense of Springer \cite[7.4.1]{Sp:LAG}.
The following are equivalent:

\begin{enumerate}
\item   The superversal group $G/F$ contains a reductive subgroup of type $\Psi$.
\item For every field extension $E/F_0$ and every
adjoint group $M/E$ of type $\D$ such that
$\varphi_M \in {\rm Im}\Bigl( H^1(E, \Gamma) \to H^1(E, \aut(\D)) \Bigr)$,
the group $M/E$ has a reductive subgroup of type $\Psi$. 
\end{enumerate}
\end{lem}


Note that there is an analogous  statement for versal groups.

We use a root datum instead of a root system so as to include also tori and the kernel of the map from the simply connected cover.

\begin{proof}  $(2) \Longrightarrow (1)$ is obvious.
Conversely,  assume that $G/F$ admits
a  $F$-subgroup $H  \subset G/F$ of type $\Psi$.
The $\aut_\Gamma(G)$--torsor $V/F$ arises as the generic fiber of
a  $\aut_\Gamma(G)$--torsor ${\mathfrak V}/X$ where $X$ is a smooth $F_0$-variety
such that $F=F_0(X)$ \cite[\S5]{Se:ci}. Define
${\mathfrak G}/X$ as the twisted $X$-form of $G^d$ by ${\mathfrak V}$.
The idea is to extend $H$ locally.
The quotient $Q=G/H$ is a $G$-variety over $F$.
According to \cite[VI$_B$.10.16]{SGA3}  there exists  an open (non empty) 
Zariski-connected subset
$U \subset X$ and a homogeneous ${\mathfrak G}$-scheme ${\mathfrak Q}/U$ such
that $Q= {\mathfrak Q} \times_U F_0(X)$.
Shrinking $U$ further if necessary, we can assume that 
the origin of $Q$ extends to a point 
$q \in  {\mathfrak Q}(U)$. 
We define the closed $U$-subgroup scheme ${\mathfrak H}:= \Stab_{\mathfrak G}(q)$.
Again, shrinking $U$ if necessary, we can assume that
${\mathfrak H}/U$ is smooth and connected.
By \cite[XIX.2.6]{SGA3}, ${\mathfrak H}/U$ is a reductive group scheme.
By Demazure's type unicity theorem \cite[XXII.2.8]{SGA3},
${\mathfrak H}$ is of type $\Psi$, i.e., all geometric fibers are of type
$\Psi$.

We can now proceed to the proof.
We are given  a field $E/F_0$ and an
adjoint group $M/E$ of type $\D$ such that
its  $*$-action $\varphi_M \in H^1(E, \aut(\D))$
comes from  $H^1(F, \Gamma)$.
By \ref{superversal},  there exists $x \in U(E)$ such
that $M$ is isomorphic to the fiber $x^* {\mathfrak G} $.
Then  $M/E$ admits as the closed  subgroup
$x^*{\mathfrak H}$ which is reductive of type $\Psi$.
\end{proof}


\begin{cor}\label{cor-rigid}
Let $G$ be a superversal group as in Definition \ref{sversal.def}, and assume that the characteristic of $F_0$ is not a torsion prime for $G$.
The following are equivalent:
\begin{enumerate}
\item   The superversal group $G/F$  is not almost abelian.
\item For every field extension $E/F_0$ and every
adjoint group $M/E$ of type $\D$ such that
$\varphi_M \in {\rm Im}\Bigl( H^1(E, \Gamma) \to H^1(E, \aut(\D)) \Bigr)$,
the group $M/E$ is not almost abelian.
\end{enumerate}
\end{cor}

\begin{proof} Again $(2) \Longrightarrow (1)$ is obvious.
Conversely,  assume that $G/F$ admits
a  proper non zero $F$-subgroup $H  \subset G$. Since $G$ is anisotropic,
$H$ is reductive. Hence Lemma \ref{rigid} applies.
\end{proof}



%Let us give a further application of
%Demazure-Grothendieck's theory of group schemes.

%

%\begin{prop}\label{lift} Let $A$ be a henselian local domain 
%with fraction field $K$ and residue field $F$.
%Let ${\mathfrak G}/A$ be a semisimple group scheme and write $G/F$ 
%for its special fiber.

%\begin{enumerate} 
%\item If $G/F$ is isotropic, so is ${\mathfrak G} \times_A K$.
%\item If $G/F$ contains a subtorus of rank $r$, so does  
%${\mathfrak G} \times_A K$.
%\item If $G/F$ contains a proper semisimple subgroup of maximal rank,
%so does  ${\mathfrak G} \times_A K$.  
%\end{enumerate}
%\end{prop}

%\begin{proof}
%If $G/F$ is isotropic, then it has a proper parabolic subgroup
%$P/F$. Let  $I$  be its geometric type and consider the
%$A$-scheme ${\mathfrak X}_I$ of parabolic subgroup schemes of
%${\mathfrak G}/A$ and of type $I$ \cite[\S{XXVI.3}]{SGA3}.
%Our hypothesis reads then ${\mathfrak X}_I(F) \not=\emptyset$.
%Since ${\mathfrak X}_I/A$ is smooth,
%${\mathfrak X}_I(A) \not=\emptyset$ by Hensel's Lemma.
%Thus  ${\mathfrak X}_I(K) \not=\emptyset$ and  ${\mathfrak G} \times_A K$
%is isotropic, which proves (1).

%Claim (2) is a special case of Grothendieck's rigidity theorem
%\cite[XI.5.8]{SGA3}. 

%To prove (3), we assume that $G/F$ contains a proper semisimple subgroup
%$H/F$ of maximal rank.  The Borel-de Siebenthal Theorem \ref{BdS}
%states that 
%the center $\mu:=Z(H)$ of $H$ is not trivial and 
%that $H=Z_G(\mu)^0$.
%Since $\mu$ is an $F$-group of multiplicative type,
%Grothendieck's theorem \cite[XI.4.1]{SGA3} applies again and provides
%an  $A$-subgroup of multiplicative type ${\bf \mu}$
%of   ${\mathfrak G}$.
%The centralizer $Z_{\mathfrak G}({\bf \mu})$ is 
%a smooth $A$-group scheme \cite[XI.5.3]{SGA3}, so we can consider
%its connected component  $Z_{\mathfrak G}({\bf \mu})^0$.
%Since the semisimpleness of the geometric  fibers of 
% $Z_{\mathfrak G}({\bf \mu})^0/A$ is an open property \cite[\S{XIX.2}]{SGA3},
%it follows that   $Z_{\mathfrak G}({\bf \mu})^0$ is a semisimple group scheme
%of the same absolute type as  $Z_G(\mu)^0$, hence it has the
%same rank as $G/F$.
%Thus    $Z_{\mathfrak G}({\bf \mu})^0 \times_A K$ is
%a proper semisimple subgroup of  ${\mathfrak G} \times_A K$
%of maximal rank.
%\end{proof}

%
%\begin{cor}\label{cor-lift} Under the hypothesis of Proposition \ref{lift},
%assume furthermore that ${\mathfrak G} \times_A K$ is almost abelian.
%Then $G/F$ is anisotropic
%and contains no nonzero proper (connected) reductive subgroups.
%\end{cor}

%\begin{borel*}\label{cor-app}
%Practically, Corollary \ref{cor-lift} applies  as follows
%when $F$ has positive characteristic. We choose a henselian DVR $A$ with residue field $F$ and
% fraction field $K$ of characteristic zero. 
%We start with a candidate $G/F$ for being almost abelian.
%%
%%By \cite[XXIV.8]{SGA3}, $G/F$ is the special fiber of a (unique)
%%semisimple group scheme  ${\mathfrak G} /A$.
%%If we know that  ${\mathfrak G} \times_A K$ is almost abelian,
%%Corollary \ref{cor-lift} permits to go down to $G/F$.
%The group $G/F$ is a twisted form of a Chevalley group 
%$G_0 \times_{\Z} F$.
%By \cite[XXIV.8.1.(ii)]{SGA3}, we have a bijection
%$H^1_\et(A, \Aut(G_0)) \cong  H^1_\et(F, \Aut(G_0))$.
%By the theory of forms, it means that  
%$G/F$ is the special fiber of a (unique)
%semisimple group scheme  ${\mathfrak G} /A$.
%If we know that  ${\mathfrak G} \times_A K$ is almost abelian,
%Corollary \ref{cor-lift} permits to go down to $G/F$.
%\end{borel*}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Rost invariant and adjoint groups}

\newcommand{\tGq}{\widetilde{G}^q}
\begin{borel*}\label{m.def}
Let $G^q$ be a quasi-split simple adjoint algebraic group, and write $\tGq$ for its simply connected cover.
The \emph{Rost invariant} of $\tGq$ is a morphism of functors $\Fields_{/F} \ra \Sets$:
\[
r_{\tGq} \!: H^1(*, \tGq) \ra H^3(*, \Zm{\delta}(2)),
\]
where $\delta$ is the \emph{Dynkin index} of $\tGq$ as defined in \cite[\S 10]{MG}.  (We remind that the group $H^3(F, \Zm{d}(2))$ has an unusual definition \cite[App.~A]{MG}.  In case the characteristic of $F$ does not divide $d$, it is the Galois cohomology group $H^3(F, \mu_d^{\otimes 2})$.)
Factor $\delta$ as $\delta_0 m$, where $m$ is maximal with the property of being relatively prime to both $\delta_0$ and the exponent of the group $Z^*(\Fsep)$ of $\Fsep$-points of the cocenter of $\tGq$.  The number $m$ is 1 exactly for $G^q$ of type $^1\!A_n, B_n$, or $C_n$ for all $n$; type $^2\!A_n$ for $n$ odd; or type ${^1\!D_n}$ or ${^2\!D_n}$ for $n \ge 4$.  The number $m$ is $\delta$ when $Z$ is trivial, i.e., for $G^q$ of type $E_8, F_4$, or $G_2$.  
All the other cases---i.e., where $m \ne 1, \delta$---are given in Table \ref{rost.table}.
\end{borel*}

\begin{table}[ht]
\begin{tabular}{|c|ccc|}\hline
type of $G^q$&$m$&Dynkin index $\delta$ of $\tGq$\upstrut{2.8}&exponent of $Z^*(\Fsep)$ \\[0.5ex] \hline
\upstrut{2.6}$^2\!A_n$ ($n$ even)&2&2&$n+1$\\
$\iiiD, \viD$&3&6&2 \\
$\oE$&2&6&3\\
$\dE$&4&12&3\\
$E_7$&3&12&2 \\ \hline
\end{tabular}
\caption{Factorizations of Dynkin indexes for simple groups.  All other quasi-split simple groups have $m = 1$ or $m = \delta$.} \label{rost.table}
\end{table}

For each extension $K/F$, the group $H^3(K, \Zm{\delta}(2))$ can naturally be written as a direct sum of its $m$ and $\delta_0$ torsion parts, i.e., as $H^3(K, \Zm{m}(2)) \oplus H^3(K, \Zm{\delta_0}(2))$.

\begin{prop} \label{rost}
Suppose the characteristic of $F$ does not divide $m$.  Then 
there is a unique invariant
\[
r_{G^q,m} \!: H^1(*, G^q) \ra H^3(*, \Zm{m}(2))
\]
such that the diagram
\[
\begin{CD}
H^1(*, \tGq) @>{r_{\tGq}}>> H^3(*, \Zm{\delta}(2)) \\
@VVV @VVV \\
H^1(*, G^q) @>{r_{G^q,m}}>> H^3(*, \Zm{m}(2))
\end{CD}
\]
commutes.
\end{prop}

%If the characteristic of $F$ does not divide $m$, then $H^3(F, \Zm{m}(2))$ is just $H^3(F, \mmu{m}^{\otimes 2})$.  If additionally $m$ divides 24---as is the case in for all the groups in Table \ref{rost.table}---then $\mmu{m}^{\otimes 2}$ is just $\Zm{m}$ \cite[\S VII, exercise 11]{KMRT} and $r_{G^q,m}$ takes values in $H^3(*, \Zm{m})$.

\begin{proof}%[Proof of Prop.~\ref{rost}]
The result is trivial if $\tGq = G^q$ (i.e., $m = \delta$) or $m = 1$, so we may assume that $m$ is given by Table \ref{rost.table}.  In particular, $m$ is a power of some prime $p$.

Fix an element $y \in H^1(K, G)$ for some extension $K/F$; we claim that there is an extension $L/K$ such that $\res_{L/K}(y)$ is in the image of $H^1(L, \tGq) \ra H^1(L, G)$ and $[L:K]$ is relatively prime to $m$.  Let $K_p$ denote the co-$p$-closure of the perfect closure of $K$.
The group $H^2_\fppf(K_p, Z)$ is zero because the exponent of $Z$ is not divisible by $p$.  The exactness of the sequence $1 \ra Z \ra \tGq \ra G^q \ra 1$ implies the existence of $L$.

Now suppose that $x_1, x_2 \in H^1(L, \tGq)$ both map to $y$.  Then there is a $z \in H^1_\fppf(L, Z)$ such that $z \cdot x_1 = x_2$.  Applying the Rost invariant, we find
\[
r_{\tGq}(x_1) = r_{\tGq}(z \cdot x_2) = r_{\tGq}(x_2)
\]
by \cite[p.~76, Lemma 7]{Gille:inv}, cf.~\cite[Remark 2.5(i)]{GQ}.  Combining this with the previous paragraph proves the existence and uniqueness of $r_{G^q,m}$ by \cite[Prop.~7.1]{G:lens} (which uses the hypothesis on the characteristic of $F$).
\end{proof}

\begin{defn}
Let $G$ be a simple algebraic group over $F$, and write $\Gbar$ for its associated adjoint group.  There is a unique class $\eta \in H^1(F, \Gbar)$ such that the group $\Gbar_\eta$ obtained by twisting $\Gbar$ by $\eta$ is quasi-split \cite[31.6]{KMRT}.  Put $m$ for the natural number defined for the quasi-split group $\Gbar_\eta$ in \ref{m.def}, and suppose that the following holds:
\begin{equation} \label{m.hyp}
\parbox{4in}{\emph{The characteristic of $F$ does not divide $m$ or $G$ has type $F_4, G_2$, or $E_8$.}}
\end{equation}

Write $\theta_\eta$ for the twisting isomorphism $H^1(F, \Gbar_\eta) \iso H^1(F, \Gbar)$ and define:
\[
r(G) := r_{\Gbar_\eta, m}\bigl(\, \theta^{-1}_\eta(0) \bigr) \quad \in H^3(F, \Zm{m}(2)) \subseteq H^3(F, \QZt).
\]
This element depends only on the isomorphism class of $G$.  If $G$ is one of the groups listed in Table \ref{rost.table}, then $m$ divides 24 and $r(G)$ takes values in $H^3(F, \Zm{m})$ \cite[\S{VII}, exercise 11]{KMRT}.

It is easy to see that $r$ is an invariant in the sense of \cite{Se:ci}, e.g., for every extension $K/F$ we have:
\[
\res_{K/F}\bigl(\,  r(G) \, \bigr)= r(G \times_F K).
\]
\end{defn}

\begin{lem} \label{versal.rost}
If $G$ is a versal  group and \eqref{m.hyp} holds, then $r(G)$ has order $m$.
\end{lem}

\begin{proof}
The claim is equivalent to the following statement in the language of Prop.~\ref{rost}: If $V$ is a 
versal $G^q$-torsor, then $r_{G^q,m}(V)$ has order $m$ in $H^3(F,\Zm{m}(2))$.  To prove this, by the specialization property of versal torsors \cite[12.3]{Se:ci}, it suffices to produce an element $y \in H^1(E, G^q)$ for some extension $E/F$ such that $r_{G^q,m}(y)$ has order $m$ in $H^3(E, \Zm{m}(2))$.  
We take $y$ to be the image of a versal $\tGq$-torsor $\tilde{V}$.  By \cite[10.8]{MG}, the element $r_{\tGq}(\tilde{V})$ has order $\delta$, and the claim follows.
\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Groups of type $E_8$} \label{E8.sec}

Besides being split or anisotropic, the possible indexes for a group $G$ of type $E_8$ are given by Table \ref{E8.index}.\begin{table}[ht]
\begin{tabular}{|c|c|} \hline 
\upstrut{3.5}\parbox{1.5in}{semisimple anisotropic kernel of $G$}&order of $r(G)$ \\[2ex] \hline
$E_7$&$\ne 1$, divides 12 \\
$D_7$&2\\
$E_6$&3 or 6\\
$D_6$&2 \\
$D_4$&2 \\ \hline
\end{tabular}
\caption{Possible invariants of a group $G$ of type $E_7$ that is neither split nor anisotropic.} \label{E8.index}
\end{table}


\begin{lem}[{Cf.~ \cite[p.~134, Prop.~5]{Ti9091}}] \label{E8.co5}
Let $G$ be a group of type $E_8$ over a $5$-special field $F$.  If $G$ is not split, then $G$ is anisotropic and every proper, nonzero reductive subgroup of $G$ is a torus of rank $4$ or $8$ or has semisimple type $A_4$ or $A_4 \times A_4$.
\end{lem}

We note that all of these possibilities can occur, see \cite[15.7]{G:lens}.

\begin{proof}
As $G$ is not split, if it is isotropic, then $r(G)$ is not zero by Table \ref{E8.index}.  But $G$ is split by a separable extension of $F$, so $r(G)$ has order dividing 5.  It follows from Table \ref{E8.index} that $G$ is anisotropic.
This implies by Table \ref{finite.table} that $G$ cannot contain tori of ranks 1, 2, or 3.  By the previous sentence, every nonzero reductive subgroup of $G$ has rank 4 or 8.  It is easy to check that any semisimple group of rank 4 over $F$ is isotropic or of type $A_4$.  For a subgroup of rank 8, we consult the list of such in \cite[Table 10]{Dynk:ssub}; in addition to $A_4 \times A_4$, we find:
\begin{enumerate}
\item $A_7 \times A_1, A_5 \times A_2 \times A_1, E_6 \times A_2, E_7 \times A_1, D_6 \times A_1^{\times 2}, D_5 \times A_3, A_3^{\times 2} \times A_1^{\times 2}$
\item $A_8, D_8, A_2^{\times 4}, D_4^{\times 2}, D_4 \times A_1^{\times 4}, A_1^{\times 8}$
\end{enumerate}
Those in list (1) do not occur because they contain tori of ranks $\ne 4, 8$.  Those in list (2) do not occur because they are isotropic over a 5-special field.
\end{proof}

\begin{lem} \label{E8.co3}
Let $G$ be a non-split group of type $E_8$ over a $3$-special field.
\begin{enumerate}
\item If $G$ is isotropic, it has rank $2$ and semisimple anisotropic kernel of type $E_6$.
\item $G$ does not contain a semisimple subgroup of type $A_4$ or $A_4 \times A_4$.
\end{enumerate}
\end{lem}

\begin{proof}
Over a 3-special field, every group of type $E_7$ is isotropic \cite[13.1]{G:lens}, so (1) follows from Table \ref{E8.index}.
Since the field is 3-special, a semisimple group of type $A_4$ or $A_4 \times A_4$ is split, contradicting the statement that $G$ has rank at most 2.
\end{proof}

\begin{thm} \label{E8.th}
In a versal group of type $E_8$, every nonzero proper reductive subgroup is a rank $8$ torus.
\end{thm}

\begin{proof}
Every group of type $E_8$ is simply connected, so the versal group $G$ is obtained by twisting the split group $G^q$ of type $E_8$ by a versal torsor.  By \cite[16.8]{MG}, the Rost invariant $r_{G^q}(G)$ has order 60 in $H^3(F, \Zm{60}(2))$.  In particular, this element is not killed by a co-3-closure nor a co-5-closure of $F$ and so $G$ is not split over such fields.  Combining Lemmas \ref{E8.co5} and \ref{E8.co3}, it suffices to show that $G$ does not contain any rank 4 tori.  

For sake of contradiction, suppose that $G$ has a rank 4 torus $S$ defined over $F$, and that an extension $L/F$ splits $S$.  Then $G$ has $L$-rank at least 4, and so $G$ is $L$-split or has semisimple anisotropic kernel a strongly inner form of $D_4$ over $L$; in either case $G$ is split by 
an extension of $F$ of degree dividing $2[L:F]$.  By \cite{Tot:E8tor}, the torsion index of the compact real Lie group $E_8$ is 2880, hence every extension that splits $G$ has dimension divisible by 2880, hence $[L:F]$ is divisible by $1440 = 2^5 \cdot 3^2 \cdot 5$.
But by Table \ref{finite.table}, every rank 4 torus over $F$ is split by an extension of degree dividing $2^7 \cdot 3^2$ or $2^3 \cdot 5$ or $2^4 \cdot 3 \cdot 5$.  This is a contradiction.
\end{proof}

Applying \ref{basic.red} gives:

\begin{cor}[{Cf.~\cite[p.~1135, Cor.~3]{Ti:deg}}] \label{E8}
Every versal group $G$ of type $E_8$ over a field of characteristic $\ne 2, 3, 5$ is almost abelian. $\hfill\qed$
\end{cor}

In the proof of Th.~\ref{E8.th}, the second paragraph amounts to a reference to Totaro's general result on the torsion index; this has the advantage of being very easy.  Alternatively, one could replace it with a reference to \cite[Lemme 5]{Ti:deg}, which says: \emph{If a group $G$ of type $E_8$ contains a regular rank $4$ torus, then $G$ is split by an extension of dimension dividing $2^7 \cdot 3^2$ or $2^3 \cdot 5$.}  This gives a strengthening of Th.~\ref{E8.th}:
\begin{thm}
If $G$ is a group of type $E_8$ such that $r(G)$ has order $60$, then every nonzero proper reductive subgroup of $G$ is a maximal torus.$\hfill\qed$
\end{thm}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\section{Tits algebras} \label{tits.sec}

Let $G$ be a semisimple algebraic group over $F$.  Tits defined certain invariants $\beta_G(\chi)$ of $G$ in \cite{Ti:R} with the following properties, cf.~\cite[\S27]{KMRT} and 
\cite[\S 4]{M:max}.  Write $Z$ for the center of a simply connected cover of $G$.  The Cartier dual $Z^*$ of the center---the \emph{cocenter}---is an \'etale group scheme.
For $\chi \in Z^*(\Fsep)$, we put $F_\chi$ for the subfield of $\Fsep$ fixed by the stabilizer of $\chi$ in $\Gal(F)$.  
The \emph{Tits algebra} $\beta_G(\chi)$ is an element of $H^2(F_\chi, \Gm)$, the Brauer group of $F_\chi$.

Let $t_G \in H^2_\fppf(F,Z)$ be the Tits class of $G$
\cite[\S 31]{KMRT}.
We note the following properties:
\renewcommand{\theenumi}{\thesection.\arabic{enumi}}
\begin{enumerate}
\item \label{Tits.comp} For all $\chi \in  Z^*(\Fsep)$, the image of $t_G$ under the composite map
\[
H^2_\fppf(F,Z) \xrightarrow{\res} H^2_\fppf(F_\chi,Z)
\xrightarrow{\chi_*}  H^2(F_\chi,\Gm)
\]
is  $\beta_G(\chi)$, see \cite[31.7]{KMRT}.

\item For $G$ of inner type, the Galois group acts trivially on $Z^*(\Fsep)$ and $F_\chi = F$ for all $\chi$.

\item \label{Tits.gal} The Galois action on $Z^*$ is ``compatible" with the Tits algebras in 
the following sense: For $\s \in \Gal(F)$, clearly $F_{\s \chi}$ equals
$\s(F_\chi)$.  Further, the natural map
\[
\s \!: H^2(F_\chi, \Gm) \ra H^2(F_{\s \chi}, \Gm)
\]
sends $\beta_G(\chi)$ to $\beta_G(\s \chi)$. This follows from \eqref{Tits.comp}.
In particular, the index of the division algebra representing
$\beta_G(\chi)$ is the same as the index of the algebra representing 
$\beta_G(\s\chi)$.

\item The Tits algebras are compatible with scalar extension in a natural way. 
If $K/F$ is an extension contained in $\Fsep$, then $K_\chi$ is the compositum of $F_\chi$ and $K$, and $\beta_{G \times K}(\chi)$ is the restriction $\res_{K_\chi/F_\chi}\beta_G(\chi)$.

\item \label{Tits.mult} For every integer $k$, the element $\res_{F_\chi/F_{k\chi}} \beta_G(k\chi)$ equals $k \cdot \beta_G(\chi)$ in $H^2(F_\chi, \Gm)$, by \eqref{Tits.comp}.
\setcounter{equation}{\value{enumi}}
\end{enumerate}

There is a number $n_G(\chi)$---depending only on $\chi$ and the absolute Killing-Cartan type of $G$---such that the index of a division algebra representing $\beta_G(\chi)$ divides $n_G(\chi)$, cf.~\cite[5.4]{Ti:R}.  It is defined as follows:
If $\chi$ is zero then $n_G(\chi)$ is defined to be $1$.  Otherwise, there is a unique Weyl orbit of minuscule weights whose restriction to $Z$ is $\chi$; write $c$ for the cardinality of the orbit.  The number $n_G(\chi)$ is the largest divisor of $c$ that is only divisible by primes dividing the order of $Z^*(\Fsep)$. 
The numbers $n_G(\chi)$ are computed in \cite[\S6]{Ti:R}, \cite{MPW1}, and \cite{MPW2}.  

Property \eqref{Tits.mult} immediately implies:
\begin{enumerate}
\setcounter{enumi}{\value{equation}}
\item \label{Tits.coprime} \emph{If $k$ is relatively prime to the exponent of $Z^*(\Fsep)$, then $\beta_G(k \chi)$ and $\beta_G(\chi)$ have the same index.}  Indeed, $\chi$ and $k \chi$ generate the same subgroup of $Z^*(\Fsep)$, hence $F_{k\chi}$ equals $F_\chi$.  The claim amounts to a standard fact about elements of the Brauer group of $F_\chi$: The index of a class $\alpha$ is equal to the dimension of the smallest separable extension of $F_\chi$ that kills $\alpha$, equivalently the subgroup generated by $\alpha$.
\setcounter{equation}{\value{enumi}}
\end{enumerate}

About the Tits algebras of versal/superversal groups,
our main tool is the following theorem of Merkurjev:

\begin{thm}[Th.~5.2 of \cite{M:max}] \label{versal.tits}
If $G$ is a versal group and $\chi$ is fixed by $\Gal(F)$, then the index of the Tits algebra $\beta_G(\chi)$ is $n_G(\chi)$. $\hfill\qed$
\end{thm}

Alternatively, one can find concrete constructions for $G$ of inner type in \cite{MPW1} and \cite{MPW2}.
We extend this theorem in \S\ref{Tits.2} below.

\renewcommand{\theenumi}{\arabic{enumi}}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Weyl group of $E_6$}  \label{oE.sec1}

This section prepares the ground for the proof of Prop.~\ref{oE}.

\setlength{\unitlength}{0.7cm}
\begin{borel*}\label{E6.roots}
We label the simple roots in a root system of type $E_6$ as in the extended Dynkin diagram
\[
\begin{picture}(7,3)
% put in the circles
\multiput(1,1)(1,0){5}{\circle*{\darkrad}}
\put(3,2.5){\circle*{\darkrad}}
\put(3,1.75){\circle*{\darkrad}}

% put in the lines
\put(1,1){\line(1,0){4}}
\put(3,1.75){\line(0,-1){.75}}

\multiput(3,1.9375)(0,0.1875){3}{\circle*{.01}}

\put(1,0.3){\makebox(0,0.4)[b]{$\alpha_1$}}
\put(2,0.3){\makebox(0,0.4)[b]{$\alpha_3$}}
\put(3,0.3){\makebox(0,0.4)[b]{$\alpha_4$}}
\put(4,0.3){\makebox(0,0.4)[b]{$\alpha_5$}}
\put(5,0.3){\makebox(0,0.4)[b]{$\alpha_6$}}
\put(3.5,1.6){\makebox(0,0.4)[b]{$\alpha_2$}}
\put(3.8,2.3){\makebox(0,0.4)[b]{$-\alpha_0$}}
\end{picture}
\]
where $\alpha_0$ denotes the highest root.  The roots
\begin{equation}\label{A23.simple}
\alpha_1, \alpha_3, \alpha_5, \alpha_6, \alpha_2, -\alpha_0
\end{equation}
are a basis for a root subsystem of type $A_2 \times A_2 \times A_2$, and we refer to it simply as $\Adt$.  Of course, all subsystems of $E_6$ of that type are conjugate under the Weyl group $W(E_6)$ \cite[Exercise VI.4.4]{Bou:g4}.
\end{borel*}

\begin{lem} \label{E6.weyl}
There is exactly one conjugacy class of elements of $W(E_6)$ of order $2$, determinant $1$, and trace $-2$.  Every element of the class normalizes a sub-root-system of type $\Adt$.
\end{lem}

Here, the determinant and trace are computed relative to the action of $W(E_6)$ on the $\Q$-vector space spanned by the roots.

\begin{proof}[Proof of Lemma \ref{E6.weyl}]
The determinant defines an exact sequence
\[
\begin{CD}
1 @>>> \G @>>> W(E_6) @>{\det}>> \pm 1 @>>> 1
\end{CD}
\]
where $\G$ is the finite simple group $U_4(2)$, cf.~\cite[Exercise VI.4.2]{Bou:g4}.  To prove the first claim, it suffices to note that there is exactly one conjugacy class of elements of $\G$ of order 2 and trace $-2$, as can be read off of the 4th line of the table on p.~27 of the Atlas \cite{atlas}.  

For the second claim, take $w$ to be the composition of the reflections in the (pairwise orthogonal) roots
\[
\begin{array}{cc}
\alpha_1 + \alpha_3 + \alpha_4 + \alpha_5&\alpha_2 + \alpha_3 + \alpha_4 + \alpha_5\\
\alpha_5 + \alpha_6&\alpha_1 + \alpha_2 + 2\alpha_3 + 2\alpha_4 + \alpha_5 + \alpha_6
\end{array}
\]
Clearly, $w$ has determinant 1.  It interchanges $\alpha_1, \alpha_3$ with $\alpha_2, -\alpha_0$ respectively and sends $\alpha_5, \alpha_6$ to $-\alpha_5, -\alpha_6$.  It follows that $w$ has trace $-2$ and stabilizes the subsystem $\Adt$.
\end{proof}

%
%\begin{borel*}
%One can construct explicitly an element $w_3$ of the Weyl group of $E_6$ that maps the roots in \eqref{A23.simple} to 
%\[
%\alpha_6, \alpha_5, \alpha_2, -\alpha_0, \alpha_3, \alpha_1
%\]
%respectively, see \cite[Lemma 5.8]{Ch:rinv}.  Furthermore, the product $w_2$ of the simple reflections in the (pairwise orthogonal) roots
%\[
%\alpha_1 + \alpha_3 + \alpha_4 + \alpha_5, \quad \alpha_2 + \alpha_3 + \alpha_4 + \alpha_5, \eand \alpha_1 + \alpha_2 + 2\alpha_3 + 2\alpha_4 + \alpha_5 + \alpha_6
%\]
%interchanges $\alpha_1, \alpha_3, \alpha_5$ with $\alpha_2, -\alpha_0, \alpha_6$ respectively.

%As 
%$w_3^3 = w_2^2 = 1$ and $w_2 w_3 = w_3^2 w_2$,
%these elements generate a copy of the symmetric group on three letters, $\sym_3$, in the stabilizer of $\Adt$ in $W(E_6)$.
%\end{borel*}

\begin{eg} \label{D4E6}
The roots $\alpha_2, \alpha_3, \alpha_4, \alpha_5$ span a sub-root-system of type $D_4$ in $E_6$, and this gives an inclusion $W(D_4) \injects W(E_6)$.  The element $-1$ of $W(D_4)$ maps to an element $w \in W(E_6)$ of order 2.  This element acts as $-1$ on the 4-dimensional subspace spanned by the roots in $D_4$ and fixes the (orthogonal) 2-dimensional subspace spanned by the fundamental weights of $E_6$ dual to $\alpha_1$ and $\alpha_6$.  It follows that $w$ has determinant 1 and trace $-2$.
\end{eg}

%\begin{lem} \label{WE6}
%The stabilizer of $\Adt$ in the Weyl group $W(E_6)$ is 
%${W(\Adt) \rtimes \sym_3}$.
%\end{lem}

%\begin{proof}
%It suffices to prove that an element $s \in W(E_6)$ that permutes the roots \eqref{A23.simple} belongs to $\sym_3$.  The element $s$ permutes the three components of $\Adt$, so after multiplying by elements of $\sym_3$ if necessary, we may assume that $s$ preserves the component with basis $\alpha_5, \alpha_6$ and even fixes those two roots.  Hence $s$ also fixes $\alpha_1 + \alpha_3 + \alpha_2 - \alpha_0$.  But the group of such elements in $W(E_6)$ is generated by the reflections in roots of $E_6$ that it contains \cite[\S 2.2]{Hum:ref}.  It is an exercise in linear algebra to see that no reflection in a root of $E_6$ fixes these three vectors.
%\end{proof}

%
%\begin{lem}\label{WE66} There exists an 
%element $w'_2$ of $W(E_6)$ of order $2$, of determinant $1$, of trace $-2$ and  which normalizes 
%the subsystem  $A_3^2$.
%\end{lem}

%\begin{proof} 
%Consider in $W(\Adt)=W(A_2) \times W(A_2) \times W(A_2) \subset W(E_6)$ the element $s=(1,s^\sharp ,1,)$ where $s^\sharp$ is the reflexion such that
%$s^\sharp(\alpha_5)= - \alpha_6$ in the second $A_2$.
%So $w'_2 := s \, w_2$  
%interchanges $\alpha_1, \alpha_3$ with $\alpha_2, -\alpha_0$
%respectively and satisfies   $w'_2(\alpha_5)=-\alpha_5$ and 
%$w'_2(\alpha_6)=-\alpha_6$.
%It is  of order $2$, of determinant $1$
%and of trace $-2$.
%\end{proof}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Groups of type $\oE$} \label{oE.sec}

Let $G$ be a group of type $\oE$.  The cocenter $Z^*$ is isomorphic to $\Zm3$.  The numbers $n_G(\chi)$ are 27 for nonzero $\chi$, and the Tits algebras  $\beta_G(1)$ and $\beta_G(2)$ have the same index by \eqref{Tits.coprime}.  We prove:
\begin{prop} \label{oE} 
Let $G$ be a group of type $\oE$ over a field $F$ of characteristic $\ne 2, 3$.

\begin{enumerate}
\item If $F$ is a $p$-special field for some prime $p$, then $G$ contains a subgroup of type
$\Adt$.

\item \cite[p.~129]{Ti9091} If $\beta_G(1)$ has index $27$, then $G$ has no nonzero, proper, closed connected $F$-subgroups other than maximal tori and possibly subgroups of semisimple type $\Adt$.
\end{enumerate}
\end{prop}

Versal groups have $\beta_G(1)$ of index 27, so (2) is not empty.  (Alternatively, an explicit construction of a group of type $\oE$ as in (2) and containing a semisimple subgroup of type $\Adt$ can be found in \cite[pp.~153--155]{MPW2}.)  We note that the statement of Prop.~\ref{oE}(2) applies equally well to groups of type $\dE$, and that there exist such groups $G$ with $\beta_G(1)$ of index 27 by Th.~\ref{sversal.tits}(2) below.

%The proposition reduces the question of whether versal groups of type $\oE$ are almost abelian to a special case of Totaro's question on zero-cycles versus rational points on homogeneous spaces from \cite{Tot:E8tor}. 

%\smallskip
Besides being split or anisotropic, the possible indexes for $G$ are given by Table \ref{E6.index}.
\begin{table}[ht]
\begin{tabular}{|c|cc|} \hline
\upstrut{3.5}\parbox{1.5in}{semisimple anisotropic kernel of $G$}&index of $\beta_G(1)$&$r(G)$ \\[2ex] \hline
$A_2 \times A_2$&3&$0$\\
$D_4$&1&$\ne 0$ \\ \hline
\end{tabular}
\caption{Possible invariants of a group $G$ of type $\oE$ that is neither split nor anisotropic.} \label{E6.index}
\end{table}
\begin{rmk}
We take this opportunity to fill a tiny gap from the classification of possible Tits indexes on p.~311 of \cite{Sp:LAG}.  That reference omits a proof that a group $G$ of type $\oE$ over a field of characteristic 3 cannot have semisimple anisotropic kernel of type $A_5$.  To see this, suppose that $G$ has such a semisimple anisotropic kernel; call it $M$.  It is isomorphic to $\SL_1(D)$ for a division algebra $D$ of index 6.  The fundamental weight $\omega_4$ corresponding to the simple root $\alpha_4$ of $G$ belongs to the root lattice of $E_6$, so the Tits algebra $\beta_G(\omega_4)$ is zero.  On the other hand, the restriction of $\omega_4$ to $M$ has Tits algebra the Brauer class of $D \ot D \ot D$.  By \cite[p.~211]{Ti:R}, these two Brauer classes are equal, so $D^{\ot 3}$ is split and $M$ is isotropic, a contradiction.  
\end{rmk}

\begin{proof}[Proof of Prop.~\ref{oE}(2)]
It suffices to prove this in the case where $F$ is 3-special.
By Table \ref{E6.index}, $G$ is anisotropic, and remains anisotropic over every cubic extension of $F$.  It follows from Table \ref{finite.table} that $G$ cannot contain $F$-tori of dimensions 1, 2, or 3, nor dimensions 4 or 5 by symmetry.  Consulting \cite[Table 10]{Dynk:ssub}, the only possible types for proper semisimple subgroups of maximal rank are $A_5 \times A_1$ and $\Adt$.  But the first type contains a 1-dimensional torus, so it cannot be the type of an $F$-subgroup of $G$.
\end{proof}

\begin{lem} \label{oE.co2}
Let $G$ be a group of type $\oE$ over a field $F$ of characteristic $\ne 2$.  If $G$ is split by an extension of $F$ of degree not divisible by $3$, then $G$ contains a subgroup of type $\Adt$.
\end{lem}

\begin{proof}
The hypothesis on $G$ implies that $G$ is split---in which case we are done---or $G$ is isotropic with semisimple anisotropic kernel of type $D_4$.

Write $E_6, D_4, G_2$ for the split simply connected groups of those types.  We view $D_4$ as a subgroup of $E_6$ via the inclusion from Example \ref{D4E6} and $G_2$ as the subgroup of $D_4$ consisting of elements fixed by the outer automorphism $\phi$ of order 3 that cyclically permutes the simple roots $\alpha_2, \alpha_3, \alpha_5$.  (Recall that the roots of $E_6$ are all relative to some fixed split maximal torus $T$.)

Write $n$ for an element of $D_4$ normalizing $T$ and representing $-1$ in the Weyl group of $D_4$.  Replacing $n$ with $n\,\phi(n)\, \phi^2(n)$, we may assume that
$n$ belongs to $G_2$.   Write $A$ for the group generated by $n$ and a maximal torus $T_2 := (T \cap G_2)^0$ in $G_2$.  The natural map $H^1(F, A) \ra H^1(F, G_2)$ is surjective, cf.~\cite[p.~1060]{ChSe}.  Further, the image of $H^1(F, G_2) \ra H^1(F, \aut(D_4))$ contains the class of the semisimple anisotropic kernel of $G$, because 8-dimensional quadratic forms in $I^3$ are similar to Pfister forms.  Combining these two observations with Tits's Witt-type theorem \cite[16.4.2]{Sp:LAG}, we deduce that the simply connected cover of $G$ is obtained by twisting $E_6$ by a 1-cocycle $\eta$ with values in $A$.  By Lemma \ref{E6.weyl}, $\eta$ stabilizes a sub-root-system of type $\Adt$, hence the claim.
\end{proof}

\begin{proof}[Proof of Prop.~\ref{oE}(1)]
Suppose that $p = 3$; otherwise the claim is Lemma \ref{oE.co2}.
The adjoint quotient of $G$ is obtained by twisting the split adjoint $E_6$ by a 1-cocycle $\eta$ with values in the normalizer $N$ of a maximal split torus $T$ such that the image of $\eta$ in $N/T$ is contained in a 3-Sylow subgroup of $W(E_6)$ of our choosing.  It suffices to note that one can find a 3-Sylow in $W(E_6)$ that normalizes the subsystem $\Adt$ from \ref{E6.roots}, cf.~\cite[Lemma 5.8]{Ch:rinv}.
\end{proof}

%

%As to the existence of a subgroup of type $\Adt$, 
%it suffices to note firstly that the obvious root subsystem of $E_6$ of type $\Adt$
%(obtained by deleting the central node from the extending Dynkin diagram) is preserved by a $3$-Sylow subgroup of the Weyl group of $E_6$, 
%cf.  \cite[Lemma 5.8]{Ch:rinv}. Then a remark by Rost 
%\cite[8.5]{MPW2} enables us to conclude that
%$G$ has a semisimple group of type $\Adt$.

%\begin{lem} \label{oE.co2}
%Let $G$ be a group of type $\oE$ over a $2$-special field $F$.  
%\begin{enumerate}
%\item It $G$ is not split, it is isotropic with anisotropic kernel
%of type $D_4$,
%\item $G$ contains a subgroup of semisimple type $\Adt$.
%\end{enumerate}

%\end{lem}

%\begin{proof}
%We assume here that $G$ is simply connected. We denoite by $G^d$
%is split form. We shall uses the inclusions of split groups
%$G_2 \subset D_4 \subset F_4 \subset G^d$? 

%\smallskip

%(1) The group $G$ is a strongly inner form of $G^d$.
%Hence we know that $G= {_zG}^d$
%with $[z] \in {\rm Im}\bigl( H^1(F, F_4) \to H^1(F, G^d) \bigr)$ 
%\cite[3.4]{G:rost}.
%Furthermore, since $F$ is a $2$-special field, we know  
%that the map $H^1(F, G_2) \to H^1(F, F_4)$ is surjective \cite[40.16]{KMRT}.
%Hence $G= {_zG}^d$
%with $[z] \in{\rm Im}\bigl( H^1(F, G_2) \to H^1(F, G^d) \bigr)$.
%Since  the connected commutant of $D_4$ inside $G^d$ 
%is a two dimensional split torus $S_2$, it follows that 
%$G= {_zG}^d$ contains ${_zS}_2=S_2$, hence is at least
%twice isotropic. 
%Table 9 enables us to conclude that $G$ is split or
%of anisotropic kernel $D_4$.

%\smallskip

%(2) Let $T_2 \subset G_2$ be a maximal split torus of $G_2$.
%Let  $A=\mu_2 \times \mu_2 \times {\bf Z}/ 2 {\bf Z} \subset N_{G_2}(T_2)$
%the standard non toral $2$-elementary subgroup of $G_2$.
%From \cite{ChSe}, we know that
%the map $H^1(F, A)
%\to H^1(F, G_2)$ is surjective. In other words, we can assume 
%that the cocycle $z$ has  value in $A$.
%The group $A$ embdeds in the standard  maximal non-toral
%$2$-elementary subgroup $(\mu_2)^4 \times {\bf Z}/ 2 {\bf Z}$ 
%of $F_4$,
%which is nothing but the standard  maximal non-toral $2$-elementary 
%subgroup of $G^d$ \cite[table I]{Griess}. So we can assume that $A \subset N_{G^d}(T^d)$ 
%where $T^d$ is a maximal split torus of $G^d$.

%We shall determine the (non-trivial) image, say $w$,  
%of the generator of ${\bf Z}/ 2 {\bf Z}$ in $W(E_6)$.
%The determinant defines an exact sequence
%$$
%1 \to \Gamma \to W(E_6) \to {\pm 1} \to 1
%$$
%where $\Gamma$ is the finite simple group $U_4(2)$.
%The element $w$ preserves the subsystem $D_4$
%and acts on $D_4$ by $-id$.
%Let $T^d$ be a maximal split torus of $G^d$ and
%let  $L^d$ be the standard Levi subgroup  of $G^d$ of type $D_4$
%and $S^d$ its connected center.Since $L^d=C_{G^d}(S^d)$,
%we have a decomposition (equivariant with respect to $w$)
%$$
%( T^d)^* \otimes_{\Z} \Q=   (S^d)^* \otimes_{\Z} \Q \oplus 
%(T^d/S^d)^* \otimes_{\Z} \Q .
%$$
%So we see that ${\rm det}(w)=1$ and ${\rm tr}(w)=-4+2=-2$. 
%So $w \in \Gamma$
%and a glance at  page 26 of the Atlas \cite{atlas}
%tells us that $w$ belongs to the conjugacy class
%$2A$ of the element $w'_2$ defined in Lemma \ref{WE66}
%It follows that $w$  (and $A$) normalizes
%a subsystem  $\Adt$ of $E_6$.
%Let $M^d \subset G^d$ be the subgroup 
%generated by the root groups $U_\alpha$ for $\alpha$
%running over the subsystem $\Adt$. Then $A \subset N_{G^d}(M^d)$.
%By twisting it follows that $G={_zG^d}$ contains 
%${_zM^d}$.
%\end{proof}

%We gather our investigations.

%
%\begin{proof} If $p \not =2,3$, $k=k^6$ and
%$G$ is split according to a result by Tits \cite[th. 2]{Ti9192}.
%So $G$ contains a subgroup of type
%$\Adt$. If $p=2$, this is the preceding Lemma.
%For $p=3$, this is Lemma \ref{oE.co3}.
%\end{proof}

%



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Groups of type $E_7$} \label{E7.sec}

In this section we consider groups $G$ of type $E_7$.  The cocenter of the simply connected cover of $G$ is $\Zm2$, and the Tits algebra $\beta_G(1)$ has index dividing $n_G(1) = 8$.  The element $r(G)$ belongs to $H^3(F, \Zm3)$.  Our goal is to prove the following:

\begin{prop} \label{E7}
Let $G$ be a group of type $E_7$ over a field $F$ of characteristic zero.
\begin{enumerate}
\item If $F$ is $p$-special, then $G$ contains semisimple subgroups of type $D_4, \Aot$, and $D_4 \times \Aot$.
\item If $\beta_G(1)$ has index $8$ and $r(G)$ is nonzero, then every proper semisimple subgroup of $G$ (if there are any) is normalized by a maximal torus of $G$ and has type $D_4, \Aot$, or $D_4 \times \Aot$.
\end{enumerate}
\end{prop}

We remark that groups of type $D_4$ and $\Aot$ can be almost abelian by Theorem \ref{main} and Example \ref{3A1}.  Also, groups $G$ as in (2) exist; for example, a versal group satisfies the hypothesis by Lemmas \ref{versal.tits} and \ref{versal.rost}.

\begin{proof}[Proof of Prop.~\ref{E7}(1)]
If the group of type $E_7$ is split (e.g., if $p \ne 2, 3$), then the result is clear.

If $F$ is 2-special, we note that the Weyl group of $E_7$ and of the standard subgroup of type $D_6 \times A_1$ have the same 2-Sylows, so every $F$-group of type $E_7$ has an $F$-subgroup of type $D_6 \times A_1$, cf.\ e.g.\ \cite[14.7]{G:lens}.  Then it suffices to note that every group of type $D_6$ contains a subgroup of type $D_4 \times D_2$, i.e., $D_4 \times A_1^{\times 2}$.

If $F$ is 3-special, then every simply connected group of type $E_7$ is obtained by twisting the split group by a 1-cocycle that normalizes the standard $D_4$ subgroup \cite[13.1]{G:lens}.  The centralizer of this $D_4$ subgroup has type $\Aot$, corresponding to the sub-root-system spanned by
\begin{gather} 
\alpha_7, \quad \label{3A1.roots}
\alpha_2 + \alpha_3 + 2\alpha_4 + 2\alpha_5 + 2\alpha_6 + \alpha_7, \quad \text{and}\\
2\alpha_1 + 2\alpha_2 + 3\alpha_3 + 4\alpha_4 + 3\alpha_5 + 2\alpha_6 + \alpha_7  \notag
\end{gather}
and the conclusion follows.
\end{proof}

\medskip

Besides being split or anisotropic, the possible indexes for $G$ are given by Table \ref{E7.index}.
\begin{table}[ht]
\begin{tabular}{|c|cc|} \hline 
\upstrut{3.5}\parbox{1.5in}{semisimple anisotropic kernel of $G$}&index of $\beta_G(1)$&$r(G)$ \\[2ex] \hline
$E_6$&1&$\ne 0$\\
$D_6$&divides 4&0 \\
$A_1 \times D_5$&2&0 \\
$A_1 \times D_4$&2&0 \\
$D_4$&1&0 \\
$\Aot$&2&0 \\[0.6ex] \hline
\end{tabular}
\caption{Possible invariants of a group $G$ of type $E_7$ that is neither split nor anisotropic.} \label{E7.index}
\end{table}

\begin{lem} \label{E7.co}
Suppose that $G$ is as in Prop.~\ref{E7}(2).  Then:
\begin{enumerate}
\item $G$ is anisotropic and every extension that splits $G$ has dimension 
divisible by $24$.
\item Let $K$ be an extension such that $G$ is $K$-isotropic.
\begin{enumerate}
\item If $3$ does not divide $[K:F]$, then $G$ has $K$-rank $1$ and $8$ divides $[K:F]$.
\item If $3$ divides $[K:F]$ but $4$ does not, then $G$ has $K$-rank $1$ and $6$ divides $[K:F]$.
\end{enumerate}
\end{enumerate}
\end{lem}



\begin{proof}
Since $\beta_G(1)$ has index 8, $G$ is anisotropic by Table \ref{E7.index}.  Further, if $G$ is split over an extension $L/F$, then $L$ must kill both $\beta_G(1)$ and $r(G)$, hence the dimension $[L:F]$ is divisible by 8 and 3 respectively.  This proves (1).

For (2a), the hypothesis on the dimension implies that $K$ does not kill $r(G)$, hence $G \times K$ must have semisimple anisotropic kernel $E_6$ by Table \ref{E7.index}.

For (2b), the hypothesis on the dimension implies that $\beta_G(1) \times K$ has index at least 4, hence the semisimple anisotropic kernel of $G\times K$ must have type $D_6$ by Table \ref{E7.index}.
\end{proof}

The lower bound ``divisible by 24" in Lemma \ref{E7.co}(1) is easily obtained.  One might hope to prove a stronger bound, but this is impossible.  For \emph{any} group $G$ of type $E_7$, the greatest common divisor of the dimensions $[L:F]$---where $L$ ranges over extensions of $F$ splitting $G$---divides 24 by \cite[6.1]{Tot:E8tor}.



\begin{proof}[Proof of Prop.~\ref{E7}(2)]
Because $G$ is anisotropic and remains so over every quadratic extension of $F$ by Lemma \ref{E7.co}(2a), it cannot contain a rank 1 torus nor an $F$-subgroup of type $A_1$.  By symmetry, $G$ contains no rank 6 tori nor any semisimple subgroup of rank 6.

Over a co-3-closure $F_3$ of $F$, the group $G$ has rank 1 by Lemma \ref{E7.co}(2a).  But a group of type $C_3$,  $G_2$, $D_n$ for $n \ne 4$, $A_n$ for $n = 3, 4, 6, 7$, or $A_1^{\times s}$ for $s \ne 3$ has $F_3$-rank $\ge 2$.  So such a group cannot be an $F$-subgroup of $G$.

A group of type $F_4$, $A_2$, or $A_5$ has rank $\ge 2$ over an extension of $F$ of degree dividing 6,  hence it cannot be an $F$-subgroup of $G$ by Lemma \ref{E7.co}(2b).  

Suppose that $H$ is a semisimple $F$-subgroup of $G$ that is not normalized by a maximal torus of $G$.  Table 34 on p.~233 of \cite{Dynk:ssub} gives a list of the possibilities for the type of $H$, using the assumption that $F$ has characteristic zero.  All of them have an isotypic component of type $A_1$, $G_2$, $F_4$, or $C_3$, so they cannot be $F$-subgroups of $G$ by the preceding observations.

Now suppose that $H$ is a semisimple subgroup of $G$ of type $A_2^{\times 2}$.  We claim that its centralizer has semisimple type $A_2$.  It suffices to prove this over an algebraic closure of $F$, where \cite{Dynk:ssub} says that all subgroup of type $A_2^{\times 2}$ are conjugate.  In particular, we may assume that $H$ is generated by the root subgroups corresponding to the highest root, $\alpha_1$, $\alpha_6$, and $\alpha_7$ as in the extended Dynkin diagram
\begin{equation} \label{E7.diag}
\begin{picture}(7.6,1.6)
% put in the circles
\multiput(0.3,0.3)(1,0){7}{\circle*{\darkrad}}
\multiput(5.55,0.3)(0.25,0){3}{\circle*{0.01}}
\put(3.3,1.3){\circle*{\darkrad}}

% put in the lines
\put(0.3,0.3){\line(1,0){5}}
\put(3.3,1.3){\line(0,-1){1}}

%    \put(5.3,0.3){\circle{\lrad}}
%    
\put(0.3,0.4){\makebox(0,0.4)[b]{$\alpha_7$}}
\put(1.3,0.4){\makebox(0,0.4)[b]{$\alpha_6$}}
\put(2.3,0.4){\makebox(0,0.4)[b]{$\alpha_5$}}
\put(3.4,-0.15){\makebox(0,0.4)[b]{$\alpha_4$}}
\put(4.3,0.4){\makebox(0,0.4)[b]{$\alpha_3$}}
\put(5.3,0.4){\makebox(0,0.4)[b]{$\alpha_1$}}
\put(3.7,1.1){\makebox(0,0.4)[b]{$\alpha_2$}}
\end{picture}
\end{equation}
The intersection of the maximal torus $T$ in $G$ with $H$ gives a maximal torus $S$ in $H$.  The centralizer $Z_G(S)$ is reductive with semisimple part generated by the root subgroups for roots $\alpha_2$ and $\alpha_4$.  Clearly, these root subgroups centralize $H$ also, proving the claim.  As $G$ has no $F$-subgroups of type $A_2$, this is a contradiction, so $G$ has no subgroups of type $A_2^{\times 2}$.

To complete the proof of (2), we consult the list of possible semisimple subalgebras of $E_7$ in Table 11 in \cite{Dynk:ssub}, and note that all the types other than $D_4$, $\Aot$, and $D_4 \times \Aot$ have rank 6 or have an isotypic component excluded by the previous arguments.
\end{proof}

%\begin{lem}
%For $G$ a group of type $E_7$ and $H$ a semisimple subgroup of type $D_4$, the group $H\cdot Z_G(H)$ is semisimple of type $D_4 \times \Aot$.
%\end{lem}

%\begin{proof}
%We may assume that $F$ is algebraically closed and fix a pinning for $G$.  Every semisimple subgroup of $G$ of type $D_4$ is conjugate by Dynkin, so we may assume that $H$ is the subgroup generated by $\alpha_2, \alpha_3, \alpha_4, \alpha_5$.  The centralizer in $G$ of the maximal torus $T$ in $H$ is reductive of type $\Aot$ generated by the roots
%\begin{gather} 
%\alpha_7, \quad \label{3A1.roots}
%\alpha_2 + \alpha_3 + 2\alpha_4 + 2\alpha_5 + 2\alpha_6 + \alpha_7, \quad \text{and}\\
%2\alpha_1 + 2\alpha_2 + 3\alpha_3 + 4\alpha_4 + 3\alpha_5 + 2\alpha_6 + \alpha_7  \notag
%\end{gather}
%(The second root is the highest root of the obvious $D_6$ sub-root-system of $E_7$ and the third root is the highest root of $E_7$.)  As these three roots are strongly orthogonal to the roots in $H$, we conclude that $Z_G(H)$ is all of $Z_G(T)$ and $H \cdot Z_G(H)$ is semisimple of type $D_4 \times \Aot$.
%\end{proof}

%\begin{borel*}
%According to Dynkin, there are two conjugacy classes of subgroups $H$ of $E_7$ of type $\Aot$ when $F$ is algebraically closed: One belongs to a maximal subgroup of type $D_4 \times \Aot$ as above and the other consists of groups such that the centralizer in $E_7$ of a maximal torus in $H$ has semisimple type $A_1^{\times 4}$.  Clearly, groups of the second type cannot be subgroups of a group of type $E_7$ satisfying \eqref{E7.hyp}.
%\end{borel*}

%\begin{cor}
%Every group of type $E_7$ over a $p$-special field $F$ contains an $F$-subgroup of type $D_4 \times \Aot$.
%\end{cor}

%\begin{proof}
%If the group of type $E_7$ is split (e.g., if $p \ne 2, 3$), then the result is clear.

%\end{proof}

\begin{borel*}
Suppose now that $G$ is a group of type $E_7$ over a field $F$ of characteristic $\ne 2$, and that $G$ contains a semisimple subgroup of type $D_4 \times \Aot$.  The $\Aot$ component is isogenous to $R_{L/F}(\SL(Q))$ for some cubic \'etale $F$-algebra $L$ and quaternion $L$-algebra $Q$, and \emph{the Tits algebra of $G$ is Brauer-equivalent to $\cores_{L/F}[Q]$}.  To see this, we examine the center of the $\Aot$ component in case $G$ is split.  Viewing $\Aot$ as generated by the roots 
from \eqref{3A1.roots}, the nonidentity elements of the center of each copy of $A_1$ are
\[
h_{\alpha_7}(-1), \quad h_{\alpha_2}(-1) \,h_{\alpha_3}(-1)\, h_{\alpha_7}(-1), \eand h_{\alpha_3}(-1) \, h_{\alpha_5}(-1) \, h_{\alpha_7}(-1)
\]
respectively, where $h_{\alpha_i} \!: \Gm \ra G$ is the cocharacter corresponding to the simple root $\alpha_i$.  In particular, the fundamental weight $\omega_7$ of $G$ restricts to the product map $\mu_2^{\times 3} \ra \mu_2$ on the center of $\Aot$.  By \cite[\S5]{Ti:R}, this proves the claim.

The previous paragraph implies that \emph{the Tits algebra of every group of type $E_7$ over a 2-special field is Brauer-equivalent to a tensor product of 3 quaternion algebras}.  Indeed, it is certainly Brauer-equivalent to $Q_1 \ot \cores_{K/F}(Q_2)$ where $Q_1, Q_2$ are quaternion algebras over $F$ and a quadratic \'etale $F$-algebra $K$ respectively.  But $\cores_{K/F}(Q_2)$ has degree 4 and exponent 2, so it is isomorphic to a tensor product of two quaternion algebras.
\end{borel*}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Maximal indexes of Tits algebras} \label{Tits.2}

The purpose of this section is extend Theorem \ref{sversal.tits} to include also the most interesting cases where the character $\chi$ is not fixed by $\Gal(F)$.  Specifically, we prove:

\begin{thm} \label{sversal.tits}
\begin{enumerate}
\item A superversal group of type $^2\!A_{n-1}$ for $n$ odd is isogenous to $\SU(D, \tau)$ for $D$ a division algebra of degree $n$ with unitary involution $\tau$.  Further, $\la^k D$ has index $\binom{n}{k}$ for $1 \le k \le n$.
\item A superversal group $G$ of type $^2\!E_6$ has Tits algebra $\beta_G(1)$ of index $27$.
\item A superversal group $G$ of type $\iiiD$ or $\viD$ has Tits algebras $\beta_G(\chi)$ of index $8$ for $\chi \ne 0$.  \end{enumerate}
\end{thm}

To prove Theorem \ref{main}, we only need the result for type $\iiiD$. 
The result for type $\viD$ is known (and we  provide here a 
different proof); it is Prop.~8.1 in \cite{KPS:generic}.

The theorem  follows easily from Lemma \ref{sversal.lem}, stated and proved below.

\begin{borel*}\label{rhod.def}
We now set up the proof of Lemma \ref{sversal.lem}.  Fix a root system $\Phi$, a set of simple roots $\D$, a subgroup $\G$ of $\aut(\D)$, and a dominant minuscule weight $\la$.  Write $G^d$ for the split simply connected group with root system $\Phi$ and $(I_\mu, \rho_\mu)$ for the irreducible representation of $G^d$ with highest weight $\mu$.  Define a representation $(V^d, \rho^d)$ of $G^d$ by setting
\[
V^d := \bigoplus_{\mu \in \G \cdot \la} I_\mu \ot I_\mu^*
\]
as a vector space,
where $G^d$ acts on $I_\mu$ via $\rho_\mu$ and \emph{acts trivially on $I_\mu^*$}.  As a representation of $G^d$, we have:
\begin{equation}\label{dimension}
(V^d, \rho^d) \cong \bigoplus_{\mu \in \G \cdot \la} \bigoplus\nolimits^{\dim I_\mu} (I_\mu, \rho_\mu).
\end{equation}


We also define a representation $\rho^A \!: \aut_\G(G^d) \ra \GL(V^d)$ for $\aut_\G(G^d)$ as in Def.~\ref{sversal.def}.  The identity component of $\aut_\G(G^d)$ is the adjoint group, and it acts via the natural representation on $I_\mu$ and $I_\mu^*$.  (The representations $\rho_\mu$ and $\rho_\mu^*$ of $G^d$ do not factor through the adjoint group, but $\rho_\mu \ot \rho_\mu^*$ does.)  For each $\gamma \in \G$, we have $\rho_\mu(\gamma^{-1} g) = \Int(a) \rho_{\gamma \mu}(g)$ for some matrix $a$ depending on $\gamma$ and $\mu$.  Combining these matrices with the natural permutation representation on the weights in $\G \cdot \la$, we find an action of $\aut_\G(G^d)$ on $V^d$.  Furthermore, for $\alpha \in \aut_\G(G^d)$ and $g \in G^d$, we have:
\begin{equation}
\rho^d(\alpha g) = \Int(\rho^A(\alpha))\rho^d(g).
\end{equation}

Now fix a 1-cocycle $z \in Z^1(F, \aut_\G(G^d))$.  We twist $G^d$ and $\rho^d$ by $z$ to obtain a group $G= {_zG}^d$ and a representation $\rho \!: G \ra  {_z\GL}(V^d)$
because  by Hilbert 90 theorem we have an isomorphism ${_z\GL}(V^d) \cong \GL(V^d)$. 

More concretely,  $G$ is the same as $G^d$ over $\Fsep$, but the Galois group $\Gal(F)$ acts on $G(\Fsep)$ via the action $\cdot_z$ defined by $\s \cdot_z g := z_\s \s(g)$.  The 1-cocycle $\rho^A(z)$ takes values in $\GL(V^d)$, and so is $\s \mapsto h^{-1}\,{^\s h}$ for some $h \in \GL(V)(\Fsep)$.  We put:
$\rho := \Int(h) \rho^d$.  Then for $g \in G(\Fsep)$, we have:
\begin{align*}
\s\rho(g) &= \Int(\s h) \rho^d(\s g) = \Int(h) \Int(\rho^A(z_\s)) \rho^d(\s g) \\
&= \Int(h) \rho^d(z_\s \s g) = \rho(\s \cdot_z g),
\end{align*}
which confirms that $\rho$ is defined over $F$ \cite[AG.14.3]{Borel}.
\end{borel*}



\begin{lem}  \label{sversal.lem}
Let $\la$ be a minuscule dominant weight of a root system $\Phi$ with set of simple roots $\D$.  Fix $\G \subset \aut(\D)$.  If 
\[
\gcd \{ \dim I_\la, {|Z^*(\Fsep)|}, {| \G \cdot \la |} \} = 1,
\]
then $\beta_G(\la)$ has index $n_G(\la)$ for every superversal group $G$ constructed from $\D$ and $\Gamma$.
\end{lem}

We define $\beta_G(\la)$ and $n_G(\la)$ where the subscript $\la$ is a weight of $\Phi$ to be $\beta_G(\chi)$ and $n_G(\chi)$ (in the notation of \S\ref{tits.sec}), where $\chi$ is the restriction of $\la$ to the center of $G$.

\begin{proof}[Proof of Lemma~\ref{sversal.lem}]
Let $H$ be a group of type $\Phi$ over a field $F$.  
The fundamental property we use is that there is a unique irreducible representation $J_{H, \la}$ of $H$ over $F$ such that the composition series for $J_{H, \la}$ over an algebraic closure of $F$ has $I_\la$ as a quotient; this follows from \cite[Th.~7.2]{Ti:R}.  This representation satisfies
\begin{equation} \label{Tits.7}
\dim J_{H, \la} = {|\!\Gal(F) \cdot \la |} \cdot \ind \beta_H(\la) \cdot \dim I_\la.
\end{equation}

Twisting the representation $\rho^d$ of $G^d$ defined in \ref{rhod.def} by a superversal torsor gives a representation $W$ of $G$, and the simple quotients in the composition series of $W$ are copies of $J_{G, \la}$.

We specialize the superversal group $G$ to a versal group $G'$ of inner type defined over a field $K'$.  The representation $W$ of $G$ specializes to a representation $W'$ of $G'$.  As $\la$ is fixed by $\Gal(K')$, the composition series for $W'$ over $K'$ has irreducible quotients various copies of $J_{G', \mu}$ for $\mu \in \G \cdot \la$.  

We view the irreducible quotients of the composition series as partitioning the weights of $W$, equivalently the weights of $W'$.  Because $G$-invariant subspaces of $W$ specialize to $G'$-invariant subspaces of $W'$, the partition coming from the composition series under $G'$ is a refinement of the partition of the composition series under $G$.  The dimension of $J_{G',\mu}$ depends only on the $\aut(\D)$-orbit of $\mu$---by \eqref{Tits.7} and \eqref{Tits.gal}---so 
$\dim J_{G,\la}$ is a multiple of $\dim J_{G',\la}$.  That is,
\[
\ind \beta_{G'}(\la) \cdot \dim I_\la \quad \text{divides} \quad {| \G \cdot \la|} \cdot \ind \beta_G(\la) \cdot \dim I_\la
\]
by \eqref{Tits.7}.  By Th.~\ref{versal.tits}, the index of $\beta_{G'}(\la)$ is the number $n_{G'}(\la)$.  Tracing through the definition of $n_{G'}(\la)$, the hypothesis on $\la$ implies that $\ind \beta_{G'}(\la)$ and $| \G \cdot \la |$ are relatively prime, so $\ind \beta_{G'}(\la)$ divides $\ind \beta_G(\la)$, i.e.,
\[
n_\la(G') = \ind \beta_{G'}(\la) \mid \ind \beta_G(\la) \mid n_G(\la).
\]
Since the numbers $n_{G'}(\la), n_G(\la)$ depend only on the root system, they are equal, and this proves the lemma.
\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Groups of type $\iiiD$ or $\viD$} \label{D4.sec}

Let $G$ be a simple group of type $D_4$ over $F$.  The cocenter of its simply connected cover is $\Zm2 \times \Zm2$ with a possibly nontrivial Galois action, and $n_G(\chi)$ is 8 for $\chi \ne 0$.

\begin{lem} \label{D4.co2}
Suppose that $G$ is a group of type $\oD$, $\iiiD$ such 
that $\beta_G(\chi)$ has index $8$ for $\chi \ne 0$.  Then:
\begin{enumerate}
\item $G$ is anisotropic.
\item If $K/F$ is a finite extension of $F$ of dimension not divisible by $8$, then the $K$-rank of $G$ is $\le 1$.
\end{enumerate}
\end{lem}

\begin{proof}
(1) is known.  For $G$ of type $\iiiD$, it is \cite[2.5]{G:iso}, and for type $\oD$ it is even easier.

For (2), suppose first that $G$ has type $\oD$, i.e., that $G$ is isogenous to $\SO(D, \s, f)$ for some central division $F$-algebra of degree 8 with quadratic pair such that the even Clifford algebra $C_0(D, \s, f)$ is isomorphic to $A \times B$ for $A$ and $B$ central division $F$-algebras of degree 8.  By the hypothesis on $K$, the index of $A \ot K$, $B \ot K$, and $D \ot K$ is in all cases at least 2.  It follows that the involution $\s$ is not $K$-hyperbolic \cite[8.31]{KMRT}, which is the claim.

For $G$ of type $\iiiD$, there is a unique cubic Galois extension $L/F$ over which $G$ is of type $\oD$.  The compositum $KL$ has dimension $[KL:L]$ dividing $[K:F]$, and in particular $[KL:L]$ is not divisible by 8.  Replacing $F$ and $K$ with $L$ and $KL$, we are reduced to the type $\oD$ case.
\end{proof}


\begin{lem} \label{D4.co3}
Let $G$ be a group of type $\iiiD$ or $\viD$ over a field of characteristic $\ne 3$.
If $r(G) \ne 0$, then $G$ is anisotropic.
\end{lem}

The hypothesis on the characteristic is here so that $r(G)$ is defined, cf.~Prop.~\ref{rost}.
Recall from Table \ref{rost.table} that $r(G)$ is an element of $H^3(F, \Zm3)$.
If $G$ has type $\oD$ (or $\dD$), then $r(G)$ is automatically zero.

\begin{proof}[Proof of Lemma \ref{D4.co3}]
$G$ is not quasi-split; suppose it is isotropic.
The paper \cite{G:iso} gives a description of $G$ by Galois descent (which also works in characteristic 2 with only cosmetic changes), and in particular $G$ is quasi-split by a separable quadratic extension $K$ of the field.  As $K$ kills $r(G)$ and the dimension of $K$ is not divisible by 3, $r(G)$ is zero.
\end{proof}

\begin{thm} \label{D4}
Let $G$ be a group of type $\iiiD$  over a field of characteristic $\ne 3$. 
If the Tits algebras $\beta_G(\chi)$ have index $8$ for nonzero $\chi$ and $r(G)$ is not zero, then every nonzero proper reductive subgroup of $G$ is a maximal torus.
\end{thm}

\begin{proof}
By Lemma \ref{D4.co3}, $G$ is anisotropic and remains anisotropic over a co-3-closure of $F$.  It follows that $G$ does not contain a rank 1 $F$-torus nor a rank 2 $F$-torus split by an extension of degree dividing 8.
By Lemma \ref{D4.co2}, $G$ contains no rank 2 $F$-tori split by an extension of degree dividing 12.  By symmetry and Table \ref{finite.table}, every nonzero $F$-torus in $G$ is maximal.

Over an algebraic closure, a proper semisimple subgroup of $G$ of rank 4 has type $A_1^{\times 4}$.  An $F$-group of this type becomes isotropic over a separable extension of $F$ of dimension a power of 2, hence by Lemma \ref{D4.co3} cannot be contained in $G$.
\end{proof}

The hypothesis on $r(G)$ was used to exclude transfers of groups of type $A_1$ from a quartic field extension $L/F$.  These are genuine dangers: The paper \cite{ChLM} shows that every isotropic group of type $\iiiD$ or $\viD$ contains a subgroup isogenous to $R_{L/F}(\SL_2)$.  

We close by proving the theorem stated in the introduction.

\begin{proof}[Proof of Th.~\ref{main}]
Let $G$ be a superversal group of type $\iiiD$ over a field $F$ of characteristic $\ne 2, 3$ (itself an extension of some base field $F_0$).
Prop.~\ref{sversal.tits} gives that $\beta_G(\chi)$ has index 8 for nonzero $\chi$.  Because a superversal group is in particular a versal group, we have $r(G) \ne 0$ by Lemma \ref{versal.rost}.  Then $G$ is almost abelian by Th.~\ref{D4} and \ref{basic.red}.

As for type $\viD$, since there are almost abelian groups of
type $\iiiD$ defined over a suitable extension of 
$F_0$, Corollary \ref{cor-rigid} implies that the superversal
group $G$ of type  $\viD$ is almost abelian.
\end{proof}


%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\begin{appendix}
%\section*{Appendix. The Borel-de Siebenthal theorem}

%\renewcommand{\thesection}{A}

%The Borel-de Siebenthal theorem is the following:
%\begin{thm}\label{BdS}
%Let $G/F$ be a reductive group and let $H/F$ be a 
%reductive subgroup of maximal rank. Then $H=Z_G(Z(H))^0$. 
%\end{thm}

%The original theorem \cite{BoSieb} is about compact Lie groups and
% specialists have known for a long time that it generalizes as stated.
%Our goal is to present here a (characteristic free) proof of that result. 

%\begin{borel*}\label{sarko}
%Let $G/F$ be a reductive group and let
%$T/F$ be a maximal torus. Let $S \subset T$ be a subgroup.  We record some observations that are well-known in the case where $S$ is a torus.  First, since $S$ is of multiplicative type,\footnote{This is not relevant?} $Z_G(S)$ is smooth \cite[XI.5.2]{SGA3}, i.e., is an algebraic group.  Further, we claim that $Z_G(S)^0$ is reductive.  To see this, we may assume that $F$ is algebraically closed, so $T$ is split.  
%Let $U_\alpha$ be the root groups
%associated to $\Phi(G,T)$.  The group $Z_G(S)^0$ is generated  by $T$ and 
%root subgroups of $Z_G(S)^0$ \cite[13.20]{Borel}. 
%But $\Phi(G, Z_G(S))= \Phi(G,T) \cap \ker( T^* \to S^*)$,
%hence $Z_G(S)^0$ is generated by $T$ and the root subgroups
%$U_\alpha$ for the $\alpha \in \Phi(G,T) \subset T^*$ 
%mapping to $0$ in $S^*$.  This shows that $Z_G(S)^0$ is reductive.
%\end{borel*}

%\smallskip

%We look at the behavior 
%of Theorem \ref{BdS} under central extensions.

%\begin{lem}\label{BdS-lem} Under the hypotheses of Theorem \ref{BdS},
%let $S$ be a central subgroup (of multiplicative type) of 
%$G$.  Then:
%\begin{enumerate}
%\item  $S \subset H$ and 
%$H/S$ is a reductive subgroup of $G/S$ of maximal rank.

%\item   If  $H/S=Z_{G/S}(Z(H/S))^0$, then  $H=Z_{G}(Z(H))^0$.
%\end{enumerate} 
%\end{lem}

%
%\begin{proof}
%(1) follows from the fact that the center of $G$ 
%is included in all maximal tori of $G$.
%As for (2), we have  an exact sequence
%of algebraic groups
%\[
%\begin{CD}
%1 @>>> S @>>> Z_G(Z(H)) @>>> Z_{G/S}(Z(H/S)).
%\end{CD}
%\]
%Since $S \subset H \subset Z_G(Z(H))$, 
%we have the following exact diagram
%$$  
%\begin{CD}
%1 @>>> S @>>> Z_G(Z(H))^0 @>>> Z_{G/S}(Z(H/S))^0  \\
%&& \mid \mid && \cup && \cup \\
%1 @>>> S @>>> H  @>>> H/S @>>> 1 .\\
%\end{CD}
%$$
%A diagram chase gives (2).
%\end{proof}

%We can now proceed to the proof of Theorem \ref{BdS}.

%\begin{proof} The lemma applied to $Z(G)$ permits to assume 
%that $G$ is semisimple  adjoint.
%We can assume furthermore  that $F$ is algebraically closed.
%We consider a  maximal (split) torus $T$  of $H$.

%\smallskip

%\noindent\emph{Reduction to  the case where $H$ is semisimple:}
%Write $S:=Z(H)^0$ for the connected center of $H$.
%Then $H \subset Z_G(S)$ and $H/S$ is a semisimple subgroup
%of $Z_G(S)/S$. If the result is known when $H$ is semisimple, then 
%we have $H/S= Z_{Z_G(S)/S}( Z(H/S))^0$. Lemma \ref{BdS-lem}
%shows that  $H= Z_{Z_G(S)}( Z(H))^0$, hence 
%$H= Z_{G}(Z(H))^0$.

%\smallskip

%\noindent\emph{Case where $H$ is a maximal proper semisimple group of $G$:}
%We choose compatible orderings on the roots systems
%$\Phi(H,T) \subsetneq \Phi(G,T)= T^*$.
%Let $A$ be the root lattice of $H$, i.e. the sublattice  of $T^*$ 
%generated by $\Phi(H,T)$. The center of $Z(H)$ is a diagonalisable group whose character group is  $Z(H)^*= T^*/A$.

%
%To show that $H=Z_G(H)^0$, we note first that
%$Z_G(H)^0$ is reductive by Lemma \ref{sarko}.
%Since $Z_G(H)^0$ contains the semisimple  group $H$,
%$Z_G(H)^0$ is necessarily semisimple.
%We claim that $Z(H)\not =1$. For sake of contradiction, 
%assume that $A= T^*$, i.e. that 
%$H$ is adjoint. Then our basis for $\Phi(H,T)$ would be a basis for 
%$\Phi(G,T)$, contradiction. So $Z(H) \not =1$
%and $Z_G(Z(H))^0$ is a proper subgroup of $G$. 
%Since $H$ is a maximal semisimple subgroup,
%we conclude that $H =Z_G(H)^0$.

%\smallskip

%
%\noindent\emph{General case:} For reasons of dimension, there is 
%a chain of semisimple groups
%\[
%H=H_0  \subsetneq H_1 \cdots  \subsetneq H_{n-1}  \subsetneq H_n=G
%\]
%such that $H_{i}$ is maximal in $H_{i+1}$ for
%$i=0, \ldots,n-1$.
%By induction on $n$, we can assume that
%\[
%\begin{array}{rclr}
%H &=&Z_{H_{n-1}}(Z(H))^0 \\ 
%&=& \Bigl( Z_G(Z(H)) \cap H_{n-1} \Bigr)^0 \\
%&=& \Bigl( Z_G(Z(H)) \cap Z_G\bigl( Z(H_{n-1}) \bigr) \Bigr)^0 
%&\text{maximal case} \\
%&\supseteq& Z_G(Z(H))^0&\text{because $Z(H_{n-1}) \subset Z(H)$}
%\end{array}
%\]

%\noindent Thus $H= Z_G( Z(H))^0$ as desired.
%\end{proof}

%\end{appendix}
\medskip


\noindent{\small{\textbf{Acknowledgments.} 
We thank Gopal Prasad for useful conversations about the global field case.
The first author's research was partially supported by National Science Foundation grant DMS-0654502.  He thanks the Institut des Hautes Etudes Scientifiques for its hospitality while the work on this article was done.}}

\medskip

\providecommand{\bysame}{\leavevmode\hbox to3em{\hrulefill}\thinspace}
\providecommand{\MR}{\relax\ifhmode\unskip\space\fi MR }
% \MRhref is called by the amsart/book/proc definition of \MR.
\providecommand{\MRhref}[2]{%
\href{http://www.ams.org/mathscinet-getitem?mr=#1}{#2}
}
\providecommand{\href}[2]{#2}
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