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\title{\large\bf Irreducible Subgroups of $GL_1(D)$ Satisfying Group Identities }
\author{R. Ebrahimian, D. Kiani, M. Mahdavi-Hezavehi\\
 ebrahimian@mehr.sharif.edu\\
 kiani@mehr.sharif.edu\\
 mahdavih@sharif.edu \\
 Department of Mathematical Sciences\\ Sharif University
of Technology\\P. O. Box 11365-9415\\Tehran, Iran}
\date{}
\begin{document}
\maketitle
\begin{abstract}
Let $D$ be a finite dimensional $ F $-central division algebra,
and $G$ be an irreducible subgroup of $D^*:=GL_1(D)$. Here we
investigate the structure of $D$ under various group identities
on $G$. In particular, it is shown that when $[D:F]=p^2$, $p$ a
prime, then $D$ is cyclic if and only if $D^*$ contains a
nonabelian subgroup satisfying a group identity.
\end{abstract}
\section {Introduction}
Let $D$ be an $F$-central division algebra of degree $n$. We
define $D$ to be {\it quasi-crossed product} (QCP for short) if it
contains a maximal subfield $L$ with a chain of fields
$F\subsetneqq K \subseteq L$ such that $K/F$ is Galois and $L/K$
is abelian Galois. If $L=K$, then $D$ is crossed product. Also,
$D$ is said to be {\it soluble} QCP if $Gal(K/F)$ is soluble. We
also recall that a subgroup $G$ of $D^*$ is {\it irreducible} if
the $F$-linear hull of $G$, $F[G]=D$. Irreducible soluble
subgroups of the multiplicative group of a divison ring were
first studied by Suprunenko in \cite{su}. Assume that $D$ is a
noncommutative finite dimensional $F$-central division algebra.
It is clear that $D^*$ is an improper irreducible subgroup of
$D^*$. More generally, if $N$ is a noncentral normal subgroup of
$D^*$, then, by Cartan-Brauer-Hua Theorem, $N$ is irreducible.
But $N$ cannot satisfy a group identity since it is known that $N$
contains a noncyclic free subgroup \cite{Go} . So, we  are
interested in proper irreducible subgroups of $D^*$ that are not
normal. To construct such subgroups of $D^*$, take a basis
$\{g_i\}_{1\leq i\leq n}$ of $D/F$ and consider the subgroup
$G=\langle g_i , 1\leq i \leq n \rangle$. Since $G$ is finitely
generated it is not normal in $D^*$ (cf. \cite{MMY}) and we
clearly have $F[G]=D$. Therefore, we can always find nonnormal
proper irreducible subgroups in $D^*$. We shall see later on that
if a nonnormal proper irreducible subgroup $G$ of $D^*$ satisfies
a group identity, then we may be able to determine some
information about the structure of $D$. For example, it is shown
that if $D^*$ contains an irreducible subgroup satisfying a group
identity, then $D$ is QCP. In particular, when $[D:F]=p^2$, $p$ a
prime, any nonabelian subgroup of $D^*$ is irreducible. In this
case, it is proved that $D$ is cyclic if and only if $D^*$
contains a nonabelian subgroup satisfying a group identity.

 \vspace {0.2 cm}
\section {Notations and conventions}
Let $ D $ be a division ring with center $ F $ and $ G $ be a
subgroup of $ D^* $. We denote by $ F[G] $ the $ F $-linear hull
of $ G $, i.e., the $ F $-algebra generated by elements of $ G $
over $ F $. We shall say that $ G $ is {\it irreducible} if $ D =
F[G] $. For any group $ G $ we denote its center by $ Z(G) $.
Given a subgroup $ H $ of $ G $, $ N_G (H) $ means the {\it
normalizer} of $ H $ in $ G $, and $\langle H,K \rangle $ the
group generated by $ H $ and $ K $, where $ K $ is a subgroup of
$ G $. We shall say that $ H $ is {\it abelian-by-finite} if
there is an abelian normal subgroup $ K $ of $ H $ such that $H/K$
 is finite. $ H $ is called {\it center-by-finite} if $ H/Z(H) $
is finite. Let $ S $ be a subset of $ D $, then the {\it
centralizer} of $ S $ in $ D $ is denoted by $ C_{D} (S) $. For
results related to central simple algebras see \cite{ro}.
\vspace{0.2 cm}
\section {Irreducible Subgroups of $D^*$}

This section deals with a few results on division algebras whose
multiplicative groups contain certain subgroups. Using these
results we eventually prove our main theorem which asserts that
if $D^*$ contains an irreducible soluble subgroup, then $D$ is
QCP. We then examine the structure of division algebras whose
multiplicative groups contain irreducible abelian-by-finite
subgroups. Using the results, it is shown that if $D$ has a
nonzero characteristic and $D^*$ contains an irreducible subgroup
satisfying a group identity, then $D$ is QCP. Finally, a
criterion is given for a division algebra $D$ of prime degree to
be cyclic. To be more precise, it is shown that $D$ is cyclic if
and only if $D^*$ contains a nonabelian subgroup satisfying a
group identity. We begin our study with the following lemma, for
a proof see \cite{super}. \vspace{0.2 cm}

{\sc Lemma A.} {\it Let $D$ be a finite dimensional $ F $-central
division algebra. If $ D $ is soluble crossed product, then $D^*$
contains an irreducible soluble subgroup.}
 \vspace{0.2cm}

 We shall also need the
following easy lemma from group theory. \vspace{0.2 cm}

{\sc Lemma 3.1.} {\it Let $G$ be a group in which $Z(G)$ is a
maximal abelian normal subgroup. Then $Z(G/Z(G))$ is trivial.}
\vspace{0.2 cm}

{\sc Proof.} Assume on the contrary that $x\in G\backslash Z(G)$
with $Z(G)x\in Z(G/Z(G))$. Using the fact that for any element
$y\in G$ we have $Z(G)xy=Z(G)yx$, we obtain $yxy^{-1}x^{-1}\in
Z(G)$. Thus, for any $y\in G$ we have $yxy^{-1}\in \langle
Z(G),x\rangle$. Hence, $\langle Z(G),x\rangle$ is an abelian
normal subgroup of $G$ properly containing $Z(G)$. This
contradiction completes the proof. \vspace{0.2 cm}\hfill$\Box$

 We are now in a position to prove the following:
 \vspace{0.2 cm}

{\sc Theorem 3.2.} {\it Let $D$ be a noncommutative finite
dimensional $ F $-central division algebra. Assume that $D^*$
contains an irreducible soluble subgroup. Then there exists an
irreducible soluble subgroup $ S $ and a noncentral abelian
normal subgroup $ A $ of $ S $ such that $ S/A $ is finite.}
\vspace{0.2 cm}

{\sc Proof.} By Lemma~3 of \cite{mah}, we know that any soluble
subgroup $ S $ of $ D^* $ is abelian-by-finite, i.e., there is an
abelian normal subgroup $ A $ in $ S $ of finite index. We may
take in our considerations $ A $ maximal when we deal with
irreducible subgroups of $ D^* $ which are clearly nonabelian.
Hence, to prove the theorem we must show that there exists an
irreducible soluble subgroup $ S $ that is not center-by-finite.
On the contrary, assume that the set $ \Sigma $ of all
irreducible soluble subgroups of $ D^* $ are center-by-finite.
Since $ D $ is of finite dimension over $ F $ we may view $ D^* $
as a linear group in $ GL_n(F) $, where $ n =[D:F] $, and
consequently each element of $ \Sigma $ is a linear group. Thus,
by Huppert-Zassenhaus Theorem (cf. \cite[p. 104]{dix}), the
soluble length $ l $ of each element $ H \in \Sigma $ is at most
$ 2n $. Using this fact and Zorn's Lemma, we conclude that every
element of $\Sigma$ is contained in a maximal element of
$\Sigma$.  Let $G$ be a maximal element of $\Sigma$. It is clear
that $F^*\subseteq G$. By our assumption $F^*$ is a maximal
abelian normal subgroup in $G$ of finite index. We claim that if
$N $ is normal in $ G$, then $Z(F[N])=F$. Using the fact that $G$
normalizes $N$ we conclude that $G$ normalizes $F[N]$. Obviously
$Z(F[N])^*G$ is an irreducible soluble subgroup of $D^*$. Also it
is easily seen that $Z(F[N])^*$ is a normal abelian subgroup of
$Z(F[N])^*G$ of finite index. Therefore, by our assumption we
have $Z(F[N])=F$. Now, $G/F^*$ as a soluble group must contain a
nontrivial abelian normal subgroup. Let $N/F^*$ be a maximal
abelian normal subgroup of $G/F^*$. We claim that $C_G(N)=F^*$.
To see this, put $M=NC_G(N)$. If $M=N$, then $C_G(N)\subseteq N$
and hence $C_G(N)=Z(N)=F^*$. Now suppose that $M\neq N$.
Considering the fact that $M $ is normal in $ G$ and that $N/F^*$
is a maximal abelian normal subgroup of $G/F^*$, we conclude that
$M/F^*$ is not abelian. The group $M/N$ is a nontrivial soluble
subgroup of $G/N$. Thus, by looking at the derived series of
$M/N$ we can find a nontrivial abelian normal subgroup $T/N$ of
$M/N$ such that $T $ is normal in $ G$. By the choice of $N/F^*$,
we obtain that $T/F^*$ is not abelian. Therefore, there exist
$z,t\in T$ such that $ztz^{-1}t^{-1}\not\in F^*$. Using the
definition of $M$, we can find $n,m\in N$ and $p,q\in C_G(N)$
such that $z=np$, $t=mq$. Therefore, we have
$ztz^{-1}t^{-1}=npmqp^{-1}n^{-1}q^{-1}m^{-1}=nmn^{-1}m^{-1}pqp^{-1}q^{-1}$.
On the other hand we have $ztz^{-1}t^{-1}\in N$ since $ T/N $ is
abelian. Thus, $pqp^{-1}q^{-1}\in N$ and hence $pqp^{-1}q^{-1}\in
N\cap C_G(N)=Z(N)=F^*$. Now, using the fact that $N/F^*$ is
abelian we obtain $nmn^{-1}m^{-1}\in F^*$ and so
$ztz^{-1}t^{-1}\in F^*$. This contradiction shows that if $N/F^*$
is a maximal abelian normal subgroup of $G/F^*$, then
$C_G(N)=F^*$ which establishes the claim. Now, set $D_1=F[N]$
and, by the above observation, we note that $Z(D_1)=F$. Since $G$
normalizes $D_1$ we see that for any $g\in G$ we may define a
natural homomorphism $f_g:D_1\longrightarrow D_1$, given by
$f_g(x)=gxg^{-1}$, for any $x\in D_1$. Hence, by Skolem-Noether
Theorem there is an element $a_g\in D_1^*$ such that
$f_g=f_{a_g}$. If $u,v\in D_1$ satisfy $f_u=f_v$, then for any
$x\in D_1$ we have $uxu^{-1}=vxv^{-1}$. Therefore, $u^{-1}v\in
Z(D_1)=F$, which shows that $u, v$ are equal modulo $F^*$, i.e.,
$F^*u=F^*v$.
 Now, for any $x\in D_1$ we have $gxg^{-1}=a_gxa_g^{-1}$, and
 hence $b_g=a_g^{-1}g\in C_D(D_1)$. The fact that $b_g$ commutes with
$a_g$ implies that $a_g$, $g$, and $b_g$ pairwise commute. Set
$A=\bigcup_{g\in G}F^*a_g$ and $B=\bigcup_{g\in G}F^*b_g$. We
claim that both $A$, $B$ are groups. To see this, it is enough to
show that for any $g,h\in G$ we have $F^*a_{g^{-1}}=F^*a_g^{-1}$,
$F^*a_ha_g=F^*a_{hg}$, $F^*b_{g^{-1}}=F^*b_g^{-1}$, and
$F^*b_hb_g=F^*b_{hg}$. We have $b_g=a_g^{-1}g$. For any $x\in
D_1$, $f_{a_{g^{-1}}}(x)=f_{g^{-1}}(x)= g^{-1}xg=
(a_gb_g)^{-1}x(a_gb_g)=a_g^{-1}xa_g=f_{a_g^{-1}}(x)$. Therefore,
$F^*a_g^{-1}=F^*a_{g^{-1}}$. Also we have  $
f_{a_{hg}}(x)=f_{hg}(x)$. Hence
$f_{a_{hg}}(x)=hgxg^{-1}h^{-1}=ha_gxa_g^{-1}h^{-1}=a_ha_gxa_g^{-1}a_h^{-1}=
(a_ha_g)x(a_ha_g)^{-1}=f_{a_ha_g}(x)$. Therefore,
$F^*a_{hg}=F^*a_ha_g$ which shows that $A$ is a group. Next
considering the fact that $a_g\in D_1$ and $b_h\in C_D(D_1)$ we
obtain $
b_hb_g=b_ha_g^{-1}g=a_g^{-1}b_hg=a_g^{-1}a_h^{-1}hg=(a_ha_g)^{-1}hg.$
Thus, since $ A $ is a group we conclude that
$F^*b_hb_g=F^*(a_ha_g)^{-1}hg= F^*a_{hg}^{-1}hg=F^*b_{hg}$,
$F^*b_g^{-1}=F^*a_gg^{-1}=F^*a_{g^{-1}}^{-1}g^{-1}=F^*b_{g^{-1}}.$
Therefore, $B$ is also a group. We claim that $B$ is a soluble
group that is normalized by $G$. To see this, consider the
epimorphism $\theta:G\longrightarrow B/F^*$ given by $ \theta(g)
= F^*b_g $ for all $g\in G$. Hence $ B/F^* $ as a homomorphic
image of a soluble group is soluble, and so $ B $ is also
soluble. Furthermore, since $a_g\in D_1$ and $B\subseteq C_D(D_1)$
for any $g\in G$ we have
$gBg^{-1}=a_gb_gBb_g^{-1}a_g^{-1}=a_gBa_g^{-1}=B$, which
establishes our claim. Therefore, $BG$ is an irreducible soluble
subgroup of $D^*$. By maximality of $G$ we conclude that
$B\subseteq G$. Now, we have $B\subseteq C_D(D_1)\cap G \subseteq
C_G(D_1)=C_G(N)=F^*$. Since $ g = a_gb_g $ we obtain $G\subseteq
D_1$ and hence $D_1=D$, i.e., $ N $ is irreducible. By our
assumption we conclude that $F^*$ is a maximal abelian normal
subgroup  of $N$. Therefore, by Lemma~3.1, $Z(N/F^*)$ is trivial,
which is a contradiction and so the result follows.
 \vspace{0.2 cm}\hfill$\Box$

To prove our main theorems, we shall need the following results,
for a proof see \cite{super}.
 \vspace{0.2 cm}

 {\sc Lemma B.} {\it Let $D$ be a finite dimensional $ F
$-central division algebra. Suppose that $K$ is a subfield of $D$
containing $F$. If $G$ is an irreducible subgroup of $D^*$ such
that $K^*\lhd G$, then $K/F$ is Galois and $G/C_G(K^*)\simeq
Gal(K/F)$.} \vspace{0.2 cm}


{\sc Lemma C.} {\it Let $D$ be a finite dimensional $ F $-central
division algebra and let $G$ be an irreducible subgroup of $D^*$.
If $K$ is a subfield of $D$ containing $F$ such that
$[G:C_G(K^*)]=[K:F]$, then $C_D(K)=F[C_G(K^*)]$.}
 \vspace{0.2 cm}

{\sc Theorem D.} {\it A noncommutative finite dimensional $ F
$-central division algebra $ D $ is abelian crossed product if
and only if there exist an irreducible subgroup $ G $ of $ D^* $
and an abelian normal subgroup $ A $ of $ G $ such that $ G/A $
is abelian. Equivalently $D^*$ contains an irreducible metabelian
subgroup.} \vspace{0.2 cm}


We are now prepared to prove one of our main results of this
section in the following form:
 \vspace{0.2 cm}

{\sc Theorem 3.3.} {\it Let $D$ be a noncommutative finite
dimensional $ F$-central division algebra. If $D^*$ contains an
irreducible soluble subgroup, then $D$ is soluble QCP.}
\vspace{0.2 cm}

{\sc Proof.} As in Theorem~3.2, assume that $ \Sigma $ is the set
of all irreducible soluble subgroups of $ D^* $. It is clear that
every element of $\Sigma$ is contained in a maximal element.
Suppose that $G$ is a maximal element of $\Sigma$. It is easily
checked that $F^*G\in \Sigma$, and by maximality of $G$ we have
$F^*G=G$. Hence $F^*\subseteq G$. Now, by Theorem~3.2, there
exists an irreducible soluble subgroup  $ S $ of $D^*$ containing
a noncentral abelian normal subgroup $ B $ of finite index. Put
$T=F(B)^*S$. Since $S$ normalizes $B$ one can easily show that
$T$ is an irreducible soluble subgroup of $D^*$. Therefore, $ T $
is contained in a maximal element $ G $ of $\Sigma$. Now, by
Lemma~3 of \cite{mah}, there exists a maximal abelian normal
subgroup $A$ in $G$ of finite index. It is clear that
$[F(A)^*:F^*]$ is infinite since otherwise $F$ would be a finite
field (cf. \cite[p. 213]{lam}), hence by Wedderburn's Theorem, we
conclude that $D$ is a field, a contradiction. Therefore, $G/F^*$
is infinite. Since $A$ is a maximal abelian normal subgroup of
finite index in $G$ we conclude that $F^*\neq A$. Set $K=F(A)$.
One may easily show that $G\subseteq N_{D^*}(K^*)$ and that
$K^*G$ is an irreducible soluble subgroup of $D^*$. Therefore, by
maximality of $G$ we have $K^*\subseteq G$. Thus, $A\subseteq
K^*\lhd G$, which shows that $A=K^*$. Now, by Lemma~B, $K/F$ is
Galois and $G/C_G(K^*)\simeq Gal(K/F)$. Therefore, $K/F$ is a
soluble Galois extension. If $ C_D(K)= K $, one may easily
conclude that $ K $ is a maximal subfield of $ D $, hence $L=K$
and so $ D $ is a soluble crossed product division algebra. So,
we may assume that $C_D(K)\neq K $. By Lemma~C, we conclude that
$F[C_G(K^*)]=C_D(K)$. Put $D_1=C_D(K)$. We have
$F[C_G(K^*)]=D_1$, hence $D_1$ contains an irreducible soluble
subgroup. On the other hand by Centralizer Theorem we have
$Z(D_1)=K$. Now, by Lemma~11 of \cite{su} $D_1^*$ contains an
irreducible metabelian subgroup. Therefore, by Lemma~D, we
conclude that $D_1$ is an abelian crossed product division
algebra. Thus, there exists a maximal subfield $ L $ of $D_1$
such that $L/K$ is abelian Galois. Now, $K/F$ is Galois and
$Gal(K/F)$ is soluble. To complete the proof, it is enough to
show that $L$ is a maximal subfield of $D$. To see this, we have
$C_D(L)\subseteq C_D(K)=D_1$, and hence $C_D(L)\subseteq
C_{D_1}(L)=L$, which completes the proof of the theorem.
 \vspace{0.2 cm}\hfill$\Box$


 As a special case of Theorem~3.3, we
obtain the following corollary, which is the main result of
\cite{tig}. \vspace{0.2 cm}

{\sc Corollary 3.4.} {\it Let $D$ be an $ F $-central division
algebra of prime degree $ p $. Then $D$ is cyclic if and only if
$D^*$ contains a nonabelian soluble subgroup.} \vspace{0.2 cm}

{\sc Proof.} the "only if" part is clear with Lemma~A. By Theorem
3.3, $D$ is soluble QCP. Thus, there exists a noncentral subfield
$K$ such that $K/F$ is Galois and $Gal(K/F)$ is soluble. Now, for
dimensional reasons, the proof is complete.
 \vspace{0.2 cm}\hfill$\Box$


The next result establishes the structure of division algebras
whose multiplicative subgroups contain irreducible
abelian-by-finite subgroups. \vspace{0.2 cm}

{\sc Proposition 3.5.} {\it Let $D$ be a noncommutative finite
dimensional $ F $-central division algebra. Assume that $ D^* $
contains an irreducible abelian-by-finite subgroup $G$. If
$CharF=p>0$, then $D$ is QCP.} \vspace{0.2 cm}

{\sc Proof.}  By our assumption, there exists an abelian normal
subgroup $ A $ in $ G $ of finite index. Since $ G $ is
nonabelian we may take $ A $ maximal. We now divide the proof
into two cases:

{\bf Case 1:} Suppose that $ A $ is noncentral in $ G $.
Therefore, $ A $ is also noncentral in $ D^* $. Set $H=F(A)^*G$.
Since $G$ normalizes $A$ one can easily see that $H$ is an
irreducible abelian-by-finite subgroup of $D^*$. Let $K=F(A)$. It
is clear that $H/K^{*}$ is finite and $K\neq F$. By Lemma~B, we
conclude that $K/F$ is Galois and $H/C_H(K^*)\simeq Gal(K/F)$. If
$ C_D(K) = K $, we conclude that $ K $ is a maximal subfield of $
D $ and $L=K$, hence $ D $ is QCP. So, we may assume that $
C_D(K) \neq K $. Now, using Lemma~C, we obtain
$F[C_H(K^*)]=C_D(K)$. Set $ D_1 = C_D(K) $. We have that  $C_H(K^*
)/K^{*}$ is a finite group. On the other hand by the Centralizer
Theorem, we have $Z(D_1)=K^*$. Hence $Z(C_H(K^*))=K^*$. Now, by
group theory, we know that the derived group $C_H(K^*)^{'}$ is a
finite group (cf. \cite[p. 443]{sc}). On the other hand
$CharF=p>0$. Now, by a theorem of \cite[p. 204]{lam}, we conclude
that $C_H(K^*)^{'}$ is cyclic. Therefore, $D_1^*$ contains an
irreducible metabelian subgroup. Hence, by Theorem~D, we conclude
that $D_1$ is an abelian crossed product division algebra. Thus,
there exists a maximal subfield $ L $ of $D_1$ such that $L/K$ is
Galois and $Gal(L/K)$ is abelian. Now, we have $C_D(L)\subseteq
C_D(K)=D_1$, and so $C_D(L)\subseteq C_{D_1}(L)=L$, which shows
that $ L $ is a maximal subfield of $ D $. Therefore, $ D $ is
QCP.

{\bf Case 2:} Assume that $ A = Z(G) $. Since $ G $ is
irreducible we conclude that $ Z(G) $ is central in $ D^* $.
Because $ A $ is maximal in $ G $ we conclude that every abelian
normal subgroup of $ G $ is contained in $ F^*$. Now, we know
that $G/Z(G)$ is finite and hence the derived group $G'$ is a
finite group (cf. \cite[p. 443]{sc}). On the other hand, 
$charF=p>0$, by a theorem of \cite[p. 204]{lam}, we conclude that
$G'$ is cyclic. Therefore, $G$ is an irreducible soluble subgroup
of $D^{*}$ and by Theorem~3.3, we obtain that $D$ is QCP, which
completes the proof. \vspace{0.2 cm}\hfill$\Box$


{\sc Corollary 3.6.} {\it Let $D$ be a noncommutative finite
dimensional division algebra of nonzero characteristic. If $ D^*
$ contains an irreducible subgroup satisfying a group identity,
then $D$ is QCP.} \vspace{0.2 cm}

{\sc Proof.} Suppose that $G$ is an irreducible subgroup of $D^*$
which satisfies a group identity. Let $\{g_1, \ldots,
g_n\}\subseteq G $ be a basis  of the vector space $D$ over the
field $Z(D)$ and set $G_{1}=\langle g_{1} \ldots g_n \rangle$.
Then $G_{1}$ is an irreducible subgroup of $D^{*}$, which is
finitely generated and satisfies a group identity. By Tits'
Alternative (cf. \cite[p. 150]{zal}), we conclude that $G_1$ is
soluble-by-finite. Now, Lemma~3 of \cite{mah} implies that $G_1$
is abelian-by-finite. Therefore, by Theorem~3.5, the result
follows. \vspace{0.2 cm}\hfill$\Box$

Finally, using above results, a criterion is given for cyclicity
of a division algebra of prime index in terms of subgroups of
$D^*$. This criterion includes those given in \cite{tig} and
provides us with various conditions on $D$ to be cyclic.
\vspace{0.2 cm}

 {\sc Theorem 3.7.} {\it Let $D$ be a finite
dimensional $ F$-central division algebra of prime degree $p$.
Then $D$ is cyclic if and only if $D^*$ contains a nonabelian
subgroup satisfying a group identity.}
 \vspace{0.2 cm}

{\sc Proof.} One way is clear  from Lemma~A. On the other hand,
assume that $D^*$ contains a nonabelian subgroup $G$ satisfying a
group identity. We may view $G$ as a linear group over $F$. By a
result of Platonov (cf. \cite[p. 149]{zal}), we may conclude that
there is a soluble normal subgroup $S$ of finite index in $G$. If
$S$ is nonabelian, the result follows from Corollary 3.4. So we
may assume that $S$ is abelian. Set $T=F^*G$ and $A=F^*S$. It is
easily seen that $A$ is a normal abelian subgroup of finite index
in $T$. We may take $A$ maximal in $T$. Two cases may occur:

{\bf Case 1:} If $A=F^*$, then $T/F^*$ is a nontrivial finite
subgroup of $D^*/F^*$ and the result follows from the Theorem of
\cite{tig}.

{\bf Case 2:} If $A \neq F^*$, then $K=F[A]$ is a maximal
subfield of $D$. Since $A$ is normal in $T$, by Lemma~B,
 we conclude that $K/F$ is Galois and it is clear that $K/F$ is
cyclic which completes the proof.

 \vspace{0.2 cm}\hfill$\Box$

 The above result provides us with various cyclicity conditions on
 subgroups of $D^*$. There are some conditions on subgroups of
 $D^*$ that imply those subgroups satisfy a group identity. We
 collect some well-known ones in the following:

{\sc Corollary 3.8.} {\it Let $D$ be an $F$-centeral division
algebra of prime degree   $p$. Then $D$ is cyclic if either of the
following conditions holds:

(a) $D^*$ contains a nonabelian soluble subgroup.

(b) $D^*$ contains a nonabelian  subgroup of a bounded exponent.

(c) $D^*$ contains a nonabelian finite subgroup.

} \vspace{0.2 cm}





The authors thank the referee for his constructive comments. They
are also indebted to the Research Council of Sharif University of
Technology for support. \vspace{0.2 cm}
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\end{document}

