Presenting Powers of Augmentation Ideals of Abelian p-Groups

Anthony Bak Department of Mathematics, University of Bielefeld 33501 Bielefeld, Germany Email: bak@mathematik.uni-bielefeld.de

Guoping Tang Department of Applied Mathematics Northwestern Polytechnical University Xi’an Shaanxi 710072 P.R. China email: tanggp@nwpu.edu.cn

Abstract

This note presents powers of the augmentation ideal of an integral group ring of an abelian p-group, generalizing results of Bak and Vavilov for elementary 2-groups, of Parmenter for elementary 3-groups, and of Tang for elementary p-groups. 2000 AMS Subject Classification 16S34, 20C05 Keywords: integral group ring, augmentation ideal

0 Introduction

Let G be a group, ZZG its integral group ring and ΔG the augmentation ideal of ZZG. The n’th power Δ^{n}(G) of ΔG is generated as an abelian group by all products (g_{1} − 1)(g_{2} − 1) ⋯ (g_{n} − 1) where g_{i} ∈ G. In [1], Bak and Vavilov considered the problem of finding a complete set of relations among these generators. The main result in [1] solves this problem when G is an elementary 2-group. In [2], Parmenter solves the problem when either G is an elementary 3-group or G is an elementary p-group where p > 3, but n ⩽ 3. In [5], Tang solves the problem when G is an elementary p-group and n is arbitrary. The current paper solves the problem when G is an abelian p-group of finite exponent and n is arbitrary.

The rest of the paper is organized as follows. In section 1, we recall well known relations among the generators (g_{1} − 1) ⋯ (g_{n} − 1) of Δ^{n}(G) when G is arbitrary or abelian and provide a family of new relations when G is an abelian p-group. In section 2, we show that the relations above account for all relations among the generators of Δ^{n}(G) when G is an abelian p-group.

Acknowledgement * The second author gratefully acknowledge the hospitality of Prof. A. Bak and the support of SFB 343, Fakultät für Mathematik, Universität Bielefeld, Germany during the preparation of the present article.

1 Old and New Relations in Δ^{n}(G)

Let G be a group, ZZ[G] its integral group ring and η: ZZ[G] → ZZ, ∑_{g} a_{g}g ↦ ∑_{g} a_{g}, the

augmentation homomorphism. The kernel Δ(G) of the homomorphism η is called the augmentation ideal of ZZ[G]. It is clear that Δ(G) is freely generated as an abelian group by the elements

[g] := g − 1, g ∈ G \ {1}

(cf. [3]). This implies that the n-th power Δ^{n}(G) := Δ(G)^{n} of the augmentation ideal Δ(G) is generated as an abelian group by the products

[g_{1}, …, g_{n}] := (g_{1} − 1) … (g_{n} − 1), g_{1}, …, g_{n} ∈ G.

There are two obvious relations which hold for any group G. The first is the normalizing relation

(N)  [g_{1}, …, g_{n}] = 0 (n ≥ 1), whenever some g_{i} = 1.

The second is that the symbol [g_{1}, g_{2}] is a 2-cocycle:

[g_{2}, g_{3}] − [g_{1}g_{2}, g_{3}] + [g_{1}, g_{2}g_{3}] − [g_{1}, g_{2}] = 0.

This implies the relation

(R)  [g_{1}, …, g_{n}] (n ≥ 2) is a 2-cocycle in g_{i−1}, g_{i}

when the other variables are fixed. If G is abelian then one has in addition to the above the symmetric relation

(S)  [g_{1}, …, g_{i}, …, g_{j}, …, g_{n}] = [g_{1}, …, g_{j}, …, g_{i}, …, g_{n}]  (n ≥ 2)

got by swapping g_{i} with g_{j} and leaving the other elements unchanged. The relations above are of course well known. When G is an abelian p-group, we shall require a new family of relations which we describe next.

For the rest of this section, let p denote some fixed natural prime and r some fixed natural number. Let s a natural number such that s ⩽ r and m a natural number such that 3 ⩽ m ⩽ p^{r} + 1. We shall define for each m and r, sequences of integers a_{1}^{(m)}, ⋯, a_{p^{r}−m+1}^{(m)} and b_{1}^{(m)}, ⋯, b_{m−2}^{(m)} and use them to describe new relations in Δ^{n}(G) for any n ⩾ m and any abelian p-group G. Let X denote an indeterminate and ZZ[X] the integral polynomial ring in X. It is easy to see that for each m, the free ZZ-module of all polynomials f(X) ∈ ZZ[X] such that degree (f(X)) ⩽ p^{r} − 1 and f(1) = 0 has as ZZ-basis the set

{(X^{i} − 1)(X − 1)^{m−2} | 1 ⩽ i ⩽ p^{r} − m + 1} ∪ {(X − 1)^{i} | 1 ⩽ i ⩽ m − 2}.

Let F_{p^{r}}^{(m)}(X) denote the polynomial

F_{p^{r}}^{(m)}(X) = (X^{p^{r}−m+2} − 1)(X − 1)^{m−2} − X^{p^{r}} + 1.

We define now the elements a_{i}^{(m)} (1 ⩽ i ⩽ p^{r} − m + 1) and b_{i}^{(m)} (1 ⩽ i ⩽ m − 2) such that

F_{p^{r}}^{(m)}(X) = ∑_{i=1}^{p^{r}−m+1} a_{i}^{(m)}(X^{i} − 1)(X − 1)^{m−2} + ∑_{i=1}^{m−2} b_{i}^{(m)}(X − 1)^{i}.

Let ∂^{i}F_{p^{r}}^{(m)}(X) denote the i’th derivative of F_{p^{r}}^{(m)}(X). For 1 ⩽ i ⩽ m − 2, one computes that ∂^{i}F_{p^{r}}^{(m)}(1) = −p^{r}(p^{r} − 1) ⋯ (p^{r} − i + 1) = b_{i}^{(m)}(i!). Thus b_{i}^{(m)} = −\binom{p^{r}}{i}.

Let ⟨h⟩ denote a group of order p^{s} generated by h. One checks straightforward that for any m the free ZZ-module Δ^{m}(⟨h⟩) has the ZZ-basis

{(h − 1)^{m−1}(h^{j} − 1) | 1 ⩽ j ⩽ p^{s} − 1}.

Since

p^{s}(h − 1) = − ∑_{i=2}^{p^{s}} \binom{p^{s}}{i}(h − 1)^{i}

it follows that for any m there exists an l such that p^{l}(h − 1) ∈ Δ^{m}(< h >). The next lemma describes how large l must be chosen.

Lemma 1.1 Let ⟨h⟩ denote a group of order p^{s} generated by h. Then

p^{l}(h − 1) ∈ Δ(< h >) ∖ Δ^{2}(< h >)  if l < s.
p^{l}(h − 1) ∈ Δ^{(p−1)(l−s+1)+1}(< h >) ∖ Δ^{(p−1)(l−s+1)+2}(< h >) if l ⩾ s.

Proof It is obvious that p^{l}(h − 1) ∈ Δ(< h >). If l < s and p^{l}(h − 1) ∈ Δ^{2}(< h >) then there are c_{1}, ⋯, c_{p^{s}−1} such that

p^{l}(h − 1) = ∑_{i=1}^{p^{s}−1} c_{i}(h − 1)(h^{i} − 1).

So

p^{l} − ∑_{i=1}^{p^{s}−1} c_{i}(h^{i} − 1) = d ∑_{i=0}^{p^{s}−1} h^{i}

for some d ∈ ℤ, since the annihilator of (h − 1) in ℤ(< h >) is ℤ(∑_{i=0}^{p^{s}−1} h^{i}). Multiplying both sides of the identity above by ∑_{i=0}^{p^{s}−1} h^{i}, we get the equation

p^{l} ∑_{i=0}^{p^{s}−1} h^{i} = d(∑_{i=0}^{p^{s}−1} h^{i})^{2} = dp^{s} ∑_{i=0}^{p^{s}−1} h^{i},

which implies that p^{l} = dp^{s}, contrary to the assumption that l < s.

Suppose that l ⩾ s. We use induction on l to prove that p^{l}(h − 1) ∈ Δ^{(p−1)(l−s+1)+1}(< h >).

Suppose l = s. Then

p^{s}(h − 1) = − ∑_{i=2}^{p^{s}} \binom{p^{s}}{i}(h − 1)^{i}.

If p = 2, it is obvious that p^{s}(h − 1) ∈ Δ^{2}(< h >). So we assume that p ⩾ 3. Since (h − 1)^{i} ∈ Δ^{p}(< h >) for all i ⩾ p, it follows that

p^{s}(h − 1) ≡ − ∑_{i=2}^{p−1} \binom{p^{s}}{i}(h − 1)^{i} = (− ∑_{i=2}^{p−1} \binom{p^{s}}{i}(h − 1)^{i−2})(h − 1)^{2} ( mod Δ^{p}(< h >))

and each of \binom{p^{s}}{2}, ⋯, \binom{p^{s}}{p−1} is divisible by p^{s}. Repeating this procedure (p−2) times more, we get that p^{s}(h − 1) ∈ Δ^{p}(< h >).

Suppose l > s. Then

p^{l}(h − 1) = −p^{l−s} ∑_{i=2}^{p^{s}} \binom{p^{s}}{i}(h − 1)^{i}.

We divide all terms p^{l−s}\binom{p^{s}}{i}(h − 1)^{i} (2 ⩽ i ⩽ p^{s}) as three parts according to i: p|i, p does not divide i and i > p as well as 2 ⩽ i ⩽ p − 1.

If p|i. Suppose p^{k}∥i (p^{k} divides i but p^{k+1} does not divides i) then p^{s−k}∥\binom{p^{s}}{i} and p^{l−k}∥p^{l−s}\binom{p^{s}}{i}. Since l − k < l by induction assumption

p^{l−s}\binom{p^{s}}{i}(h − 1)^{i} ∈ Δ^{(p−1)(l−k−s+1)+1}(< h >)Δ^{i−1}(< h >) ⊆ Δ^{(p−1)(l−k−s+1)+i}(< h >).

Note that if l − k < s the statement above is obviously true. Since p^{k}∥i (k ⩾ 1) we have

(p − 1)(l − k − s + 1) + i ⩾ (p − 1)(l − k − s + 1) + p^{k} ⩾ (p − 1)(l − s + 1) + 1

and equals hold if and only if i = p^{k} and p^{k} − 1 = k(p − 1), i.e., equals holds if and only if k = 1. Thus we have proved that if p|i then

p^{l−s}\binom{p^{s}}{i}(h − 1)^{i} ∈ Δ^{(p−1)(l−s+1)+1}(< h >).

In fact p|i and p ≠ i then

p^{l−s}\binom{p^{s}}{i}(h − 1)^{i} ∈ Δ^{(p−1)(l−s+1)+2}(< h >).

If p does not divide i and i > p then p^{l−1}|p^{l−s}\binom{p^{s}}{i} and by induction assumption, one gets

p^{l−s}\binom{p^{s}}{i}(h − 1)^{i} ∈ Δ^{(p−1)(l−s)+1}(< h >)Δ^{(i−1)}(< h >) ⊆ Δ^{(p−1)(l−s+1)+1}(< h >).

Hence

p^{l}(h − 1) ≡ −p^{l−s} ∑_{i=2}^{p−1} \binom{p^{s}}{i}(h − 1)^{i} ( mod Δ^{(p−1)(l−s+1)+1}(< h >)).

Since p^{l}∥p^{l−s}\binom{p^{s}}{i} for i from 2 to p − 1 the formula above implies that p^{l}(h − 1) ∈ Δ^{(p−1)(l−s+1)+1}(< h >).

Next we prove that if l ⩾ s

p^{l}(h − 1) ∉ Δ^{(p−1)(l−s+1)+2}(< h >).

From the proof of p^{l}(h − 1) ∈ Δ^{(p−1)(l−s+1)+1}(< h >) above it is easy to see that

p^{l−s}\binom{p^{s}}{i}(h − 1)^{i} ∈ Δ^{(p−1)(l−s+1)+2}(< h >)

for all i ≠ p and i > 1. So one has

p^{l}(h − 1) ≡ −p^{l−s}\binom{p^{s}}{p}(h − 1)^{p} ( mod Δ^{(p−1)(l−s+1)+2}(< h >)).

If p^{l}(h−1) ∈ Δ^{(p−1)(l−s+1)+2}(< h >) then p^{l−s}\binom{p^{s}}{p}(h−1)^{p} ∈ Δ^{(p−1)(l−s+1)+2}(< h >). There are c_{1}, ⋯, c_{p^{s}−1} such that

p^{l−s}\binom{p^{s}}{p}(h − 1)^{p} = ∑_{i=1}^{p^{s}−1} c_{i}(h − 1)^{(p−1)(l−s+1)+1}(h^{i} − 1).

Since ℤ(< h >) has no non-zero nilpotent elements( [4], or by an elementary inductive argument), it follows that

p^{l−s}\binom{p^{s}}{p}(h − 1) = ∑_{i=1}^{p^{s}−1} c_{i}(h − 1)^{(p−1)(l−s)+1}(h^{i} − 1) ∈ Δ^{(p−1)(l−s)+2}(< h >).

Since p^{s−1}∥\binom{p^{s}}{p} one can write \binom{p^{s}}{p} = p^{s−1}u for some u ∈ ℤ with (u, p) = 1. There are v, w ∈ ℤ such that uv + pw = 1. However p^{l}(h − 1) ∈ Δ^{(p−1)(l−s+1)+1}(< h >) ⊆ Δ^{(p−1)(l−s+1)+2}(< h >) it follows that

p^{l−1}(h − 1) = p^{s−1}uv(h − 1) + p^{l}w(h − 1) ∈ Δ^{(p−1)(l−s)+2}(< h >).

Thus we have proved that

p^{l}(h − 1) ∈ Δ^{(p−1)(l−s+1)+2}(< h >)  implies  p^{l−1}(h − 1) ∈ Δ^{(p−1)(l−s)+2}(< h >).

This implies further that p^{s−1}(h − 1) ∈ Δ^{2}(< h >), a contrary to what we have prove that p^{s−1}(h − 1) ∉ Δ^{2}(< h >). So the proof is complete.

We use the computation of the b_{i}^{(m)}’s in the definition of F_{p^{r}}^{(m)}(X) to define a further sequence b_{i,1}^{(m)}, ⋯, b_{i,p^{s}−1}^{(m)} of integers for some s ⩽ r.

If h ∈ G of order p^{s} (s ⩽ r), pick an l ⩾ 0 such that p^{l+r}(h − 1) ∈ Δ^{m−1}(< h >). The smallest such an l is \{\frac{m−2}{p−1}\} + s − r − 1 where \{\frac{m−2}{p−1}\} is the smallest integer bigger than or equals to \frac{m−2}{p−1}. By Lemma 1.1 (p − 1)(l + r − s + 1) + 1 ⩾ m − 1. Suppose that p^{k}∥i then (p − 1)k ⩽ i − 1 and (p − 1)(l + r − s − k + 1) + 1 ⩾ m − 1 − (p − 1)k ⩾ m − i, so by Lemma 1.1, one has p^{l}\binom{p^{r}}{i}(h − 1) ∈ Δ^{m−i}(< h >) for any 1 ⩽ i ⩽ m − 2.

Define the integers b_{i,j}^{(m)} (1 ⩽ i ⩽ m − 2, 1 ⩽ j ⩽ p^{s} − 1)

by

−p^{l}\binom{p^{r}}{i}(h − 1) = ∑_{j=1}^{p^{s}−1} b_{i,j}^{(m)}(h − 1)^{m−i−1}(h^{j} − 1).

Note that if we let h = 1, the equation above holds trivially.

Let G be an abelian p-group. It follows from the definition of F_{p^{r}}^{(m)}(X) that for any element g ∈ G of order p^{r}, we have the identity

(g^{p^{r}−m+2} − 1)(g − 1)^{m−2} = ∑_{i=1}^{p^{r}−m+1} a_{i}^{(m)}(g^{i} − 1)(g − 1)^{m−2} + ∑_{i=1}^{m−2} b_{i}^{(m)}(g − 1)^{i}

in Δ^{m−1}(G). Multiplying both sides of the identity above by p^{l}(h−1), we get the equation

p^{l}(g^{p^{r}−m+2} − 1)(g − 1)^{m−2}(h − 1) = ∑_{i=1}^{p^{r}−m+1} p^{l}a_{i}^{(m)}(g^{i} − 1)(g − 1)^{m−2}(h − 1) +

∑_{i=1}^{m−2} ∑_{j=1}^{p^{s}−1} b_{i,j}^{(m)}(g − 1)^{i}(h − 1)^{m−i−1}(h^{j} − 1)

in Δ^{m}(G), i.e.

p^{l}[g^{p^{r}−m+2}, \underbrace{g, ⋯, g}_{m−2}, h] = ∑_{i=1}^{p^{r}−m+1} p^{l}a_{i}^{(m)}[g^{i}, \underbrace{g, ⋯, g}_{m−2}, h] +

∑_{i=1}^{m−2} ∑_{j=1}^{p^{s}−1} b_{i,j}^{(m)}[\underbrace{g, ⋯, g}_{i}, \underbrace{h, ⋯, h}_{m−i−1}, h^{j}].

Therefore for any n ⩾ m, we get the relation

p^{l}[g^{p^{r}−m+2}, \underbrace{g, ⋯, g}_{m−2}, h, g_{m+1}, ⋯, g_{n}] = (T_{p^{s},p^{r}}^{m})

∑_{i=1}^{p^{r}−m+1} p^{l}a_{i}^{(m)}[g^{i}, \underbrace{g, ⋯, g}_{m−2}, h, g_{m+1}, ⋯, g_{n}] +

∑_{i=1}^{m−2} ∑_{j=1}^{p^{s}−1} b_{i,j}^{(m)}[\underbrace{g, ⋯, g}_{i}, \underbrace{h, ⋯, h}_{m−i−1}, h^{j}, g_{m+1}, ⋯, g_{n}]

in Δ^{n}(G), where l = \{\frac{m−2}{p−1}\} + s − r − 1.

Setting m = 3 and p = 2 in the relation (T_{p,p}^{m}), we get the relation (T) in [1], in the equivalent form found in Lemma 3.1 of [1]. Setting m = 3 and letting p ⩾ 3 in the relation (T_{p,p}^{m}), we get the relation denoted by (T_{p}) in [2]. The relation (T_{p,p}^{m}) is just the relation (T_{m}) in [5].

2 A Presentation of Δ^{n}(G) for G p-Abelian

The goal of this section is to prove the following result.

Theorem 2.1 Let G be an abelian p-group of exponent p^{k}. Then the n’th power Δ^{n}(G) of the augmentation ideal ΔG of the integral group ring ZZG is a free abelian group on the symbols [g_{1}, ⋯, g_{n}] = (g_{1} − 1) ⋯ (g_{n} − 1), g_{i} ∈ G, modulo the relations

(N)  if n = 1
(N), (R), (S)  if n = 2
(N), (R), (S), (T_{p^{s},p^{r}}^{m}),  3 ⩽ m ⩽ n, s ⩽ r ⩽ k if 3 ⩽ n ⩽ p^{k}
(N), (R), (S), (T_{p^{s},p^{r}}^{m}),  3 ⩽ m ⩽ p^{k} + 1, s ⩽ r ⩽ k if p^{k} < n.

Proof Let Δ̃^{n}(G) denote the free abelian group on the symbols [g_{1}, ⋯, g_{n}] = (g_{1} − 1) ⋯ (g_{n} − 1), g_{i} ∈ G, modulo the relations in Theorem 2.1. Then there is a surjective map φ from Δ̃^{n}(G) to Δ^{n}(G). It is sufficient for us to prove that φ is also bijective.

It suffices to prove the theorem in the special case that G is finitely generated. Hence the order |G| of G is finite. Since Δ^{n}(G) is a free abelian group with rank |G| − 1, any |G| − 1 generators of Δ^{n}(G) will be a basis of it. If we can prove that Δ̃^{n}(G) is free then its rank must be bigger than or equals to |G| − 1. If we can prove that Δ̃^{n}(G) can be generated by not more than |G| − 1 generators, then it is a free abelian group with rank |G| − 1 and isomorphism to Δ^{n}(G).

The proof will be by induction on the number of generators of G, which is denoted by μ(G).

Suppose G is cyclic. Let g be a generator of G of order p^{k}. One verifies easily that the image under φ of the set B^{n}(G) := \{[\underbrace{g, ⋯, g}_{n−1}, g^{j}] | 1 ⩽ j ⩽ p^{k} − 1\} is a ZZ-basis for

Δ^{n}(G). It suffices to show that any generator [g^{i_{1}}, ⋯, g^{i_{n}}] of Δ̃^{n}(G) can be exhibited using just the relations in the theorem, as an integral linear combination of the elements in B^{n}(G). If n = 1, the result is trivial. We proceed now by induction on n. Suppose n > 1. Suppose that one of the exponents i_{j} is 1. Applying (S), we can assume that j = 1, i.e. i_{1} = 1. By the induction assumption, we can write [g^{i_{2}}, ⋯, g^{i_{n}}] as a linear combination of the elements \{[\underbrace{g, ⋯, g}_{n−2}, g^{j}] | 1 ⩽ j ⩽ p^{k} − 1\}. Thus [g, g^{i_{2}}, ⋯, g^{i_{n}}] is a

linear combination of the elements in \{[\underbrace{g, ⋯, g}_{n−1}, g^{j}] | 1 ⩽ j ⩽ p^{k} − 1\}. We proceed now by induction on i_{1}. Suppose i_{1} > 1. By (R), we can write

[g^{i_{1}}, ⋯, g^{i_{n}}] = [g^{i_{1}−1}, g^{i_{2}}, ⋯, g^{i_{n}}] + [g, g^{i_{1}+i_{2}−1}, g^{i_{3}}, ⋯, g^{i_{n}}] − [g, g^{i_{2}}, ⋯, g^{i_{n}}] >

By the induction assumption, we can write each of the generators on the right hand side of the equation above, as a linear combination of the basis elements. This completes the proof when G is cyclic.

Suppose the number of generators of G is more than one. Write G = ⟨g⟩ ⊕ H where g is an element of G of order p^{k}.

If n = 1, the conclusion of the theorem is trivial. No induction on the number of generators of G is required. So we suppose for the rest of the proof that n ⩾ 2.

We claim that the relation (R) along with the relation (S) allows one to write any generator as a linear combination of generators [g_{1}, ⋯, g_{n}] such that each g_{j} is either a power of g or is in H. Let [g^{i}h_{1}, g^{j}h_{2}, −] denote a typical generator. The relation (R) shows that

[g^{i}h_{1}, g^{j}h_{2}, −] = [g^{i+j}h_{1}, h_{2}, −] + [g^{i}h_{1}, g^{j}, −] − [g^{j}, h_{2}, −].

Applying (R) to the first 2 summands on the right hand side, we obtain

[g^{i+j}h_{1}, h_{2}, −] = [g^{i+j}, h_{1}h_{2}, −] + [h_{1}, h_{2}, −] − [g^{i+j}, h_{1}, −]

and

[g^{i}h_{1}, g^{j}, −] = [h_{1}, g^{i+j}, −] + [g^{i}, g^{j}, −] − [h_{1}, g^{i+j}, −].

The claim follows from the 3 equations above and (S). Applying (R) to a generator [g^{i}, g^{j}, −], we obtain

[g^{i}, g^{j}, −] = [g^{i+j−1}, g, −] + [g^{i}, g^{j−1}, −] − [g^{j−1}, g, −].

It follows now that any generator of Δ̃^{n}(G)(n ⩾ 2) can be written using only the relations (R) and (S) as a linear combination of generators of the form

[\underbrace{g^{i}, g, ⋯, g}_{j}, h_{j+1}, ⋯, h_{n}]

where 0 ⩽ i ⩽ p^{k} − 1 and 0 ⩽ j ⩽ n. Moreover from (N), it follows that we can assume additionally that 1 ⩽ i ⩽ p^{k} − 1.

Suppose n = 2. The subgroup of Δ̃^{2}(G) generated by [g^{i}, g], 1 ⩽ i ⩽ p^{k} − 1 is denoted by Δ̃^{2}(< g >)_{G}. There is a sequence

Δ^{2}(< g >) ⟶ Δ̃^{2}(< g >) ⟶ Δ̃^{2}(< g >)_{G} ⟶ Δ^{2}(< g >)

and the first map is an isomorphism, the second one is surjective, the third is induced from Δ̃^{2}(G) ⟶ Δ^{2}(G). The composition of these maps is identity, so the second map is injective, hence isomorphic. i.e., Δ̃^{2}(< g >)_{G} and is with rank p^{k} − 1. Similarly, the subgroup of Δ̃^{2}(G) generated by [g^{i}, h], 1 ⩽ i ⩽ p^{k} −1, h ∈ H ∖{1} is (p^{k} −1)(|H|−1) and the subgroup of Δ̃^{2}(G) generated by [h_{1}, h_{2}], h_{1}, h_{2} ∈ H, can be generated by (|H| − 1) elements since it is an image of Δ̃^{2}(H), which is an free abelian group of rank (|H|−1) by induction assumption. Thus Δ̃^{2}(G) is generated by not more than p^{k} − 1 + (p^{k} − 1)(|H| − 1) + (|H| − 1) = |G| − 1 elements. The proof is complete in this case.

Suppose n ⩾ 3. Let [\underbrace{g^{i}, g, ⋯, g}_{j}, h_{1}, ⋯, h_{n−j}] be a generator such that i > 1 and

n − j ⩾ 2. To shorten notation, let (g^{i}, h_{1}, h_{2}) = [\underbrace{g^{i}, g, ⋯, g}_{j}, h_{1}, ⋯, h_{n−j}]. Applying (R)

and (S) several times, we can write (g^{i}, h_{1}, h_{2}) as linear combination

(g^{i}, h_{1}, h_{2}) = (h_{1}, g^{i−1}g, h_{2}) = (h_{1}g^{i−1}, g, h_{2}) +
(h_{1}, g^{i−1}, h_{2}) − (g^{i−1}, g, h_{2}) = (g^{i−1}, g, h_{1}h_{2}) +

(g, h_{1}, h_{2}) − (g^{i−1}, g, h_{1}) + (g^{i−1}, h_{1}, h_{2}) − (g^{i−1}, g, h_{2}) = (∗)

(g^{i−1}, g, h_{1}h_{2}) + (g^{i−1}, h_{1}, h_{2}) + (g, h_{1}, h_{2}) −
(g^{i−1}, g, h_{1}) − (g^{i−1}, g, h_{2})

This implies that Δ̃^{n}(G) can be generated by

(i) [\underbrace{g^{i}, g, ⋯, g}_{n}], 1 ⩽ i ⩽ p^{k} − 1.

(ii) [h_{1}, ⋯, h_{n}], h_{i} ∈ H ∖ {1}, 1 ⩽ i ⩽ n.

(iii) [\underbrace{g, ⋯, g}_{j}, h_{1}, ⋯, h_{n−j}], 1 ⩽ j ⩽ n − 2, h_{1}, ⋯, h_{n−j} ∈ H ∖ \{1\}.

(iv) [\underbrace{g^{i}, g, ⋯, g}_{n−1}, h], where 1 ⩽ i ⩽ p^{k} − 1 and h ∈ H ∖ \{1\}.

Set [\underbrace{g, ⋯, g}_{i}]Δ̃^{n−i}(H) to be the subgroup of Δ̃^{n}(G) generated by [\underbrace{g, ⋯, g}_{j}, h_{1}, ⋯, h_{n−j}],

where 1 ⩽ j ⩽ n − 2 and h_{1}, ⋯, h_{n−j} ∈ H ∖ {1}. Then there is a sequence

Δ^{n−i}(H) ⟶ Δ̃^{n−i}(H) ⟶ [\underbrace{g, ⋯, g}_{i}]Δ̃^{n−i}(H) ⟶ (g − 1)^{i}Δ^{n−i}(H) ⟶ Δ^{n−i}(H).

The first map is an isomorphism by induction assumption, the second one is surjective, the third is induced from Δ̃^{n}(G) ⟶ Δ^{n}(G). The composition of these maps is identity, so the second map is injective, hence isomorphic, i.e., [\underbrace{g, ⋯, g}_{i}]Δ̃^{n−i}(H) ≅ Δ̃^{n−i}(H) and

is a free abelian group with rank |H| − 1 by induction assumption.

Set [\underbrace{g^{i}, g, ⋯, g}_{n−1}]Δ̃(H) to be the subgroup of Δ̃^{n}(G) generated by [\underbrace{g^{i}, g, ⋯, g}_{n−1}, h], where

1 ⩽ i ⩽ p^{k} − 1 and h ∈ H. One can get for each i (1 ⩽ i ⩽ p^{k} − 1) [\underbrace{g^{i}, g ⋯, g}_{n−1}]Δ̃(H) is

isomorphic to Δ̃(H) and is a free abelian group with rank |H| − 1.

Similarly, the subgroup of Δ̃^{n}(G) generated by the first kind generators [\underbrace{g^{i}, g, ⋯, g}_{n}],

1 ⩽ i ⩽ p^{k} − 1 is a free abelian group with rank p^{k} − 1, and we denote it by Δ̃^{n}(< g >) also. The subgroup of Δ̃^{n}(G) generated by the second kind generators [h_{1}, ⋯, h_{n}], h_{i} ∈ H ∖ {1}, 1 ⩽ i ⩽ n is isomorphic to Δ̃^{n}(H) and is a free abelian group with rank |H| − 1.

Since Δ̃^{n}(G) is generated by elements (i)-(iv) it implies that

Δ̃^{n}(G) = Δ̃^{n}(H) + ∑_{i=1}^{n−2}[\underbrace{g, ⋯, g}_{i}]Δ̃^{n−i}(H) + ∑_{i=1}^{p^{k}−1}[\underbrace{g^{i}, g, ⋯, g}_{n−1}]Δ̃(H) + Δ̃^{n}(< g >)

and a free abelian group because each summand of it is free. The rank of Δ̃^{n}(H) is |H|−1 by induction assumption and the rank of Δ̃^{n}(< g >) is p^{k} − 1. We want to prove that the rank of

∑_{i=1}^{n−2}[\underbrace{g, ⋯, g}_{i}]Δ̃^{n−i}(H) + ∑_{i=1}^{p^{k}−1}[\underbrace{g^{i}, g, ⋯, g}_{n−1}]Δ̃(H)

is not more than (p^{k} − 1)(|H| − 1). Before proving it, we prove the following Lemma.

Lemma 2.2 Suppose that A is a free abelian group generated by x_{1}, ⋯, x_{s}, y_{1}, ⋯, y_{t} and for each y_{i} (1 ⩽ i ⩽ t) there is m_{i} ⩾ 0 such that

p^{m_{i}}y_{i} ∈< x_{1}, ⋯, x_{s}, y_{1}, ⋯, y_{i−1} > .

Then the rank of A is not more than s.

Proof Since p^{m_{1}}y_{1} ∈< x_{1}, ⋯, x_{s} > there are c_{1}, ⋯, c_{s} ∈ ℤ such that p^{m_{1}}y_{1} = ∑_{i=1}^{s} c_{i}x_{i}. After canceling a power of p we can assume that some c_{i}, say c_{1}, is coprime with p, hence p^{m_{1}}. There are u, v ∈ ℤ such that uc_{1}+vp^{m_{1}} = 1. y_{1} = uc_{1}y_{1}+vp^{m_{1}}y_{1} = c_{1}(uy_{1}+ vx_{1})+vc_{2}x_{2} +⋯+vc_{s}x_{s} and x_{1} = uc_{1}x_{1} +vp^{m_{1}}x_{1} = p^{m_{1}}(uy_{1} +vx_{1})−uc_{2}x_{2} −⋯−uc_{s}x_{s}. Thus < x_{1}, ⋯, x_{s}, y_{1} > can be generated by uy_{1} + vx_{1}, x_{2}, ⋯, x_{s}. The Lemma is proved by iterating this procedure.

We continue to the proof of Theorem 2.1. If n ⩽ p^{k}. Set

A_{l} = ∑_{i=1}^{n−2}[\underbrace{g, ⋯, g}_{i}]Δ̃^{n−i}(H) + ∑_{i=1}^{l}[\underbrace{g^{i}, g, ⋯, g}_{n−1}]Δ̃(H), 1 ⩽ l ⩽ p^{k} − 1.

By induction assumption, the rank of A_{p^{k}−n+1} is not more than (p^{k} − 1)(|H| − 1). By (T_{p^{s},p^{k}}^{n}) relations a multiple of some power of p on generator [\underbrace{g^{p^{k}−n+2}, g, ⋯, g}_{n−1}, h] in

[\underbrace{g^{p^{k}−n+2}, g, ⋯, g}_{n−1}]Δ̃(H) is in A_{p^{k}−n+1}. In general, by (T_{p^{s},p^{k}}^{n−i+1}) relations a multiple of

some power of p on generator [\underbrace{g^{p^{k}−n+1+i}, g, ⋯, g}_{n−1}, h] in [\underbrace{g^{p^{k}−n+1+i}, g, ⋯, g}_{n−1}]Δ̃(H) is in

A_{p^{k}−n+i}, 1 ⩽ i ⩽ n − 2. Thus the rank of

A_{p^{k}−1} = ∑_{i=1}^{n−2}[\underbrace{g, ⋯, g}_{i}]Δ̃^{n−i}(H) + ∑_{i=1}^{p^{k}−1}[\underbrace{g^{i}, g, ⋯, g}_{n−1}]Δ̃(H)

is not more than (p^{k} − 1)(|H| − 1) by Lemma 2.2

If n ⩾ p^{k} + 1. Set

B_{l} = \{ \begin{matrix}∑_{i=1}^{l}[\underbrace{g, ⋯, g}_{i}]Δ̃^{n−i}(H), & if 1 ⩽ l ⩽ n − 2 \\ ∑_{i=1}^{n−2}[\underbrace{g, ⋯, g}_{i}]Δ̃^{n−i}(H) + ∑_{i=1}^{l−n+2}[\underbrace{g^{i}, g, ⋯, g}_{n−1}]Δ̃(H), & if n − 1 ⩽ l ⩽ p^{k} + n − 2. \end{matrix}

By induction assumption, the rank of B_{p^{k}−1} is not more than (p^{k} − 1)(|H| − 1). By (T_{p^{s},p^{k}}^{p^{k}+1}) relations a multiple of some power of p of generator [\underbrace{g, ⋯, g}_{p^{k}}]Δ̃^{n−p^{k}}(H) is in

B_{p^{k}−1}. In general, by (T_{p^{s},p^{k}}^{p^{k}+1}) relations a multiple of some power of p of generator

[\underbrace{g, ⋯, g}_{p^{k}+i}, h_{1}, ⋯, h_{n−p^{k}−i}] of [\underbrace{g, ⋯, g}_{p^{k}+i}]Δ̃^{n−p^{k}−i}(H) is in B_{p^{k}+i−1}, 0 ⩽ i ⩽ n − 2 − p^{k}. By

(T_{p^{s},p^{k}}^{p^{k}+1}) relations a multiple of some power of p on generator [\underbrace{g, ⋯, g}_{n−1}, h] in [\underbrace{g, ⋯, g}_{n−1}]Δ̃(H)

is in B_{n−2}. In general, by (T_{p^{s},p^{k}}^{p^{k}+2−i}) relations a multiple of some power of p on generator

[\underbrace{g^{i}, g, ⋯, g}_{n−1}, h] in [\underbrace{g^{i}, g, ⋯, g}_{n−1}]Δ̃(H) is in B_{n−3+i}, 1 ⩽ i ⩽ p^{k} − 1. Thus the rank of

B_{n+p^{k}−3} = ∑_{i=1}^{n−2}[\underbrace{g, ⋯, g}_{i}]Δ̃^{n−i}(H) + ∑_{i=1}^{p^{k}−1}[\underbrace{g^{i}, g, ⋯, g}_{n−1}]Δ̃(H)

is not more than (p^{k} − 1)(|H| − 1) by Lemma 2.2

References

[1] Anthony Bak and Nikolai Vavilov, Presenting powers of augmentation ideals and Pfister forms, K-Theory 20 (2000), 299 – 310

[2] M. M. Parmenter, A basis for powers of the augmentation ideal, Algebra Colloquium, to appear

[3] I. B. S. Passi, Group rings and their augmentation ideals, Lecture Notes in Math. Vol 715 (1979)

[4] Sudarshan K. Sehgal, Topics in Group Rings, Marcel Dekker, New York, 1978 [5] Guoping Tang, Presenting powers of augmentation ideals of Elementary p-groups

K-Theory, to appear