On a Question of Karpilovsky

Guoping Tang ^{∗}

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

Abstract.

Let G be a finite elementary group. Let Δ^{n}(G) denote the n’th power of the augmentation ideal Δ(G) of the integral group ring ℤG. The paper gives an explicit basis of the quotient group Q_{n}(G) = Δ^{n}(G)/Δ^{n+1}(G) and as a consequence settles a question of Karpilovsky for this particular class of groups.

2000 AMS Subject Classification: 16S34, 20C05 Keywords: integral group ring, augmentation ideal.

Let G be a group, ℤG its integral group ring and Δ(G) the augmentation ideal of ℤG. The n’th power Δ^{n}(G) of Δ(G) is generated as an abelian group by the products (g_{1} − 1)(g_{2} − 1) ⋯ (g_{n} − 1), g_{i} ∈ G for all i. It is well known that if G is a finite group of order r, then Δ^{n}(G) is a free ℤ-module of rank r − 1 for any n ≥ 1 [5, P.122]. Recall that a group is called p-elementary where p is a natural prime number, if it is abelian of exponent p. A group is called elementary, if it is a sum of its p-elementary subgroups taken over all primes p. In [5, P121] Passi proved that if G is a p-elementary group of rank k, then the quotient Q_{n}(G) = Δ^{n}(G)/Δ^{n+1}(G) is a p-elementary group of rank (p^{k} − 1)/(p − 1) for n ≥ (p − 1)(k − 1) + 1. F.Bachmann and L.Grünenfelder [1](see also A.W.Hales and I.B.S.Passi [2]) proved that for a finite abelian group G, there exists a number N such that for all n ≥ N, Q_{n}(G) is isomorphic to Q_{n+1}(G). G. Karpilovsky posed in [3] the problem of determining the isomorphism classes of the groups Q_{n}(G) (n ≥ 1) for a finite abelian group G. This paper solves the question for finite elementary groups G. Our proof will use an explicit basis of Δ^{n}(G) given by Parmenter in [4] for a p-elementary group G to get an explicit 𝔽_{p} basis for Q_{n}(G). The result for an arbitrary finite elementary group G follows from the case of p-elementary groups, because if G = ⊕_{p}G_{p} then Q_{n}(G) = ⊕_{p}Q_{n}(G_{p}).

∗The author gratefully acknowledges the hospitality of Prof. A.Bak and the support of SFB 343, Fakultät für Mathematik, Universität Bielefeld, Germany as well as Mathematics Section, International Centre for Theoretical Physics, Trieste, Italy during the preparation of the present article.

In [4] Parmenter proved the following theorem.

Theorem 1 Suppose G = H⊕ < g > is a p-elementary group. Then a ℤ-basis B^{n}(G) of Δ^{n}(G) can be given as follows:

If G =< g > set B^{n} < g >= {(g − 1)^{n−1}(g^{i} − 1)|i = 1, …, p − 1}. For G arbitrary define B^{n}(G) as follows:

i) If n = 1,  B^{1}(G) = {g − 1|g ∈ G, g ≠ 1}.
ii) If 2 ⩽ n ⩽ p, B^{n}(G) = ⋃_{i=0}^{n−2} B^{n−i}(H)(g − 1)^{i} ⋃

⋃_{i=1}^{p−n+1} B^{1}(H)(g − 1)^{n−2}(g^{i} − 1) ⋃ B^{n} < g > .

iii) If n > p,  B^{n}(G) = ⋃_{i=0}^{p−1} B^{n−i}(H)(g − 1)^{i} ⋃ B^{n} < g > .

This result will be used to prove the following main result.

Theorem 2 Suppose G =< g_{1} > ⊕ … ⊕ < g_{k} > is a p-elementary group of rank k. Let

S_{n,k} = \{(α_{1}, α_{2}, …, α_{k}) ∈ ℤ^{k}| \begin{matrix}α_{i} ≥ 0, α_{1} + α_{2} + ⋯ + α_{k} = n. \\ If α_{j} ≠ 0, then α_{i} < p for all i > j \end{matrix} \}

and let

T_{n,k} = {(g_{1} − 1)^{α_{1}}(g_{2} − 1)^{α_{2}} ⋯ (g_{k} − 1)^{α_{k}} + Δ^{n+1}(G)|(α_{1}, α_{2}, …, α_{k}) ∈ S_{n,k}} .

Then T_{n,k} is an 𝔽_{p} basis for Q_{n}(G).

Proof. Using the relation

(xy − 1) = (x − 1)(y − 1) + (x − 1) + (y − 1)

and the relation

x^{m} − 1 = ∑_{i=1}^{m} \binom{m}{i}(x − 1)^{i},

which results from the binomial expansion of ((x−1)+1)^{m}, where x, y, z ∈ G, one gets that

{(g_{1} − 1)^{α_{1}}(g_{2} − 1)^{α_{2}} ⋯ (g_{k} − 1)^{α_{k}} + Δ^{n+1}(G)|α_{i} ≥ 0, ∑_{i=1}^{k} α_{i} = n}

is a generating set of Q_{n}(G). Since

(h − 1)(g − 1)^{p} ≡ (h − 1)^{p}(g − 1) (mod Δ^{p+2}(G)) [5, P.120]

for any g, h ∈ G, it follows that

T_{n,k} = {(g_{1} − 1)^{α_{1}}(g_{2} − 1)^{α_{2}} ⋯ (g_{k} − 1)^{α_{k}} + Δ^{n+1}(G)|(α_{1}, α_{2}, …, α_{k}) ∈ S_{n,k}}

is a generating set of Q_{n}(G). Since g^{p} = 1 for any g ∈ G, it follows from the binomial expansion of ((g − 1) + 1)^{p} that p(g − 1) = − ∑_{i=2}^{p} \binom{p}{i}(g − 1)^{i} . Thus Q_{n}(G) is an 𝔽_{p}-vector space with generating set T_{n,k}.

Now we prove that the elements in T_{n,k} are linearly independent over 𝔽_{p}. This will complete the proof that T_{n,k} is an 𝔽_{p} basis for Q_{n}(G). Suppose u is an integral linear combination of elements (g_{1} − 1)^{α_{1}}(g_{2} − 1)^{α_{2}} ⋯ (g_{k} − 1)^{α_{k}}, where (α_{1}, α_{2}, …, α_{k}) ∈ S_{n,k}. It suffices to show that if u ∈ Δ^{n+1}(G), then all coefficients in this combination can be divided by p. The proof is by induction on k.

Suppose k = 1. Suppose that u = a(g_{1} − 1)^{n} for some a ∈ ℤ and that u ∈ Δ^{n+1}(< g_{1} >). By Theorem 1 there exist a_{1}, …, a_{p−1} ∈ ℤ such that

u = a(g_{1} − 1)^{n} = ∑_{i=1}^{p−1} a_{i}(g_{1} − 1)^{n}(g_{1}^{i} − 1).

Hence

(g_{1} − 1)^{n}(a − ∑_{i=1}^{p−1} a_{i}(g_{1}^{i} − 1)) = 0.

Using the well known fact that ℤG contains no nonzero nilpotent elements, we conclude that

(g_{1} − 1)(a − ∑_{i=1}^{p−1} a_{i}(g_{1}^{i} − 1)) = 0.

Thus there exists b ∈ ℤ such that

a − ∑_{i=1}^{p−1} a_{i}(g_{1}^{i} − 1) = b ∑_{i=0}^{p−1} g_{1}^{i},

because any annihilator of (g_{1} − 1) has this form. Multiplying both sides of the equation above by 1 + g_{1} + … + g_{1}^{p−1}, one gets

a ∑_{i=0}^{p−1} g_{1}^{i} = pb ∑_{i=0}^{p−1} g_{1}^{i}.

Thus a = pb can be divided by p.

Suppose now by induction on k that the statement is true for k − 1. We prove that it is also true for k. To simplify notation, we denote G = H⊕ < g >, where H =< g_{1} > ⊕ … ⊕ < g_{k−1} > and g = g_{k}. Since u is an integral linear combination of elements (g_{1} − 1)^{α_{1}}(g_{2} − 1)^{α_{2}} ⋯ (g_{k} − 1)^{α_{k}}, where (α_{1}, α_{2}, …, α_{k}) ∈ S_{n,k}, we can express u as

1) u = u_{n} + u_{n−1}(g − 1) + ⋯ + u_{1}(g − 1)^{n−1} + u_{g}, if n ⩽ p,

2) u = u_{n} + u_{n−1}(g − 1) + ⋯ + u_{n−p+1}(g − 1)^{p−1} + u_{g}, if n ≥ p,

where each u_{n−i} is a linear combination of terms (g_{1}−1)^{α_{1}}(g_{2}−1)^{α_{2}} ⋯ (g_{k−1}− 1)^{α_{k−1}} such that ∑_{j=1}^{k−1} α_{j} = n − i, (α_{1}, …, α_{k−1}) ∈ S_{n−i,k−1}, and u_{g} = a(g − 1)^{n} for some a ∈ ℤ. Since u is in Δ^{n+1}(G), it can be expressed using the basis B^{n+1}(G) in Theorem 1 as a sum

3) \begin{matrix}u = & v_{n+1} + v_{n}(g − 1) + ⋯ + v_{2}(g − 1)^{n−1}+ \\ & + ∑_{i=1}^{p−n} v′_{i}(g − 1)^{n−1}(g^{i} − 1) + v_{g}, & if n < p, \end{matrix}

4) u = v_{n+1} + v_{n}(g − 1) + ⋯ + v_{n−p+2}(g − 1)^{p−1} + v_{g}, if n ≥ p,

where each v_{i} ∈ Δ^{i}(H), v_{g} ∈ Δ^{n+1} < g > and each v′_{i} ∈ Δ(H).

It is easy to show that u_{g} = v_{g}. Namely since u − u_{g} and u − v_{g} are annihilated by ∑_{h∈H} h, it follows that u_{g} ∑_{h∈H} h = v_{g} ∑_{h∈H} h. But since the subring ring ℤ < g > contains no zero divisors of elements of the subring ℤH, it follows that u_{g} = v_{g}. Since u_{g} = a(g − 1)^{n} = v_{g} ∈ Δ^{n+1} < g >, we obtain from the case k = 1 above that a is divisible by p.

Subtracting u_{g} = v_{g} from u and comparing the coefficients of g^{p−1}, g^{p−2}, …, g^{0}, we get

v′_{j} = 0, for j = 1, …, p − n, and u_{i} = v_{i+1}, for i = 1, 2, …, n, if n < p,
u_{i} = v_{i+1}, for i = n − p + 1, …, n,  if n ≥ p.

Since u_{i} is a linear combination of terms (g_{1} −1)^{α_{1}}(g_{2} −1)^{α_{2}} ⋯ (g_{k−1} −1)^{α_{k−1}} with ∑_{j=1}^{k−1} α_{j} = i and v_{i} ∈ Δ^{i}(H), it follows from the induction assumption on k that each u_{i} has all coefficients divisible by p. □

We tackle now the question of comparing the rank of the 𝔽_{p} vector spaces Q_{n}(G) (n ≥ 1) for a fixed p-elementary group G. By Theorem 2, T_{n,k} is an 𝔽_{p} basis of Q_{n}(G). Let f(n, k) denote the order of the set T_{n,k}. f(n, k) has the following properties: a) f(n, 1) = 1, for any n ≥ 1.

When n ⩽ p, the number of elements in T_{n,k} with α_{k} = i is f(n − i, k − 1) for any 0 ⩽ i ⩽ n − 1 and there is only one element in T_{n,k} with α_{k} = n. Hence

b) f(n, k) = f(n, k − 1) + … + f(1, k − 1) + 1, if n ⩽ p.

From b) it follows that

f(n, k) = f(n, k − 1) + f(n − 1, k), if n ⩽ p,

and further

c) f(n, k) = \binom{n+k−1}{k−1}, if n ⩽ p.

When n > p, the number of elements in T_{n,k} with α_{k} = i is f(n − i, k − 1) for any 0 ⩽ i ⩽ p − 1 and there is only one element in T_{n,k} with α_{k} = n. It is easy to see that there are no elements in T_{n,k} with α_{k} = i for p ⩽ i ⩽ n − 1. Hence

d) \begin{matrix}f(n, k) & = f(n, k − 1) + ⋯ + f(n − p + 1, k − 1) + 1 \\ & = f(n, k − 1) + f(n − 1, k) − f(n − p, k − 1), if n > p. \end{matrix}

If n ≥ (p−1)(k−1)+1, consider the set T′_{n,k} of all elements in T_{n,k} with α_{1} ≠ 0. For elements of T′_{n,k}, the α_{i}’s where 2 ⩽ i ⩽ k can be chosen freely in set {0, 1, …, p−1}. So there is a bijection from the set {(α_{2}, …, α_{k})|0 ⩽ α_{i} < p} to the set T′_{n,k}. So the number of elements in T_{n,k} with α_{1} ≠ 0 is p^{k−1}. In general, the number of elements in T_{n,k} with α_{1} = α_{2} = … = α_{i−1} = 0 and α_{i} ≠ 0 is p^{k−i} for i = 1, 2, …, k. Thus we have

e)  \begin{matrix}f(n, k) & = p^{k−1} + p^{k−2} + … + p + 1 \\ & = (p^{k} − 1)/(p − 1),  if n ≥ (p − 1)(k − 1) + 1. \end{matrix}

The conclusion of e) implies the following result of Passi.

Theorem 3 If G is a p-elementary group of rank k, then for n ≥ (p−1)(k− 1)+1, the quotient Q_{n}(G) = Δ^{n}(G)/Δ^{n+1}(G) is a p-elementary group of rank (p^{k} − 1)/(p − 1).

There is an injective map φ from T_{n,k} to T_{n+1,k}:

(g_{i} − 1)^{α_{i}}(g_{i+1} − 1)^{α_{i+1}} … (g_{k} − 1)^{α_{k}} + Δ^{n+1}(G) ↦
(g_{i} − 1)^{α_{i}+1}(g_{i+1} − 1)^{α_{i+1}} … (g_{k} − 1)^{α_{k}} + Δ^{n+2}(G)

where α_{i} ≠ 0 but α_{1} = … = α_{i−1} = 0. So f(n + 1, k) ≥ f(n, k). If n ≥ (p − 1)(k − 1) + 1, then both sets T_{n,k} and T_{n+1,k} have the same order (p^{k} − 1)/(p − 1) by e). Hence φ is a bijection. If n ⩽ (p − 1)(k − 1), there exists α_{2}, …, α_{k} ∈ {0, 1, …, p − 1} such that ∑_{i=2}^{k} α_{i} = n. Thus (g_{1} − 1)(g_{2} − 1)^{α_{2}} … (g_{k} − 1)^{α_{k}} + Δ^{n+2}(G) ∈ T_{n+1,k}, but it has no pre-image in T_{n,k}. In this case, the map φ is not surjective. So it is a bijective if and only if n ⩾ (p − 1)(k − 1) + 1. Hence

f) f(n + 1, k) ⩾ f(n, k), with equality if and only if n ⩾ (p − 1)(k − 1) + 1.

Thus we can answer Karpilovsky’s question for all finite p-elementary groups for all primes p.

Theorem 4 If G is a p-elementary group of rank k, then the groups

Q_{1}(G), Q_{2}(G), …, Q_{(p−1)(k−1)+1}(G)

are pairwise nonisomorphic and for any n, Q_{n}(G) is isomorphic to one of these groups.

The result above allows us to answer Karpilovsky’s question for any finite elementary group G, because if G_{p} denotes the p-Sylow subgroup of G then by [5, P117], Q_{n}(G) = ⊕_{p}Q_{n}(G_{p}).

References

[1] F.Bachmann and L.Grünenfelder, The periodicity in the graded ring associated with an integral group ring, J.Pure Appl.Algebra 5(1974), 253-264.

[2] A.W.Hales and I.B.S.Passi, The augmentation quotients of finite Abelian p-groups, Contemporary Mathematics, Vol.93, 167-171, (1989)

[3] G. Karpilovsky, Commutative Group Algebras, Monographs and Textbooks in Pure and Applied Mathematics, Vol.78, Marcel Dekker,Inc. New York, 1983

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

[5] I.B.S. Passi, Group Rings and their Augmentation Ideals, Lecture Notes in Math. 1979, Vol.715