Beiträge zur Algebra und Geometrie Contributions to Algebra and Geometry Volume 43 (2002), No. 2, 513-520.

Another Counterexample to a Conjecture of Zassenhaus

M. Hertweck^{∗}

Mathematisches Institut B, Universität Stuttgart Pfaffenwaldring 57, D-70569 Stuttgart, Germany e-mail: hertweck@mathematik.uni-stuttgart.de

Abstract. A metabelian group G of order 1440 is constructed which provides a counterexample to a conjecture of Zassenhaus on automorphisms of integral group rings. The group is constructed in the spirit of [8]. An augmented automorphism of ℤG which has no Zassenhaus factorization is given explicitly (this was already done in [7] for a group of order 6720), but this time only a few distinguished group ring elements are used for its construction, carefully exploiting certain congruence relations satisfied by powers of these elements.

1. Introduction

Let G be a finite group, and denote its integral group ring by ℤG. A group basis of ℤG is a subgroup H of the group of units of ℤG of augmentation 1 such that ℤG = ℤH and |G| = |H|. H. Zassenhaus conjectured that each two group bases of ℤG are conjugate by a unit of ℚG (i.e., “rationally conjugate”). However, Roggenkamp and Scott constructed in [8] a counterexample to this conjecture.

Note that, by the Skolem-Noether theorem, the Zassenhaus conjecture asserts that G can be mapped onto any group basis of ℤG by a central ring automorphism of ℤG (an automorphism fixing the center element-wise). We shall say that an augmentation preserving automorphism α of ℤG has a Zassenhaus factorization if it is the composition of a group automorphism of G (extended to a ring automorphism) and a central automorphism (cf. [11, p. 327]).

∗The research was supported by Deutsche Forschungsgemeinschaft 0138-4821/93 $ 2.50 \begin{matrix}c \\ ◯ \end{matrix} 2002 Heldermann Verlag

Roggenkamp and Scott produced a metabelian group G of order 2880 such that there is an augmentation preserving automorphism α of ℤG which has no Zassenhaus factorization. Then, G and its image Gα are group bases of ℤG which are not rationally conjugate. Their construction of the automorphism α is explicit in the semilocal situation. To show that their example is also a global counterexample, they developed some kind of a general theory— using Picard groups and Milnor’s Mayer Vietoris sequence. This work is excellently outlined in [11].

Subsequently, Klingler [7] constructed a global automorphism explicitly. (However, he changed slightly the definition of the group G—replacing a normal subgroup of order 3 by a normal subgroup of order 7—in order to write ℤG as a “multiple pullback”, taken over semisimple finite skew group rings. Then, he essentially used the fact that certain elements of these rings, which, roughly speaking, correspond to elementary matrices, can be lifted to units of the factors.)

Since then, Blanchard [1, 2] constructed counterexamples in the semilocal case, using the idea from [11, Section 2] (see also [4, 2.1]).

In this paper, another counterexample is given:

Theorem. There is a metabelian group G of order 1440 (= 2^{5} · 3^{2} · 5) and an automorphism α of ℤG which has no Zassenhaus factorization.

The group G has been designed to satisfy a certain group-theoretical obstruction, in the very same way as Roggenkamp and Scott constructed their group. The elementary construction of the automorphism involves only a few elements of ℤG, carefully exploiting certain congruence relations satisfied by powers of these elements.

Finally, it should be remarked that a weaker question simply asks whether any two group bases of an integral group ring ℤG are isomorphic (this is the so-called “isomorphism problem for integral group rings”). There are some exciting positive results, but also a counterexample has been found (see [3, 4]).

2. The group-theoretical obstruction

The group G is the semidirect product G = (M × N × Q) ⋊ W, where • W = ⟨w : w^{8}⟩ ⋊ (⟨b : b^{2}⟩ × ⟨c : c^{2}⟩), with w^{b} = w^{−1} and w^{c} = w^{5};

• M = ⟨m : m^{5}⟩, N = ⟨n : n^{3}⟩ and Q = ⟨q : q^{3}⟩;

• C_{W} (m) = ⟨wc, b⟩, C_{W} (n) = ⟨w^{2}, b, c⟩ and C_{W} (q) = ⟨w, b⟩ (these are subgroups of index

2 in W). For a group X, let Aut_{c}(X) denote the group of class-preserving automorphisms of X, and write Out_{c}(X) = Aut_{c}(X)/Inn(X). It is well-known that the group W has a non-inner, class-preserving automorphism δ of order 2 (see [12]), which maps c to w^{4}c, and b, w stay fixed. (We remark that Out(W) ≅ C_{2} × C_{2}.) Here, an extension of this automorphism to an automorphism σ of G will be considered, defined by

σ : \{ \begin{matrix}c ↦ w^{4}c \\ b, w, m, n, q stay fixed \end{matrix} .

Let ϵ = \frac{1}{2}(1 + w^{4}), a central idempotent of the group algebra ℚG.

As in [11], write

G_{3} = G/M, G_{5} = G/N and H = G/MN.

Note that we may identify G_{3} with NQW, G_{5} with MQW, and H with QW.

Lemma 2.1. The automorphism σ induces an inner automorphism of ℤ[1/2]H, given by conjugation with the unit

μ = ϵ + (1 − ϵ)(w + w^{−1}) ∈ ℤ[1/2]⟨w⟩.

Proof. It follows from (1 − ϵ)(w + w^{−1})^{2} = 2(1 − ϵ) that μ is a unit, and it is easily checked that xμ = μ(xσ) for the given generators x of H. □

Together with the next lemma, this shows that, at the group-theoretic level, [σ] is a “candidate obstruction” in the sense of [11, p. 330]. Write x ≁ y to indicate that group elements x and y are not conjugate.

Lemma 2.2. We have Out_{c}(G_{3}) = 1 and Out_{c}(G_{5}) = 1.

Proof. Let ϕ ∈ Aut_{c}(NQW); we have to show ϕ ∈ Inn(NQW). We may assume that wϕ = w, bϕ = b and either cϕ = c or cϕ = w^{4}c. From qw ≁ q^{−1}w it follows that qϕ = q. If cϕ = c, then nb ≁ n^{−1}b implies that ϕ = id. So assume that cϕ = w^{4}c. From [n, c] = 1 it follows that nc and (nc)ϕ are conjugate in W. As c^{w} = w^{4}c = cϕ and C_{W} (c) = C_{W} (n), it follows that nϕ = n^{w} = n^{−1}, yielding the contradiction nb ≁ n^{−1}b = (nb)ϕ.

Similar for ϕ ∈ Aut_{c}(MQW). Again, we may assume that wϕ = w, bϕ = b and either cϕ = c or cϕ = w^{4}c. From qw ≁ q^{−1}w it follows that qϕ = q. If cϕ = c, then mwc ≁ m^{−1}wc implies that ϕ = id. So assume that cϕ = w^{4}c. From [m, wc] = 1 it follows that mwc and (mwc)ϕ are conjugate in W. As wc and w^{5}c (= (wc)ϕ) are not conjugate in C_{W} (m), mϕ = m^{−1}. From [mqb, b] = 1 it follows that mqb and (mqb)ϕ (= m^{−1}qb) are conjugate in W. Hence m^{x} = m^{−1} for some x ∈ C_{W} (q) ∩ C_{W} (b) = ⟨w^{4}, b⟩ ⊆ C_{W} (m), a contradiction. □

We shall need the following simple observation from [8].

Remark 2.3. Let x and y be elements of a group with y^{x} = y^{−1}. Then (y − y^{−1})x = −x(y − y^{−1}). □

For a group X, write X̂ for the sum of its elements. The (two-sided) ideal generated by group ring elements s, t, … will be denoted by (s, t, …). The quotient

Λ = ℤG/(M̂, N̂)

is the projection on a factor of ℚG (to which all blocks having neither M nor N in their kernel belong).

Though the next lemma is not really needed in the construction of the automorphism α of ℤG, a short proof is included. (It could also be proved character-theoretically, analyzing the inertia groups of the characters of Λ, just as in [8]). It already shows that the group G provides a semilocal counterexample (cf. [11, p. 333–34]).

Lemma 2.4. The automorphism σ induces a central automorphism of Λ.

Proof. We have to show that σ induces an inner automorphism of ℚΛ. On Λ, the group elements m and n correspond to, roughly speaking, primitive fifth and third roots of unity, respectively. Hence the images u and v of m − m^{−1} and n − n^{−1} in ℚΛ are units. It follows from Remark 2.3 that both units normalize Λ, and that on Λ,

conj(u) : \{ \begin{matrix}c ↦ −c \\ w ↦ −w \\ b, m, n, q stay fixed \end{matrix} ,

conj(v) : \{ \begin{matrix}w ↦ −w \\ b, c, m, n, q stay fixed \end{matrix} ,

conj(uv) : \{ \begin{matrix}c ↦ −c \\ b, w, m, n, q stay fixed \end{matrix} .

Note that ℚΛ = ϵℚΛ ⊕ (1 − ϵ)ℚΛ. As w^{4} = −1 on (1 − ϵ)Λ, it follows that σ on (1 − ϵ)ℚΛ is given by conjugation with (1 − ϵ)uv. Since σ induces the identity on ϵℚΛ, this proves the lemma. □

3. The automorphism α

The quotient Λ = ℤG/(M̂, N̂) is the projection of ℤG on a factor of ℚG. The projection of ℤG on the complementary factor is the image Γ of ℤG under the natural map ℤG → ℤG_{3} ⊕ ℤG_{5}. Hence there are pull-back diagrams

[figure: Γ; ℤG_{5}; ℤG_{3}; ℤH] and [figure: ℤG; Γ; Λ; Ω] .

Note that N and M can be viewed as normal subgroups of G_{3} and G_{5}, respectively. As in [11], let

Λ_{3} = ℤG_{3}/(N̂) and Λ_{5} = ℤG_{5}/(M̂).

Roggenkamp and Scott proved in [8, Section 2] (see also [3, 4]) that the ring Ω has the form

Λ_{3}/5Λ_{3} ⊕ Λ_{5}/3Λ_{5}.

Let

ν = 1 + (1 − w^{4})(1 + w + w^{−1})
= ϵ + (1 − ϵ)(3 + 2(w + w^{−1})).

From (1 − ϵ)(w + w^{−1})^{2} = 2(1 − ϵ) it follows that ν is a unit of ℤ⟨w⟩, with inverse ν^{−1} = ϵ + (1 − ϵ)(3 − 2(w + w^{−1})).

From ν^{3} = ϵ + (1 − ϵ)(99 + 70(w + w^{−1})) it follows that

\begin{matrix}ν^{3} ≡ ϵ − (1 − ϵ) = w^{4} & mod 5ℤ⟨w⟩, \\ ν^{3} ≡ μ = ϵ + (1 − ϵ)(w + w^{−1}) & mod 3ℤ[1/2]⟨w⟩. \end{matrix}

Given a normal subgroup Y of a finite group X, there is a well-known pull-back diagram of rings

[figure: ℤX; ℤX/(Ŷ); ℤX/Y; (ℤ/|Y |ℤ) (X/Y)] .

Note that the pull-back diagram describing Γ can be extended to the right-hand side by a diagram of this type.

The automorphism σ of G induces automorphisms of G_{5} and H, which, for simplicity, shall be denoted by the same symbol. Recall that σ = conj(μ) on H, so σ = conj(ν^{3}) on 𝔽_{3}H, and there is γ ∈ Aut(Γ), inducing σ on ℤG_{5} and a central automorphism β on ℤG_{3}, as shown below.

[figure: γ; β; conj(ν^{3})]
[figure: Γ; ℤG_{3}; Λ_{3}]

[figure: σ; conj(μ); conj(μ)]
[figure: ℤG_{5}; ℤH; 𝔽_{3}H]

As ν^{3} ≡ w^{4} mod 5 and w^{4} ∈ Z(G),

γ induces the identity on Λ_{3}/5Λ_{3}, and induces σ on Λ_{5}/3Λ_{5}.  (1)

On Λ, the group element mn corresponds to a “primitive 15-th root of unity”, so the image t of mn − (mn)^{−1} in Λ is a unit in Λ (cf. [13, Proposition 2.8]). Binomial expansion gives (mn−(mn)^{−1})^{15} = 1−( ∑_{i=1}^{14}(−1)^{i}\binom{15}{i}(mn)^{2i})−1, and for 1 ≤ i ≤ 14 it follows from \binom{15}{1} ≡ 1 mod 2 and \binom{15}{i} + \binom{15}{i+1} = \binom{16}{i+1} ≡ 0 mod 2 that \binom{15}{i} ≡ 1 mod 2. Hence (mn − (mn)^{−1})^{15} ≡ 1 + M̂N̂ ≡ 1 mod 2 and consequently t^{15} ∈ 1 + 2Λ.

Recall from Lemma 2.4 the definition of the automorphisms conj(u) and conj(v) of Λ. Note that Λ can be written as a pull-back

[figure: Λ; (1 − ϵ)Λ; ϵΛ; \overline{Λ}] ,

where 2 = 0 in \overline{Λ} since 2ϵ ∈ Λ. The automorphism conj(v)conj(t^{15}) of Λ induces an automorphism ϕ of (1 − ϵ)Λ, which in turn induces the identity on \overline{Λ}. It follows that there is an automorphism λ of Λ which induces ϕ on (1 − ϵ)Λ, and the identity on ϵΛ.

The elements mn − (mn)^{−1} and n − n^{−1} map to the same element in ℤG_{3}, and it follows that the inner automorphism conj(t) and the automorphism conj(v) induce the same automorphism (of order 2) on (1 − ϵ)Λ_{3}/5Λ_{3}. Similarly, mn − (mn)^{−1} and m − m^{−1} map to the same element in ℤG_{5}, so that conj(t) and conj(u) induce the same automorphism (of order 2) on (1 − ϵ)Λ_{5}/3Λ_{5}. As conj(v)conj(u) and σ induce the same automorphism of (1 − ϵ)Λ (see the proof of Lemma 2.4), it follows that

λ induces the identity on Λ_{3}/5Λ_{3}, and induces σ on Λ_{5}/3Λ_{5}.  (2)

Using the given description of ℤG as a pull-back, it follows from (1) and (2) that there is an automorphism α of ℤG, inducing γ on Γ and λ on Λ.

As γ induces σ on ℤG_{5} (which implies that α preserves the augmentation), and a central automorphism on ℤG_{3}, it follows from Lemma 2.2 and the discussion in [11, p. 327–28] that α has no Zassenhaus factorization, and the theorem is proved.

4. Conjugacy of Sylow subgroups

Let G be a finite group, H a group basis of ℤG, and fix a prime p. It has been shown in [6] that if G is solvable, then Sylow p-subgroups of G and H are conjugate in the units of ℚG.

One might also ask if the Sylow p-subgroups are conjugate in the units of ℤ_{p}G, where ℤ_{p} are the p-adic integers.

For nilpotent groups, this problem is solved by Theorem 2 of [14], except for the determination of a global obstruction in a certain genus class group. Here we provide an example, based on the group W and the unit ν.

Example 4.1. Let G = W × C, with C a cyclic group of order 3. Then there is a group basis H of ℤG such that Sylow 2-subgroups of G and H are not conjugate in the units of ℤ_{2}G.

Proof. Recall from the beginning of Section 2 that W has a non-inner class-preserving automorphism δ, given by conjugation with the unit μ of ℤ[1/2]⟨w⟩. As ν^{3} ≡ μ mod 3, it follows that there is an augmentation preserving automorphism α of ℤG which induces δ on ℤG/C ≅ ℤW and is given by conjugation with the image of ν^{3} on ℤG/(Ĉ), as indicated below.

α ^{}δ
ℤG  \overline{}ℤW

[figure: ℤG/(Ĉ); 𝔽_{3}W]
[figure: conj(ν^{3}); conj(μ)]

In ℤ_{2}G, there are central idempotents ϵ = \frac{1}{3} Ĉ and η = 1 − ϵ, and W α is conjugate in the units of ℤ_{2}G to {ϵ(xδ) + ηx : x ∈ W }. Assume that this group is conjugate to W within the units of ℤ_{2}G. Then there is an automorphism β of W such that either δβ^{−1} induces an inner automorphism of the group ring ϵℤ_{2}G ≅ ℤ_{2}W or β induces an inner automorphism of the group ring ηℤ_{2}G ≅ ℤ_{2}[ζ]W, where ζ is a primitive third root of unity. As δ is a non-inner group automorphism, this contradicts a result of D. B. Coleman (see [5, 2.6]). □

However, it should be remarked that it is not known whether the projections of the Sylow p-subgroups of G and H on the principal block of ℤ_{p}G are conjugate within the units of the block. This is a special case of Scott’s defect group (conjugacy) question (see [9, p. 267], [10]).

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520 M. Hertweck: Another Counterexample to a Conjecture of Zassenhaus

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Received March 26, 2001