A COUNTEREXAMPLE TO THE ISOMORPHISM PROBLEM FOR INTEGRAL GROUP RINGS

MARTIN HERTWECK

Abstract. Let X be a finite group, and denote its integral group ring by ℤX. A group basis of ℤX is a subgroup Y of the group of units of ℤX of augmentation 1 such that ℤX = ℤY and |X| = |Y |. An example of a finite group X is given such that ℤX has a group basis which is not isomorphic to X. A main ingredient is the existence of a subgroup G of X which possesses a non-inner automorphism which becomes inner in the integral group ring ℤG.

The question whether a finite group X is determined by its integral group ring ℤX is known as the ‘isomorphism problem for integral group rings’. It was first posed by Graham Higman in his thesis ([3]). Investigating when nonisomorphic groups have isomorphic group algebras was stimulated by R. Brauer (Problems 2, 2^{∗} in [1]). In this paper, a counterexample to the isomorphism problem is given. This should be seen as a contribution to the program proposed by Brauer.

In order to state our result we introduce some notation. For a finite group G, and a commutative ring R with 1, we write U(RG) for the unit group of the group ring RG, and V(RG) for the group of normalized units (augmentation 1). A group basis of RG is a subgroup H of V(RG) with RG = RH and |G| = |H|. Thus the isomorphism problem asks whether there is only one isomorphism class of group bases of ℤX. We shall prove

Theorem A. There is a finite group G with a non-inner group automorphism τ, and t ∈ V(ℤG), such that gτ = g^{t} for all g ∈ G. The group G has order 2^{25} · 97^{2}, a normal Sylow 97-subgroup, and is metabelian.

Theorem B. There is a finite solvable group X, which is a semidirect product of a normal subgroup G and a cyclic subgroup ⟨c⟩, such that

1. There is a non-inner group automorphism τ of G, and t ∈ V(ℤG), such that gτ = g^{t} for all g ∈ G.

Date: February 3, 2001.

1991 Mathematics Subject Classification. Primary 20C05, 20C10; Secondary 16U60.

2 MARTIN HERTWECK

2. (tc)^{2} = c^{2}, that is, the element c inverts the element t.

3. The group Y = ⟨G, tc⟩ is a group basis of ℤX which is not isomorphic to X.

4. The group X has order 2^{21} · 97^{28}, a normal Sylow 97-subgroup, and derived length 4.

We remark that the group G given in Theorem B is different from the group G given in Theorem A. The latter group is, for the most part, simpler in structure, and provides a good introduction to the group G used in Theorem B. Beyond this expository advantage, the group G in Theorem A shows that the latter result can be achieved with a group of derived length 2, whereas we have been forced in Theorem B to use a G with derived length 3. (This makes the group X in Theorem B have derived length 4, and we have been unable to find a counterexample to the isomorphism problem with derived length 3.) Nevertheless, it will be clear that the main ingredients for Theorem B already appear in the proof of Theorem A, with many of the details entirely parallel. As some compensation to increasing the derived length, we are able to use a slightly smaller Sylow 2-subgroup for the group G in Theorem B.

The paper is organized as follows. In Section 1, a short summary on the ‘normalizer problem’, for which Theorem A provides a counterexample, is given. In Section 2, an obstruction theory to the isomorphism problem is described which provides the general framework the counterexample fits in, and also the basic idea which shows how the normalizer problem comes into play is given. Section 3 contains a description of some pullback diagrams for integral group rings which are relevant to the following constructions. Certain congruences satisfied by units in the integral group ring of a cyclic group of order 8 are established in Section 4. It is here that 97 is chosen, as the smallest prime satisfying an arithmetic condition. The congruences obtained are used to construct the unit t occurring in Theorem A explicitely. Section 6 is devoted to the proof of Theorem B. In Section 7, the embedding of the group basis Y into ℤX with regard to the Zassenhaus conjecture is discussed. Section 8 deals with an example of K. W. Roggenkamp and A. Zimmermann, explaining some of the difficulties one faces when dealing with the normalizer problem. Finally, in Section 8 a construction of M. Mazur is generalized so that it applies to group rings of finite groups too.

THE ISOMORPHISM PROBLEM FOR INTEGRAL GROUP RINGS 3

Our notation is mostly standard. For group elements x, y we set x^{y} = y^{−1}xy. By conj(y) we denote any homomorphism of the form x ↦ x^{y}. We shall use the bar convention, that is, a homomorphic image Kα of a group K is often denoted by K̄ (or K̃ or K^{∗}), and kα, for some k ∈ K, is written as k̄ (or k̃ or k^{∗}). This will be done without comment.

1. The normalizer problem

Theorem A provides a counterexample to the ‘normalizer problem’, i.e. the question whether N_{U}(G) = G · Z(U), where U = U(ℤG), holds for any finite group G. (It has been formulated in [4, 3.7], and subsequently as Problem 43 in [16]. Some of the results given in this section can be found in [16, §9].) Because of its impact on the isomorphism problem, we give a brief account of the current state of the subject under consideration.

Let G be a finite group, and R an integral domain of characteristic 0. If no prime divisor of |G| is invertible in R, we shall say that R is G-adapted. We shall identify group automorphisms of G with the R-algebra automorphisms of RG normalizing the group basis G. We define

Aut_{R}(G) = {ϕ ∈ Aut(G) : ϕ = conj(u) for some u ∈ U(RG)}.

Note that Aut_{R}(G) contains Aut_{c}(G), the group of conjugacy class preserving automorphisms of G. The quotient Aut_{R}(G)/Inn(G), denoted by Out_{R}(G), has two natural interpretations. It is the kernel of the natural map Out(G) → Out(RG), and it is isomorphic to N_{U}(G)/G·Z(U), where U = U(RG). In particular, Out_{R}(G) = 1 means that there are no ‘non-obvious’ units in RG which normalize G. In this context, mainly G-adapted coefficient rings are of interest, since Aut_{ℚ}(G) = Aut_{c}(G) by the Noether-Deuring theorem. So the normalizer problem really deals with the question whether Out_{R}(G) = 1 holds for a G-adapted ring R.

A positive result, due to J. Krempa (see [4, 3.2]), states that Out_{ℤ}(G) is an elementary abelian 2-group, for any finite group G. It is known that each prime divisor of the order of Aut_{c}(G) also divides |G|. Hence Out_{ℤ}(G) = 1 if G is of odd order, and this remains true if G has a normal Sylow 2-subgroup. For a group G whose generalized Fitting subgroup is a p-group, and an integral domain S in which p is not invertible, we have Out_{S}(G) = 1 ([2, 4.2.4]). The last two results

4 MARTIN HERTWECK

depend on an extension of a result of D. B. Coleman (see [4, 2.6]), stating that for ϕ ∈ Aut_{S}(G), and a Sylow p-subgroup P of G, there is γ ∈ Inn(G) with ϕ|_{P} = γ|_{P}.

A basic example of a G-adapted ring is ℤ_{π(G)}, the intersection of the localizations ℤ_{(p)} with p in π(G), the set of prime divisors of |G|.

M. Mazur established in [6] a connection of the normalizer problem to the isomorphism problem for infinite groups (see also Section 9). This might have inspired K. W. Roggenkamp and A. Zimmermann, who realized that the question whether Out_{ℤ_{π(G)}}(G) is trivial or not can be attacked by using some kind of a general theory, and furnished examples of finite, three-step abelian groups G with Out_{ℤ_{π(G)}}(G) ≅ C_{2}, and hence also with Out_{O}(G) ≅ C_{2} for some suitable chosen ring O of integers in an algebraic number field ([12]). However, in Section 8 it is shown that Out_{ℤ}(G) = 1 for their groups G.

We note, but do not include a proof, that Out_{ℤ_{π(G)}}(G) is always an abelian group, and that any finite abelian group A is isomorphic to Out_{ℤ_{π(G)}}(G) for some group G, in contrast to the R = ℤ case.

We remark that the normalizer problem is a problem about certain central units. For if t ∈ N_{U(RG)}(G), we may take a cyclic group ⟨x⟩ of the same order as τ := conj(t) and form the semidirect product G_{τ} := G ⋊_{τ} ⟨x⟩ with respect to τ, i.e. g^{x} = g^{t} for all g ∈ G. Then tx^{−1} is a central unit in RG_{τ}.

Finally, we point out that we are really interested in ‘multiplicative cocycles’ with values in N_{U(RG)}(G), and not only in the apparently ‘small’ quotient Out_{R}(G) (cf. end of Section 2).

2. The isomorphism problem

We do not attempt to give a survey of the isomorphism problem. For that, we refer to [14, 8, 11, 16]. We only outline some of the theory which is crucial for the understanding of the counterexample, and give the basic idea which eventually led to it.

Prerequisites are not required, except for the normal subgroup correspondence, which ensures that for certain normal subgroups N of a finite group G (for example those which are uniquely determined by their order), the kernel of the natural homomorphism ℤG → ℤG/N is characteristic.

THE ISOMORPHISM PROBLEM FOR INTEGRAL GROUP RINGS 5

The following far reaching result, which we shall call the F^{∗}-Theorem, was obtained in 1987 by L. L. Scott when collaborating with K. W. Roggenkamp ([15], [11, VI.1]). Up to now, there is no published account. The main ingredients of the proof, which depends on the strong results of A. Weiss on permutation modules ([17], appendix of [16]), are sketched in [7, 8].

A p-adic ring is the integral closure of the p-adic integers ℤ_{p} in a finite extension field of the p-adic field ℚ_{p}.

F^{∗}-Theorem. Let G be a finite group which has a normal p-subgroup N containing its own centralizer, C_{G}(N) ⊆ N. Let R be a p-adic ring. Then for any augmented automorphism α of RG which stabilizes the kernel of the natural map RG → RG/N, the groups Gα and G are conjugate in the units of RG.

The stabilization of the kernel is automatic if N = O_{p}(G) and the automorphism α comes from an automorphism of ℤG, since then the normal subgroup correspondence is available.

A conjecture of H. Zassenhaus asserts that each two group bases of an integral group ring ℤG are rationally conjugate. Though this conjecture is not true in general ([9]), it has had a big influence on the development of the isomorphism problem. The F^{∗}-Theorem confirms this conjecture for solvable groups G with O_{p′}(G) = 1 for some prime p, the case we shall be interested in.

For a solvable group G, the F^{∗}-Theorem applies to limit seriously the number of (isomorphism classes of) groups which can occur as group bases of ℤG, and the remaining possibilities can be desribed explicitely in terms of G. This is the content of the obstruction theory from [5], which began in [9]. Since it has paved the way for the construction of the counterexample to the isomorphism problem, we briefly describe it.

Let G be a finite solvable group, and H a group basis of ℤG. The basic result of the theory is that H can be described by a 1-cocycle ρ of a certain Čech cohomology class depending only on G, with H ≅ G if and only if ρ is a coboundary.

We describe the cocycle in more detail. Let R, S and T be subsets of π(G), the set of prime divisors of |G|. Put N_{S} = ⟨O_{p′}(G) : p ∈ S⟩ and G_{S} = G/N_{S}. For S ⊆ T let ϕ_{T}^{S} : ℤG_{S} → ℤG_{T} be the natural homomorphism. Then G is isomorphic to the projective limit Ĝ = \underleftarrow{lim}\{G_{S}, ϕ_{T}^{S} \}. If the image of H in ℤG_{S} is denoted

6 MARTIN HERTWECK

by H_{S}, the projective limit Ĥ = \underleftarrow{lim}\{H_{S}, ϕ_{T}^{S} \} can be formed, and Ĥ ≅ H just as before. Since the kernel of the ϕ_{T}^{S} are characteristic, G ≅ H if and only if Ĝ ≅ Ĥ as projective limits.

By the F^{∗}-Theorem, there is a family ρ = {ρ_{S,T} }_{S⊆T} of class preserving automorphisms ρ_{S,T} ∈ Aut_{c}(G_{T}) with ρ_{R,T} = ρ̄_{R,S} · ρ_{S,T} for all R ⊆ S ⊆ T on the common quotient Ḡ = G_{T}, so that Ĥ ≅ G(ρ) := \underleftarrow{lim}\{G_{S}, ϕ_{T}^{S} · ρ_{S,T} \}.

The family of all ρ arising in this way becomes a Čech cohomology set if it is assumed that ρ is a coboundary if and only if there is a family σ = (σ_{S})_{S} of automorphisms σ_{S} ∈ Aut(G_{S}) such that for S ⊆ T the automorphism σ_{S} induces an automorphism σ_{S,T} of G_{T} and ρ_{S,T} = σ_{S,T} · σ_{T}^{−1}. (The family {ρ_{S,T} }_{S⊆T} is constructed originally by a similar formula, but using conjugation isomorphisms between G_{S} and H_{S} in place of the automorphisms σ_{S}.) Then, as expected, Ĝ ≅ Ĥ if and only if ρ is a coboundary.

For our purposes, the special case where the order of the Fitting subgroup of G is divided by exactly two primes p and r suffices. Then, let N_{1} = O_{p′}(G) and N_{2} = O_{r′}(G). Since N_{1} ∩ N_{2} = 1, G is a pullback as shown below. The cocycle ρ is a class preserving automorphism δ_{c} of G/N_{1}N_{2}, and H is isomorphic to a pullback G(δ_{c}), also shown below.

[figure: G; G/N_{2}; G/N_{1}; π_{2}; G/N_{1}N_{2}; π_{1}; G(δ_{c}); G/N_{2}; G/N_{1}; π_{2}; G/N_{1}N_{2}; π_{1} · δ_{c}]

The group-theoretical obstruction is that G(δ_{c}) ≅ G if and only if there are automorphisms σ_{1} ∈ Aut(G/N_{1}) and σ_{2} ∈ Aut(G/N_{2}) which induce automorphisms σ̄_{1} and σ̄_{2} of Ḡ = G/N_{1}N_{2} such that δ_{c} = σ̄_{1} · σ̄_{2}^{−1}.

The question remains how to realize δ_{c}. In other words, how can a group basis H with H ≅ G(δ_{c}) be constructed?

We briefly explain the idea which led to the counterexample. For simplicity, we shall assume that N_{1}N_{2} has a complement K in G, so δ_{c} ∈ Aut(K). The map μ : K → K, defined by kδ_{c} = k · μ(k) for all k ∈ K, is a multiplicative cocycle, i.e. satisfies μ(xy) = μ(x)^{y} · μ(y) for all x, y ∈ K.

Let μ̃ : K → V(ℤN_{1}) be another multiplicative cocycle. Then k ↦ k · μ̃(k) defines a homomorphism from K to K(μ̃) := {k · μ̃(k) : k ∈ K} ≤ V(ℤKN_{1}). The

THE ISOMORPHISM PROBLEM FOR INTEGRAL GROUP RINGS 7

crucial observation now is that, if μ and μ̃ are related by [μ̃(k)^{−1} · μ(k), N_{1}] = 1 for all k ∈ K, then H := ⟨N_{1}N_{2}, K(μ̃)⟩ is a group basis isomorphic to G(δ_{c}). This is because the automorphism β ∈ Aut(ℤKN_{1}) = Aut(ℤG/N_{2}), defined by xβ = x for all x ∈ N_{1} and (k · μ̃(k))β = kδ_{c} for all k ∈ K, maps the image of H to the image of G and induces on ℤK the automorphism δ_{c}. Furthermore, the images of H and G in ℤG/N_{1} coincide.

The connection with the normalizer problem is obvious, for if μ and μ̃ are as above, then τ_{k} = conj(μ(k))|_{N_{1}} ∈ Aut_{ℤ}(N_{1}) for k ∈ K since it is given by conjugation with t_{k} = μ̃(k). This is a substantial family of automorphisms, and one might expect one of them might not be inner on grounds of chance alone. If some τ_{k} is a non-inner group automorphism, then Out_{ℤ}(N_{1}) ≠ 1, and we have a counterexample to the normalizer problem. However, we have to admit that we do not know whether this is a necessary condition for G and G(δ_{c}) being non-isomorphic. Finally, we point out that the cocycle condition implies that an element k of order 2 inverts the unit t_{k} = μ̃(k).

3. Some pullback diagrams

In this section, we introduce pullback diagrams describing the integral group ring ℤG of a finite group G. They show up in the presence of normal subgroups of G. Let K ⊴ G. The following notation is used throughout the paper:

K^{+} = ∑_{k∈K} k, ϵ_{K} = \frac{1}{|K|}K^{+} and η_{K} = 1 − ϵ_{K}.

Clearly, ϵ_{K} and η_{K} are central idempotents of ℚG. If K is cyclic, K = ⟨x⟩ say, put ϵ_{x} = ϵ_{K} and η_{x} = η_{K}. We write I(K) for the augmentation ideal of ℤK, so I(K)G is the kernel of the natural map ℤG → ℤG/K. The (two-sided) ideal generated by group ring elements a, b, … will be denoted by (a, b, …).

Associated with K ⊴ G is the following well-known pullback diagram of rings,

ℤG  \overline{}ℤG/K
(∗)  .
[figure: ℤG/(K^{+}); (ℤ/|K|ℤ) (G/K)]

Let L, Q ⊴ G. Denote by Γ the image of ℤG under the natural map κ : ℤG → ℤG/L ⊕ ℤG/Q. The kernel of this map is T = I(L)G ∩ I(Q)G. Since Γ can be viewed as the projection of ℤG on a summand of ℚG, there is a projection Λ on the complementary summand, namely Λ = ℤG/S with S = (L^{+}, Q^{+}). So there are pullback diagrams

[figure: ℤG; Γ; (∗∗); Λ; Ω; and; Γ; ℤG/Q; (∗ ∗ ∗)  .; ℤG/L; ℤG/LQ]

If L and Q are of coprime order, the quotient Ω in (∗∗) has a particular nice description:

Theorem 1 (Roggenkamp, Scott [9]). Let G be a finite group with normal subgroups L and Q of coprime order. With notation as above, there is a commutative diagram with exact rows (the maps being the natural ones):

[figure: 0; T; ℤG; κ; ℤG/L ⊕ ℤG/Q; 0; T; ‖; ℤG/S; π; \frac{ℤG/L}{(Q^{+}, |L|)} ⊕ \frac{ℤG/Q}{(L^{+}, |Q|)}; 0]

Proof. Since (|L|, |Q|) = 1, the map π is surjective. It remains to show that T ≅ S + T/S = ker(π), that is, S + T = (S + I(L)G) ∩ (S + I(Q)G). The inclusion “⊆” is obvious. From

|Q| · (S + I(L)G) · (1 − η_{L}η_{Q}) = |Q| · S + I(L)G · Q^{+} ⊆ S and
|Q| · (S + I(L)G) · η_{L}η_{Q} = I(L)G · (|Q| − Q^{+}) ⊆ T,

it follows that |Q| · (S + I(L)G) ⊆ S + T. The same holds with roles of L and Q reversed. This shows the inclusion “⊇” since (|L|, |Q|) = 1.

Remark 1. The diagram (∗ ∗ ∗) fits into a larger diagram, for it can be extended to the right and to the bottom by a diagram of type (∗). This has the following interesting consequence.

Assume that there is τ ∈ Aut_{ℤ}(G) which is minimal in the sense that there are x ∈ G/L and y ∈ G/Q such that τ induces the inner group automorphisms conj(x) and conj(y) on G/L and G/Q, respectively. Note that z = x̄ȳ^{−1} is central in the common quotient Ḡ = G/LQ. If τ induces an inner automorphism of Γ, there must be central units u and v of ℤG/L and ℤG/Q, respectively, such that z = ūv̄^{−1} in ℤG/LQ. Using the extended diagram, it is easily seen that this happens if there is a central unit c ∈ ℤG/LQ such that c ≡ 1 mod |L| and c ≡ z mod |Q|. (Use the right-hand extension to construct a central unit v in ℤG/Q with v̄ = c^{−1}, and note that z̄v̄ ≡ 1 mod |Q|.) Examples of such units are given in the next section.

4. Units of the group ring ℤC_{8}

Let ⟨w⟩ be a cyclic group of order 8. It is known (see [16, 10.8]) that the normalized unit group of the integral group ring ℤ⟨w⟩ is generated by w and the unit ν = 2 − w^{4} + (1 − w^{4})(w + w^{−1}). Perhaps it is more convenient to write ν in the form

ν = ϵ_{w^{4}} + η_{w^{4}}(3 + 2(w + w^{−1}))

—recall our notation for idempotents from Section 3. Then η_{w^{4}}(w + w^{−1})^{2} = 2η_{w^{4}} shows that ν corresponds to the unit 3 + 2\sqrt{2} of ℤ[\sqrt{2}]. A crucial observation is given by the following proposition.

Proposition 1. Let x, y ∈ ℤ with x^{2} − 2y^{2} = 1. Let p be a prime, and m, n ∈ ℕ such that p^{m}|x and 2^{n}|y. Then

ϑ = ϵ_{w^{4}} + η_{w^{4}}(x + y(w + w^{−1}))

is a unit in ℤ⟨w⟩, the inverse given by ϑ^{−1} = ϵ_{w^{4}} +η_{w^{4}}(x−y(w+w^{−1})). Moreover,

ϑ^{2} ≡ w^{4} mod p^{m} and ϑ^{2} ≡ 1 mod 2^{n}.

Proof. It follows from an easy pullback diagram argument that ϑ is unit in ℤ⟨w⟩, and the congruences follow from

ϑ^{2} = ϵ_{w^{4}} + η_{w^{4}}[x^{2} + 2y^{2} + 2xy(w + w^{−1})]
= ϵ_{w^{4}} + η_{w^{4}}[2x^{2} − 2 + 1 + 2xy(w + w^{−1})]
= 1 + (x^{2} − 1)(1 − w^{4}) + (1 − w^{4})xy(w + w^{−1})
= 1 + 2y^{2}(1 − w^{4}) + xy(1 − w^{4})(w + w^{−1})
= w^{4} + x^{2}(1 − w^{4}) + xy(1 − w^{4})(w + w^{−1}).

Proposition 2. Let x, y ∈ ℤ. Let p be an odd prime, and m, n ∈ ℕ such that p^{m}|x and 2^{n}|y. Then (x+y\sqrt{2})^{p} = x′ +y′\sqrt{2} for some x′, y′ ∈ ℤ with p^{m+1}|x′ and 2^{n}|y′.

Proof. Just look at the binomial expansion

\begin{matrix}x′ = \binom{p}{1}2^{(p−1)/2}xy^{p−1} + \binom{p}{3}2^{(p−3)/2}x^{3}y^{p−3} + … + \binom{p}{p−2}2x^{p−2}y^{2} + \binom{p}{p}x^{p}, \\ y′ = \binom{p}{0}2^{(p−1)/2}y^{p} + \binom{p}{2}2^{(p−3)/2}x^{2}y^{p−2} + … + \binom{p}{p−3}2x^{p−3}y^{3} + \binom{p}{p−1}x^{p−1}y. \end{matrix}

10 MARTIN HERTWECK

We will be interested in the case where ν^{k} ≡ 1 mod 8, and ν^{k} ≡ w^{4} mod q for some power q of an odd prime. Since (3 + 2\sqrt{2})^{4} = 577 + 8 · 51\sqrt{2}, Proposition 1 (applied with x = 577 and y = 8 · 51) tells us that ν^{8} ≡ 1 mod 8. We have checked that p = 97 is the smallest prime p for which there is k ∈ ℕ such that (3 + 2\sqrt{2})^{k} = pr + 8s\sqrt{2} for some r, s ∈ ℤ (take k = 12).

Corollary 1. Let q be some natural power of the prime 97. Then

θ := ν^{12q} = ϵ_{w^{4}} + η_{w^{4}}(qr_{0} + 8s_{0}(w + w^{−1}))

for some r_{0}, s_{0} ∈ ℤ. Moreover, qr_{0} ≡ 1 mod 8 (1a) (qr_{0})^{2} − 2(8s_{0})^{2} = 1 (1b) θ^{2} ≡ w^{4} mod q (1c) θ^{2} ≡ 1 mod 8 (1d)

Proof. (1a) follows from (3 + 2\sqrt{2})^{4} ≡ 1 mod 8, and the rest is obtained from the foregoing discussion together with Propositions 1 and 2.

5. Proof of Theorem A

A Sylow 2-subgroup P of G is a semidirect product

P = (⟨u : u^{32}⟩ × ⟨v : v^{4}⟩ × ⟨w : w^{8}⟩ × ⟨h : h^{256}⟩) ⋊ (⟨a : a^{64}⟩ × ⟨g : g^{2}⟩), with the action of a and g defined by

• u ∈ Z(P);

• v^{a} = u^{16}v, h^{a} = h^{5} and w^{a} = u^{4}w;

• v^{g} = v^{−1}u^{−8}, [w, g] = 1 and [h, g] = 1. Let K = ⟨u, v, g, h, a^{8}w⟩. Then K ⊴ P and P/K = ⟨ā⟩ ≅ C_{64}. Let Q = 𝔽_{97} ⊕ 𝔽_{97}. The group G is the semidirect product G = Q ⋊ P with [Q, K] = 1, and a acts (faithfully) on Q by multiplication with the matrix A = [ \begin{matrix}0 & 1 \\ ζ & 0 \end{matrix} ], where ζ is a (fixed) primitive 32^{nd}-root of unity of 𝔽_{97}. Clearly G is metabelian, of order 2^{25} · 97^{2}.

The reader interested in checking all details of Theorem B might wish at this point to familiarize himself or herself with the description of the group G used in that result, since the proof of Theorem B from Claim 5 forward will repeat and use some of the facts we are about to establish for our G.

THE ISOMORPHISM PROBLEM FOR INTEGRAL GROUP RINGS 11

An automorphism τ ∈ Aut(G) is given by aτ = u^{16}a and xτ = x for all x ∈ ⟨Q, u, v, w, g, h⟩. Note that

τ|_{P} = conj(w^{4})|_{P} .

It follows that τ is a non-inner automorphism: Otherwise τ = conj(x) for some x ∈ P since P τ = P. But then x ∈ Z(P)w^{4} ∩ C_{P} (Q) = ⟨u, h^{64}⟩w^{4} ∩ K = ∅, which is impossible.

Let

L = ⟨u^{4}⟩, a normal subgroup of G of order 8.

Using the description of ℤG given by Theorem 1, we will construct a unit t ∈ V(ℤG) with τ = conj(t). Thereby, three elements, κ_{q}, κ_{8} and θ, play an important role, and they shall be described next.

The element κ_{q} will not come by surprise. The group G has been designed so that Clifford theory can be applied to show that τ induces an inner automorphism on η_{Q} · ℤ[\frac{1}{97}]G (cf. [12]), and the conjugating element can be given explicitely.

Let I be the set of primitive idempotents of ℤ[\frac{1}{97}]Q except ϵ_{Q}, the trivial idempotent. The elements of I correspond to the nontrivial cyclic subgroups of Q. Since the matrix A does not have eigenvalues over the prime field, a fixes no idempotent of I, and it follows easily that I is the disjoint union of sets E and F with E^{a} = F, and E^{x} = E for all x ∈ ⟨Q, K, a^{2}⟩. Denote by E^{+} and F^{+} the sum of the idempotents in E and F, respectively, and let q = 97^{2}, the order of Q. Then, with

κ_{q} = q(E^{+} + u^{16}F^{+}) ∈ ℤ⟨Q, u^{16}⟩, (2a)

it follows that

xκ_{q} = κ_{q}(xτ) for all x ∈ G (2b) κ_{q} ∈ I(Q)G (2c) κ_{q} ≡ q mod (I(L)G, Q^{+}) (2d) κ_{q}^{2} = q^{2} − q · Q^{+} (2e)

In particular, τ is given by conjugation with the unit ϵ_{Q}w^{4} + η_{Q}κ_{q} of the group ring ℤ[\frac{1}{97}]⟨Q, u^{16}, w^{4}⟩. Note that also τ = conj(s), where

s = ϵ_{u^{16}} + η_{u^{16}}(u^{4}v + u^{−4}v^{−1}), (3)

12 MARTIN HERTWECK

a unit in ℤ[\frac{1}{2}]⟨u, v⟩ since s^{2} = ϵ_{u^{16}} + 2η_{u^{16}}. Hence we already know that τ becomes semilocally inner, that is, τ ∈ Aut_{ℤ_{π(G)}}(G) (see [10, 1.2.10]).

Let e_{i} be the idempotents of ℚL defined by u^{16}e_{1} = −e_{1}, u^{8}e_{2} = −e_{2} and u^{4}e_{3} = −e_{3}, i.e.

e_{1} = η_{u^{16}}, e_{2} = ϵ_{u^{16}}η_{u^{8}} and e_{3} = ϵ_{u^{8}}η_{u^{4}}. (4)

The element

κ_{8} = 8[e_{1}(u^{4}v + u^{−4}v^{−1}) + e_{2}(u^{2} + u^{−2}) + e_{3}(u + u^{−1})] ∈ ℤ⟨u, v⟩ (5a)

has the properties

xκ_{8} = κ_{8}(xτ) for all x ∈ G (5b) κ_{8} ∈ I(L)G (5c) κ_{8}^{2} = 2 · 8^{2} − 16 · L^{+} (5d)

Let r_{0}, s_{0} and θ ∈ ℤ⟨w⟩ be defined as in Corollary 1. Since

κ_{8}κ_{q} = κ_{q}κ_{8}, (6)

(r_{0}κ_{q} + s_{0}κ_{8})(r_{0}κ_{q} − s_{0}κ_{8}) ≡ 1 mod (L^{+}, Q^{+})

by (1b), (2e) and (5d). Hence the image λ of r_{0}κ_{q} + s_{0}κ_{8} in Λ = ℤG/(L^{+}, Q^{+}) is a unit, and τ induces the inner automorphism conj(λ) of Λ by (2b) and (5b).

Next, a unit γ of the pullback Γ, as shown in (∗ ∗ ∗) of Section 3, will be constructed such that τ induces the inner automorphism conj(γ) of Γ (the final result is illustrated by the figure below). By (1c), there is a unit γ_{1} of ℤG/L which maps to 1 in ℤG/L/(Q^{+}), and to the image of w^{4}θ^{2} in ℤG/LQ. Since

(7) the images of w and θ in ℤG/LQ are central elements,

γ_{1} is a central unit. (Thus conjugation by γ_{1} on G/L agrees with τ, which induces the identity there). By (1d), the element w^{4}θ^{2} maps to the image of w^{4} in (ℤ/8ℤ) (G/LQ). Since (qr_{0}w^{4} + s_{0}κ_{8})(qr_{0}w^{4} − s_{0}κ_{8}) ≡ 1 mod (L^{+}) by (1b) and (5d), the image of qr_{0}w^{4} + s_{0}κ_{8} in ℤG/Q/(L^{+}) is a unit, which, by (1a) and (5c), also maps to the image of w^{4} in (ℤ/8ℤ) (G/LQ). Hence there is a unit γ_{2} ∈ ℤG/Q which maps to the image of w^{4}θ^{2} in ℤG/LQ, and to the image of qr_{0}w^{4} + s_{0}κ_{8} in ℤG/Q/(L^{+}). By (5b), and the fact that τ induces the inner automorphism conj(w̄^{4})

THE ISOMORPHISM PROBLEM FOR INTEGRAL GROUP RINGS 13

of G/Q, τ induces the inner automorphism conj(γ_{2}) of ℤG/Q. The desired unit γ is the unit of Γ which maps to γ_{1} in ℤG/L, and to γ_{2} in ℤG/Q.

[figure: γ; γ_{2}; qr_{0}w^{4} + s_{0}κ_{8}]
∈  ∈
[figure: Γ; ℤG/Q; ∈; ℤG/Q/(L^{+})]
[figure: γ_{1}; w^{4}θ^{2}; w^{4}; \begin{matrix}qr_{0}≡1 mod 8 \\ κ_{8}∈I(L)G \end{matrix}]
∈ ∈ ∈
[figure: ℤG/L; ℤG/LQ; θ^{2}≡1 mod 8; (ℤ/8ℤ) (G/LQ)]
[figure: 1; 1]
∈  ∈
[figure: ℤG/L/(Q^{+}); (ℤ/qℤ) (G/LQ); θ^{2}≡w^{4} mod q]

From Theorem 1, and (1a), (2c), (2d) and (5c), it follows that the units λ and γ give rise to a unit t of ℤG. This is illustrated in the next figure.

ℤG ∋ t  (γ_{1}, γ_{2}) ∈ Γ

[figure: Λ ∋; r_{0}κ_{q} + s_{0}κ_{8}; image calculated using κ_{8}∈I(L)G, κ_{q}≡q mod (I(L)G,Q^{+}), qr_{0}≡1 mod 8, and κ_{q}∈I(Q)G.; (1, s_{0}κ_{8}); evident; ∈ \frac{ℤG/L}{(Q^{+}, 8)} ⊕ \frac{ℤG/Q}{(L^{+}, q)}]

Altogether, we have shown that τ = conj(t), and Theorem A is proved.—We close this section with the following remark. The units t and t^{−1} are given by

\begin{aligned}t & = η_{L}η_{Q}(r_{0}κ_{q} + s_{0}κ_{8}) + ϵ_{L}η_{Q} + η_{L}ϵ_{Q}(qr_{0}w^{4} + s_{0}κ_{8}) + ϵ_{L}ϵ_{Q}w^{4}θ^{2}, \\ t^{−1} & = η_{L}η_{Q}(r_{0}κ_{q} − s_{0}κ_{8}) + ϵ_{L}η_{Q} + η_{L}ϵ_{Q}(qr_{0}w^{4} − s_{0}κ_{8}) + ϵ_{L}ϵ_{Q}w^{4}θ^{−2}. \end{aligned} (8)

Let σ be the automorphism of ⟨u, v, w⟩ which maps each x to x^{5}. Then κ_{8}σ = −κ_{8}, κ_{q}σ = κ_{q} and θσ = θ^{−1}, so it is natural to ask whether σ can be extended to an automorphism σ̂ of G with tσ̂ = t^{−1}. This, however, is impossible since ⟨w⟩ operates on Q faithfully through multiplication with the scalar ζ^{−4}.

6. Proof of Theorem B

The group X. A Sylow 2-subgroup P of X is a semidirect product

P = (⟨u : u^{32}⟩ × ⟨v : v^{4}⟩ × ⟨w : w^{8}⟩) ⋊ (⟨a : a^{128}⟩ × ⟨b : b^{2}⟩ × ⟨c : c^{8}⟩),

the operation of a, b and c given by • u^{a} = u, v^{a} = u^{16}v and w^{a} = u^{4}w;

14 MARTIN HERTWECK

• x^{b} = x^{−1} and x^{c} = x^{5} for all x ∈ ⟨u, v, w⟩. The element a acts “the same” on ⟨u, v, w⟩ as before, but has larger order. Note that the element u is no longer central, because of the inverting action of b.

The normal Sylow 97-subgroup Q of X is the direct product of normal subgroups N and M of X, defined as follows. Let D = (⟨d_{3}⟩ × ⟨d_{2}⟩) ⋊ ⟨d_{1}⟩ ≅ C_{97}^{(2)} ⋊ C_{97} with d_{2}^{d_{1}} = d_{3}d_{2} and [d_{3}, d_{1}] = 1, and let R = D^{(2)} (the direct product of two copies of D). Then N = R^{(4)}. The group M is the additive group of the finite field 𝔽_{97^{4}}.

Next, the action of P on Q is described. The largest normal 2-subgroup of X is O_{2}(X) = C_{P} (Q) = ⟨u, v, c^{2}⟩. Let X̄ = X/O_{2}(X) (this notation is fixed throughout the rest of this section). Then P̄ = ⟨ā⟩ × H̄ with H = ⟨w, b, c⟩ and H̄ ≅ C_{8} ⋊ Aut(C_{8}), see the proof of Claim 1 below.

The subgroup H of P centralizes M, and a operates on M via multiplication with a (fixed) primitive 128^{th}-root of unity of 𝔽_{97^{4}}.

An automorphism δ ∈ Aut(D) of order 64 is given by

δ : \{ \begin{matrix}d_{1} ↦ d_{2}^{19} \\ d_{2} ↦ d_{1} \\ d_{3} ↦ d_{3}^{−19} \end{matrix} ,

since the relation (d_{2}δ)^{d_{1}δ} = d_{1}^{d_{2}^{19}} = d_{3}^{−19}d_{1} = d_{3}δ · d_{2}δ is satisfied. From 19^{16} ≡ −1 mod 97 it follows that

δ^{16} : \{ \begin{matrix}d_{1} ↦ d_{1}^{19^{8}} \\ d_{2} ↦ d_{2}^{19^{8}} \\ d_{3} ↦ d_{3}^{−1} \end{matrix} and δ^{32} : \{ \begin{matrix}d_{1} ↦ d_{1}^{−1} \\ d_{2} ↦ d_{2}^{−1} \\ d_{3} ↦ d_{3} \end{matrix} ,

so δ has order 64, and an automorphism ρ ∈ Aut(R) of order 128 is defined by (x, y)ρ = (y, xδ) for all x, y ∈ D.

The operation of P on N is defined by (r_{1}, r_{2}, r_{3}, r_{4})^{a} = (r_{1}ρ, r_{2}ρ, r_{3}ρ, r_{4}ρ),

(r_{1}, r_{2}, r_{3}, r_{4})^{w} = (r_{4}ρ^{64}, r_{1}, r_{2}, r_{3}),
(r_{1}, r_{2}, r_{3}, r_{4})^{b} = (r_{1}, r_{4}ρ^{64}, r_{3}ρ^{64}, r_{2}ρ^{64}),
(r_{1}, r_{2}, r_{3}, r_{4})^{c} = (r_{1}, r_{2}ρ^{64}, r_{3}, r_{4}ρ^{64}),

for all (r_{1}, …, r_{4}) ∈ N. This can be done since an (irreducible) representation Ψ of H̄ over 𝔽_{97} is given by

w̄Ψ = [\begin{matrix}0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ −1 & 0 & 0 & 0 \end{matrix}] , b̄Ψ = [\begin{matrix}1 & 0 & 0 & 0 \\ 0 & 0 & 0 & −1 \\ 0 & 0 & −1 & 0 \\ 0 & −1 & 0 & 0 \end{matrix}] , c̄Ψ = [\begin{matrix}1 & 0 & 0 & 0 \\ 0 & −1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & −1 \end{matrix}] ,

THE ISOMORPHISM PROBLEM FOR INTEGRAL GROUP RINGS 15

and if for h ∈ H, all entries in the matrix h̄Ψ which are equal to −1 are replaced by ρ^{64}, then h acts on N = R^{(4)} by “multiplication” from the right with this modified matrix.

The group X is now completely fixed. We will need the following facts. The automorphism δ^{16} induces fixed-point free automorphisms on Z(D) and on D/Z(D), and δ^{32} induces the identity on Z(D) and inverts the elements of D/Z(D). Analogous statements hold for ρ^{32} and ρ^{64}, and it follows that

(9a) a^{32} induces fixed-point free automorphisms on Z(Q) and on Q/Z(Q), (9b) a^{64} induces the identity on Z(N).

Let R̃ = R/Z(R) ≅ C_{97}^{(4)}, and let f_{1}, …, f_{4} ∈ R with R̃ = ⟨f̃_{1}, …, f̃_{4}⟩. Then N/Z(N), as 𝔽_{97}H̄-module, is the direct sum of the four 4-dimensional modules ⟨(r̃_{1}, …, r̃_{4}) : r_{i} ∈ ⟨f_{k}⟩⟩, k = 1, …, 4, all of them being isomorphic to the faithful irreducible module corresponding to the representation Ψ. Hence

(9c) EndH̄(N/Z(N)) ≅ Mat_{4}(𝔽_{97}).

Finally, we remark that X′ = Q⟨u^{2}, v^{2}, w^{2}⟩ and X′′ = N, so the derived length of X is 4.

The group-theoretical obstruction. In this paragraph, we give a nontrivial cocycle δ_{c} (in the meaning of Section 2). Let X^{∗} = X/QO_{2}(X), and note that we may identify X^{∗} with P̄ = ⟨ā⟩ × H̄. We exhibit a class preserving, non-inner automorphism of H̄:

Claim 1. There are automorphisms ζ_{c} and ψ of H̄, the first defined by w̄ζ_{c} = w̄, b̄ζ_{c} = b̄ and c̄ζ_{c} = w̄^{4}c̄, the second by w̄ψ = w̄c̄, b̄ψ = b̄c̄ and c̄ψ = c̄. The outer automorphism group of H̄ is ⟨ζ_{c}, ψ⟩/Inn(H̄), and isomorphic to C_{2} × C_{2}. The automorphism ζ_{c} is a class preserving automorphism.

Proof. Omit the bars temporarily, so H = ⟨w⟩ ⋊ (⟨b⟩ × ⟨c⟩) ≅ C_{8} ⋊ (C_{2} × C_{2}) with w^{b} = w^{−1} and w^{c} = w^{5}. The automorphisms ζ_{c} and ψ of H are defined by

ζ_{c} : \{ \begin{matrix}c ↦ w^{4}c \\ b ↦ b \\ w ↦ w \end{matrix} and ψ : \{ \begin{matrix}c ↦ c \\ b ↦ bc \\ w ↦ wc \end{matrix} .

16 MARTIN HERTWECK

Both automorphisms are of order 2, and [ζ_{c}, ψ] = conj(w^{2}c). The non-trivial conjugacy classes of H are as follows.

classes of elements of order 2 : {w^{4}}, ⟨w^{4}⟩c, ⟨w^{2}⟩b, ⟨w^{2}⟩wb, ⟨w^{2}⟩bc;
classes of elements of order 4 : ⟨w^{4}⟩w^{2}, ⟨w^{4}⟩w^{2}c, ⟨w^{2}⟩wbc;
classes of elements of order 8 : ⟨w^{2}⟩w, ⟨w^{2}⟩wc.

It is easily checked that ζ_{c} is a central, non-inner automorphism of H, and that ψ is a non-central automorphism of H. Let θ ∈ Aut(H); it remains to show that θ ∈ ⟨ζ_{c}, ψ⟩Inn(H). Modifying θ by an inner automorphism, it is possible to get either 1) wθ = w and cθ = w^{4}c, or 2) wθ = wc and cθ = c. Replacing θ by θζ_{c} in the first case and by θψ in the second yields wθ = w and cθ = c. The class of c is fixed by θ, since it is the only class of length 2 consisting of elements of order 2. Moreover, the class of wb is fixed by θ, since it is the only class of elements of order 2 and length 4, whose elements do not commute with the fixed point c of θ. Consequently w(bθ) = (wb)θ ∈ ⟨w^{2}⟩wb and bθ ∈ ⟨w^{2}⟩b, and it follows that θ ∈ ⟨ζ_{c}⟩Inn(H).

There is an automorphism δ_{c} = id × ζ_{c} of X^{∗}, defined by āδ_{c} = ā and h̄δ_{c} = h̄ζ_{c} for all h ∈ H. The next two claims establish that δ_{c} is a nontrivial cocycle.

Claim 2. Let φ ∈ Aut(P). Assume that φ fixes O_{2}(X), so φ induces an automorphism φ̄ of P̄. Assume further that φ̄ fixes H̄. Then φ̄|H̄ is an inner automorphism.

Proof. Suppose that φ̄|H̄ = ψ or φ̄|H̄ = ζ_{c}ψ. Then b̄φ̄ = b̄c̄, but the order of b is 2, and the order of any preimage of b̄c̄ in P is ≥ 8, a contradiction. By Claim 1, we may assume that φ̄|H̄ = ζ_{c}, and have to reach a contradiction. There are r_{i}, s_{i}, t_{i} ∈ ℤ with

aφ = u^{r_{1}}v^{r_{2}}c^{2r_{3}}w^{4r_{4}}a^{2r_{5}+1}, bφ = u^{s_{1}}v^{s_{2}}c^{2s_{3}}b, cφ = u^{t_{1}}v^{t_{2}}w^{4}c^{2t_{3}+1}.

Let n, m ∈ ℤ. Then c^{n}u^{m} = u^{m·13^{n}}c^{n} and (u^{m}c^{n})^{2} = u^{m(1+13^{n})}c^{2n}. Since 2∥1+13^{n}, it follows that the order of u^{m}c^{n} is the maximum of the orders of u^{m} and c^{n}. Hence c^{4}φ = (u^{t_{1}}c^{2t_{3}+1})^{4} implies that u^{t_{1}}c^{2t_{3}+1} has order 8, and (i) 4|t_{1}. The relation bφ · cφ = cφ · bφ gives

u^{s_{1}−t_{1}·13^{2s_{3}}} v^{s_{2}−t_{2}}w^{4}bc^{2s_{3}+2t_{3}+1} = u^{t_{1}+s_{1}·13^{2t_{3}+1}}v^{s_{2}+t_{2}}w^{4}bc^{2s_{3}+2t_{3}+1}, so (ii) 2|t_{2} and

s_{1}\underbrace{(1 − 13^{2t_{3}+1})}_{4∥} ≡ \underbrace{t_{1}}_{4| (by (i))} \underbrace{(1 + 13^{2s_{3}})}_{2|} mod 32,

which implies that (iii) 2|s_{1}. The relation bφ · aφ = aφ · bφ gives

u^{s_{1}−r_{1}·13^{2s_{3}}} v^{s_{2}−r_{2}}w^{4r_{4}}a^{2r_{5}+1}bc^{2s_{3}+2r_{3}} = u^{r_{1}+s_{1}·13^{2r_{3}} +16s_{2}}v^{s_{2}+r_{2}}w^{4r_{4}}a^{2r_{5}+1}bc^{2s_{3}+2r_{3}},

so (iv) 2|r_{2} and

\underbrace{s_{1}}_{2| (by (iii))} \underbrace{(1 − 13^{2r_{3}})}_{8|} ≡ r_{1}\underbrace{(1 + 13^{2s_{3}})}_{2∥} + 16s_{2} mod 32,

which implies (v) 8|r_{1}. From (i)–(v) it follows that the r_{i}, t_{i} may be chosen so that

aφ = u^{8r_{1}}v^{2r_{2}}c^{2r_{3}}w^{4r_{4}}a^{2r_{5}+1} and cφ = u^{4t_{1}}v^{2t_{2}}w^{4}c^{2t_{3}+1}.

The element aφ commutes with u^{4}, v^{2} and c, but not with w^{4}. This yields the contradiction [aφ, cφ] ≠ 1.

The next claim is the place where we need that Q is non-abelian (see [5, 4.12]).

Claim 3. Let φ ∈ Aut(X̄), and denote the induced automorphism on X^{∗} by φ^{∗}. Then H̄φ^{∗} = H̄, and φ^{∗}|H̄ ∉ ζ_{c} · Inn(H̄).

Proof. We may modify φ by an inner automorphism so that P̄φ = P̄. Clearly, the center M̄ × Z(N̄) of Q̄ is φ-invariant. The element ā^{64} inverts the elements of M̄, and centralizes Z(N̄) by (9b). Furthermore, ā^{64}φ = ā^{64} as P̄ = ⟨ā⟩ × H̄ and H̄ has exponent 8. Hence M̄φ = M̄ and H̄φ = CP̄(M̄)φ = CP̄(M̄) = H̄, and the first part of the claim is proved.

Moreover, w̄^{4}ā^{64}φ = w̄^{4}ā^{64}, and w̄^{4}ā^{64} centralizes N̄ and inverts the elements of M̄, so N̄φ = N̄. In particular, φ induces an automorphism of N̄H̄.

Let X̃ = X/O_{2}(X)Z(Q). We consider the elementary abelian normal subgroup Ñ as 𝔽_{97}H̃-module. Note that we may identify H̃ with H̄.

By (9c), EndH̃(Ñ) ≅ Mat_{4}(𝔽_{97}). Also End(R̃) ≅ Mat_{4}(𝔽_{97}), and for any μ ∈ Aut(R̃) there is β ∈ Aut(ÑH̃) defined by xβ = x for all x ∈ H̃ and

(r̃_{1}, …, r̃_{4})β = (r̃_{1}μ, …, r̃_{4}μ) for all r_{i} ∈ R. (⋆)

Hence for any β ∈ Aut(ÑH̃) fixing H̃ element-wise, which implies β|Ñ ∈ AutH̃(Ñ), there is μ ∈ Aut(R̃) so that (⋆) holds.

The automorphism ζ_{c} ∈ Aut_{c}(H̄) can be extended to an automorphism ζ̂_{c} ∈ Aut(ÑH̃). Indeed, with

Z_{c} = (w̄ + w̄^{−1})Ψ = [\begin{matrix}0 & 1 & 0 & −1 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ −1 & 0 & 1 & 0 \end{matrix}] ∈ GL(4, 97)

we have h̄Ψ · Z_{c} = Z_{c} · h̄ζ_{c}Ψ for all h ∈ H, so an automorphism ζ̂_{c} ∈ Aut(ÑH̃) is defined by xζ̂_{c} = xζ_{c} for all x ∈ H̃ and

(r̃_{1}, …, r̃_{4})ζ̂_{c} = (r̃_{2}r̃_{4}^{−1}, r̃_{1}r̃_{3}, r̃_{2}r̃_{4}, r̃_{1}^{−1}r̃_{3}) for all r_{i} ∈ R.

Now, assume that φ^{∗}|H̄ = ζ_{c}. The automorphism φ induces an automorphism α of ÑH̃, and by the above, there is μ ∈ Aut(R̃) such that

(r̃_{1}, …, r̃_{4})α = (r̃_{2}r̃_{4}^{−1}μ, r̃_{1}r̃_{3}μ, r̃_{2}r̃_{4}μ, r̃_{1}^{−1}r̃_{3}μ) for all r_{i} ∈ R.

For any r, s ∈ R with r̃ = s̃μ, (s̄, 1, 1, 1)φ ∈ (1, r̄, 1, r̄^{−1})·Z(N̄) and [(s̄, 1, 1, 1), N̄]φ = [(1, r̄, 1, r̄^{−1}), N̄]. Hence ⟨[(s̄, 1, 1, 1), N̄]φ : s ∈ R⟩ = ⟨[(1, r̄, 1, r̄^{−1}), N̄] : r ∈ R⟩, but the group on the left hand side equals (R̄′, 1, 1, 1)φ, a group of order |R′| ≠ 1, and the group on the right hand side equals (1, R̄′, 1, R̄′), a group of order |R′|^{2}. This is a contradiction, so φ|H̄ ≠ ζ_{c}.

We are ready to formulate the final result of this paragraph. Recall that δ_{c} = id × ζ_{c} ∈ Aut(X^{∗}). From now on, we let X̆ = X/Q. Note that X is the pullback

[figure: X; X̄; = X/O_{2}(X); X/Q =; X̆; X^{∗}] .

Claim 4. Let B be a group basis of ℤX. Assume that there exist normalized ring automorphisms φ_{1} ∈ Aut_{n}(ℤX̄) and φ_{2} ∈ Aut_{n}(ℤX̆), each, respectively, mapping the image of X onto the image of B. Assume further that φ_{2} induces an automorphism φ_{2}^{∗} of ℤX^{∗}. The automorphism φ_{1} induces an automorphism φ_{1}^{∗} of ℤX^{∗}. Hence φ_{1}^{∗}φ_{2}^{∗−1} is a group automorphism of X^{∗}. If φ_{1}^{∗}φ_{2}^{∗−1} = δ_{c}, with δ_{c} defined as above, then B and X are not isomorphic.

Proof. By the normal subgroup correspondence, φ_{1} induces an automorphism φ_{1}^{∗} of ℤX^{∗}. Assume that there is α ∈ Aut(ℤX) with Xα = B. Also by the normal subgroup correspondence, α induces automorphisms α_{1} ∈ Aut(ℤX̄) and α_{2} ∈ Aut(ℤX̆), which on their part induce (equal) automorphisms α_{1}^{∗} and α_{2}^{∗} of ℤX^{∗}.

THE ISOMORPHISM PROBLEM FOR INTEGRAL GROUP RINGS 19

Let σ_{i} = φ_{i}α_{i}^{−1}. Then σ_{1} ∈ Aut(X̄) and σ_{2} ∈ Aut(X̆), and σ_{i} induces an automorphism σ_{i}^{∗} of X^{∗}. Assume that φ_{1}^{∗}φ_{2}^{∗−1} = δ_{c}. Then

σ_{1}^{∗}σ_{2}^{∗−1} = φ_{1}^{∗}α_{1}^{∗−1}α_{2}^{∗}φ_{2}^{∗−1} = φ_{1}^{∗}φ_{2}^{∗−1} = δ_{c}.

By Claim 3, σ_{1}^{∗} fixes H̄, and σ_{1}^{∗}|H̄ ∉ ζ_{c} · Inn(H̄). So σ_{2}^{∗} fixes H̄, too, and σ_{2}^{∗}|H̄ ∈ Inn(H̄) by Claim 2. Consequently σ_{1}^{∗}σ_{2}^{∗−1}|H̄ ≠ ζ_{c} and σ_{1}^{∗}σ_{2}^{∗−1} ≠ δ_{c}, a contradiction.

The normalizer problem, again. Now parts 1 and 2 of Theorem B will be proved. The group G is the subgroup ⟨Q, u, v, w, a, b⟩ of X, so X = G ⋊ ⟨c⟩. Let S = ⟨u, v, w, a, b⟩, a Sylow 2-subgroup of G. The automorphism τ ∈ Aut(G) is defined by xτ = x for all x ∈ ⟨Q, u, v, w, b⟩ and aτ = u^{16}a. Note that

τ|_{S} = conj(w^{4})|_{S}.

It follows that τ is a non-inner group automorphism since

w^{4}Z(S) ∩ C_{S}(Q) = w^{4}⟨u^{16}, v^{2}, a^{8}⟩ ∩ ⟨u, v⟩ = ∅.

A unit t ∈ V(ℤG) will be constructed with gτ = g^{t} for all g ∈ G, and t^{c} = t^{−1}. (One might check that τ = conj(s), with s defined by (3), so τ ∈ Aut_{c}(G). As before τ can already be seen to be semilocally inner.)

Claim 5. Let I be the set of the central primitive idempotents of ℚQ except the trivial one. Then each element of the coset āH̄ fixes no idempotent from I, i.e. I is the disjoint union of sets E and F with Eā = F, and Ex̄ = E for all x ∈ ⟨H, a^{2}⟩.

Proof. The following is a list of mutual orthogonal central idempotents of ℚQ, which add up to η_{Q}.

1. ϵ_{U}η_{Z(Q)}, where U is a subgroup of index 97 in Z(Q);

2. ϵ_{V} η_{Q}, where V is a subgroup of index 97 in Q containing Z(Q). Assume that there is h ∈ H such that āh̄ fixes one these idempotents. Then, as Q/Z(Q) and Z(Q) are semisimple 𝔽_{97}⟨āh̄⟩-modules, one of these modules contains a 1-dimensional 𝔽_{97}⟨āh̄⟩-submodule. However, as 𝔽_{97}^{×} ≅ C_{3} ×C_{32} and (āh̄)^{32} = ā^{32}, this submodule is a trivial ⟨ā^{32}⟩-module, contradicting (9a). Consequently, each element of the coset āH̄ fixes no idempotent from I. Since ā operates on the orbits in I under the operation of H̄, I is a disjoint union of sets E and F as claimed.

20 MARTIN HERTWECK

Choose sets E and F with these properties. As in the proof of Theorem A, let q = 97^{28} be the order of the normal Sylow 97-subgroup Q of G, and define the element κ_{q} by (2a). Let L = ⟨u^{4}⟩, a normal subgroup of order 8 in G, define the idempotents e_{i} as in (4) and the element κ_{8} as in (5a). Finally, let θ ∈ ℤ⟨w⟩, as well as r_{0} and s_{0}, be as in Corollary 1.

Then equations (2), (5), (6) and (7) from Section 5 hold. Applying the same careful pullback diagram arguments as in the proof of Theorem A, we can define units λ, γ, and finally a unit t of ℤG (again, t and t^{−1} are given by equation (8)), so that τ = conj(t).

The elements θ, κ_{8} and κ_{q} have the following additional properties (see remark at the end of section 5; the last equation follows from Claim 5):

θ^{c} = θ^{−1}, κ_{8}^{c} = −κ_{8} and κ_{q}^{c} = κ_{q}.

It follows that t^{c} = t^{−1}, and parts 1 and 2 of Theorem B are proved.

Completing the proof of Theorem B. From the above, it follows that Y = ⟨G, tc⟩ is a finite subgroup of V(ℤX) of the same order as X, so Y is a group basis of ℤX.

Recall that we write X̄ = X/O_{2}(X), X̆ = X/Q and X^{∗} = X/QO_{2}(X).

The next claim, which is already obvious from the preceding constructions, shows in combination with Claim 4 that the groups X and Y are not isomorphic, and all parts of Theorem B are proved.

Claim 6. There are automorphisms 1. φ_{1} ∈ Aut(ℤX̄), defined by x̄φ_{1} = x̄ for all x ∈ K := ⟨Q, w, a, b⟩ and c̄φ_{1} = t̄c̄.

2. φ_{2} ∈ Aut(ℤX̆), defined by x̆φ_{2} = x̆ for all x ∈ S = ⟨u, v, w, a, b⟩ and c̆φ_{2} =

w̆^{4}t̆c̆. Both automorphisms map the image of X onto the image of Y, and induce automorphisms of ℤX^{∗}. The automorphism φ_{1}^{∗}φ_{2}^{∗−1} ∈ Aut(ℤX^{∗}) is induced from the group automorphism δ_{c} = id × ζ_{c} of X^{∗}.

Proof. We have X̄ = K̄ ⋊ ⟨c̄⟩. The element t̄ centralizes K̄, because τ induces the identity on K̄. The element t̄c̄ has the same order as c̄, namely 2. Hence φ_{1} is an automorphism. We have X̆ = S̆ ⋊ ⟨c̆⟩. The element w̆^{4}t̆ centralizes S̆, because the automorphism of X̆ induced by τ is given by conjugation with w̆^{4}. The element w̆^{4}t̆c̆ has the same order as c̆, namely 8. Hence φ_{2} is an automorphism. Obviously φ_{1}^{∗}φ_{2}^{∗−1} = δ_{c}.

7. Concluding remarks

For the moment, let G be an arbitrary finite group. We have already noted in Section 2 that the conjecture of Zassenhaus, which asserts that each two group bases of ℤG are rationally conjugate, is not true in general. However, one can formulate a ‘block version’ of this conjecture, namely that for any group basis H of ℤG, the projections of G and H on each block of the rational group algebra ℚG are conjugate within the units of the block. (Incidentally, it is already a serious question whether the projections on the block are isomorphic.) No counterexamples are known to this conjecture. This doesn’t contradict Theorem B due to the fact that the center of X is not cyclic (it contains ⟨u^{16}, v^{2}⟩). Indeed, the following remark shows that the group X provides no counterexample.

Remark 2. The projections of X and Y on each block of the rational group algebra ℚX are conjugate within the units of the block. Conceptionally, this follows from a semilocal analysis of the counterexample involving central automorphisms, one of ℚP, and one of ℚT, where T is the inertia group of E^{+}, which induce on each block a group automorphism. Here, we content ourselves with listing explicitely conjugating units.

In ℚX, there is a decomposition ∑_{i=1}^{6} f_{i} = 1 with central idempotents

f_{1} = ϵ_{L}ϵ_{Q}, f_{2} = ϵ_{L}η_{Q}, f_{3} = η_{L}ϵ_{Q},
f_{4} = ϵ_{u^{16}}η_{L}η_{Q}, f_{5} = η_{u^{16}}ϵ_{v^{2}}η_{Q}, f_{6} = η_{u^{16}}ϵ_{u^{16}v^{2}}η_{Q}.

We give rational group ring elements μ_{i}, μ′_{i} such that f_{i}μ′_{i}μ_{i} = f_{i} and f_{i}μ′_{i}Xμ_{i} = f_{i}Y. Note that

f_{1}tc = f_{1}w^{4}θ^{2}c, f_{2}tc = f_{2}c, f_{3}tc = f_{3}(w^{4}qr_{0} + s_{0}κ_{8})c,
f_{4}tc = f_{4}(qr_{0} + s_{0}κ_{8})c, f_{i}tc = f_{i}(qr_{0}(E^{+} − F^{+}) + s_{0}κ_{8})c (i = 5, 6).

We may take μ_{1} = θ^{−1}, μ′_{1} = θ and μ_{2} = μ′_{2} = 1. Somewhat harder to find are

μ_{3} = \frac{4s_{0}}{qr_{0} − 1} − \frac{1}{32}w^{4}κ_{8} and μ′_{3} = 16s_{0} + \frac{qr_{0} − 1}{8}w^{4}κ_{8}.

22 MARTIN HERTWECK

Let

β_{±} = ±\frac{4s_{0}}{qr_{0} − 1} − \frac{1}{32}κ_{8} and β′_{±} = ±16s_{0} + \frac{qr_{0} − 1}{8}κ_{8}.

Again it is tedious but straightforward to check that η_{L}·β′_{±}β_{±} = η_{L} and η_{L}·β′_{±}cβ_{±} = η_{L} · (qr_{0} ± s_{0}κ_{8})c, and that we may take

μ_{4} = β_{+}, μ′_{4} = β′_{+},

μ_{5} = [β_{+}(ϵ_{v} + (u^{4} + u^{−4})η_{v})E^{+}] − [β_{−}(η_{v} + (u^{4} + u^{−4})ϵ_{v})F^{+}],

μ′_{5} = [β′_{+}(ϵ_{v} + \frac{1}{2}(u^{4} + u^{−4})η_{v})E^{+}] − [β′_{−}(η_{v} + \frac{1}{2}(u^{4} + u^{−4})ϵ_{v})F^{+}],

μ_{6} = [β_{+}(ϵ_{u^{8}v} + (u^{4} + u^{−4})η_{u^{8}v})E^{+}] − [β_{−}(η_{u^{8}v} + (u^{4} + u^{−4})ϵ_{u^{8}v})F^{+}],

μ′_{6} = [β′_{+}(ϵ_{u^{8}v} + \frac{1}{2}(u^{4} + u^{−4})η_{u^{8}v})E^{+}] − [β′_{−}(η_{u^{8}v} + \frac{1}{2}(u^{4} + u^{−4})ϵ_{u^{4}v})F^{+}].

Remark 3. The automorphisms φ_{1} and φ_{2} defined in Claim 6 are central automorphisms (in accordance with the F^{∗}-Theorem). The automorphism φ_{1} is given by conjugation with ω̄_{1}, where

ω_{1} = ϵ_{Q}θ^{−1}[ϵ_{w^{4}} + η_{w^{4}}(w + w^{−1})] + η_{Q}

is a unit in ℚ⟨Q, w⟩, with ω_{1}^{−1} = ϵ_{Q}θ[ϵ_{w^{4}}+η_{w^{4}} \frac{1}{2}(w+w^{−1})]+η_{Q}. The automorphism φ_{2} is given by conjugation with ω̆_{2}, where ω_{2} = ϵ_{L}θ^{−1} + η_{L}μ_{3}, with μ_{3} defined as above.

Remark 4. One might ask whether there are counterexamples of odd order, since the way the counterexample is produced, it must be necessarily of even order. As indicated in Section 1, this limitation drops if one works over a semilocal coefficient ring, so in principle it should be possible to construct counterexamples X of odd order, at least for X-adapted coefficient rings, using the ideas of this paper.

One might also try to construct non-isomorphic groups X and Y (possibly of odd order) such that an isomorphism RX ≅ RY, R = ℤ_{π(X)}, can be proved working ‘a prime at a time’ (for the group X and Y of Theorem B this is possible and leads to more insight). This helps to separate the representation-theoretic and the number-theoretic aspects of the problem, and makes it more transparent.

8. To the example of Roggenkamp and Zimmermann

A. Zimmermann and K. W. Roggenkamp constructed in [12] a finite group G and a non-inner automorphism α of G which becomes inner in ℤ_{π(G)}G. In this section it is proved that α does not induce an inner automorphism of ℤG. The group G is given as a semidirect product

G = (M_{1} × M_{2} × ⟨a : a^{8}⟩) ⋊ Q_{8} ⋊ ⟨σ : σ^{2}⟩.

The group M_{i} is an elementary abelian normal subgroup of order p_{i}^{2}, where p_{1} and p_{2} are two distinct odd primes. The group Q_{8} = ⟨b, c⟩ is a quaternion group of order 8 which acts faithfully on M_{1} and M_{2}, and on the normal subgroup ⟨a⟩ via a^{b} = a^{−1} and a^{c} = a^{5}. The involution σ commutes with M_{1}, inverts the elements of M_{2}, and acts on ⟨a, b, c⟩ via a^{σ} = a, b^{σ} = b and c^{σ} = a^{4}c.

The automorphism α is defined by xα = x^{σ} for all elements x of the Sylow 2-subgroup P = ⟨a, b, c, σ⟩, and xσ = x for all x ∈ M_{1}M_{2}.

Let π_{i} : ℤM_{i}P → ℤQ_{8} be the ring homomorphisms induced by the natural homomorphisms M_{i}P → M_{i}P/K_{i}, where K_{i} = M_{i}⟨a, σ⟩. It is shown that α induces a non-inner automorphism of the pullback of the pair {π_{1}, π_{2}}, which clearly implies that α induces a non-inner automorphism of ℤG. Assume, by the way of contradiction, that α induces an inner automorphism of the pullback of {π_{1}, π_{2}}. Since α is given by conjugation with σ on ℤM_{1}P, and by conjugation with c^{2}σ on ℤM_{2}P, there are central units z_{i} of V(ℤM_{i}P) with z_{1}π_{1} · z_{2}π_{2} = c^{2}. The crucial point here is that the center of V(ℤQ_{8}) equals the center ⟨c^{2}⟩ of Q_{8} (by a well-known result of G. Higman, even V(ℤH) = H for a Hamiltonian 2-group). So either z_{1}π_{1} = c^{2} or z_{2}π_{2} = c^{2}. The class sums C of ℤM_{1}P with Cπ_{1} a multiple of c^{2} are precisely the class sums of the elements of the coset K_{1}c^{2}. However, the length of each of these class sums is divisible by the order of M_{1}. Hence the coefficient of c^{2} in z_{1}π_{1} (viewed as a linear combination of elements of Q_{8}) is divisible by the order of M_{1}, so z_{1}π_{1} = 1 and z_{2}π_{2} = c^{2}. Let P̄ = P/⟨a⟩ = ⟨σ̄⟩ × Q̄_{8}, and look at the commutative diagram

[figure: ℤM_{2}P; ℤM_{2}P̄; ℤP; κ_{2}; ℤP̄; κ_{1}]

Again, the center of V(ℤP̄) is ⟨c̄^{2}, σ̄⟩, the center of P̄. Moreover, π_{2} factors through ℤP̄, so either z̄_{2} = c̄^{2} or z̄_{2} = c̄^{2}σ̄ in ℤP̄. Each class sum of ℤM_{2}P̄ whose image under κ_{1} is a multiple of c̄^{2} has length divisible by the order of M_{2}, so z̄_{2} = c̄^{2}σ̄.

24 MARTIN HERTWECK

On the other hand, as Z(P) = ⟨a^{4}, c^{2}⟩, each class sum of ℤP whose image under κ_{2} is a multiple of c̄^{2}σ̄ has even length, so z̄_{2} ≠ c̄^{2}σ̄. This provides a contradiction.

9. A remark to a construction of Mazur

M. Mazur established in [6] the following connection of the normalizer problem to the isomorphism problem for infinite groups. For any finite group G and τ ∈ Aut(G) we might form the semidirect product G_{τ} = G ⋊_{τ} ⟨c⟩ of G by an infinite cyclic group ⟨c⟩ with g^{c} = gτ for all g ∈ G (that is, c acts on G via τ). Then G_{τ} ≅ G_{id} if and only if τ ∈ Inn(G), and for a G-adapted ring R, RG_{τ} ≅ RG_{id} if and only if τ ∈ Aut_{R}(G). This idea together with the example from [12] yield a counterexample to the isomorphism problem for polycyclic groups [13]. A modified construction works for finite groups too, see the proposition below. However, it should be remarked that condition (ii) of this proposition seems to be difficult to verify in practice.

Proposition 3. Let G be a finite group. Let τ ∈ Aut(G) and t ∈ V(ℤG) such that gτ = g^{t} for all g ∈ G. Assume that there is γ ∈ Aut(G) such that

(i) tγ = t^{−1}; (ii) No conjugate of γ in Aut(G) differs from an automorphism of the coset τ⟨γ⟩

only by an inner automorphism. Let ⟨c⟩ be a finite cyclic group whose order is divisible by the orders of γ and τγ, so the groups G_{γ} = G ⋊_{γ} ⟨c⟩ and G_{τγ} = G ⋊_{τγ} ⟨c⟩ can be defined, where c acts on G via γ and τγ, respectively. Then the integral group rings ℤG_{γ} and ℤG_{τγ} are isomorphic. However, the order of c can be chosen such that the groups G_{γ} and G_{τγ} are not isomorphic.

Proof. Consider tc as an element of ℤG_{γ}. From (i) it follows that (tc)^{2} = c^{2}. In particular, tc is of finite order.

Assume that tc or c is of odd order. Then t^{2} = 1 since t^{tc} = t^{−1} = t^{c} by (i). As t normalizes G, it follows that ⟨G, t⟩ is a finite subgroup of V(ℤG). By (ii), τ is a non-inner automorphism of G, so ⟨G, t⟩ has order strictly greater then |G|. But it is well-known that this is impossible, so tc and c have the same order.

It follows that G_{τγ} is isomorphic to the subgroup ⟨G, tc⟩ of V(ℤG_{γ}), so the integral group rings ℤG_{γ} and ℤG_{τγ} are isomorphic.

Also, it follows that γ and τγ have the same order n (say). Denote the order of

G by m, let c be of order nm^{2}, and suppose that an isomorphism φ : G_{γ} → G_{τγ}

exists. Let K = ⟨c^{nm}⟩, a central subgroup of G_{γ} of order m. Then Kφ = K as

c^{n}φ ∈ G × ⟨c^{n}⟩. Since Gφ ⊆ G × K, there is θ ∈ Aut(G) with gφ ∈ gθ · K for

all g ∈ G. Let cφ = xc^{r} for some x ∈ G and r ∈ ℕ. Note that r is odd since φ

is surjective. For each g ∈ G, gγφ ∈ gγθ · K, as well as gγφ = (g^{c})φ = gφ^{cφ} ∈

(gθ)^{x}(τγ)^{r} · K. Consequently γ · θ = θ · conj(x) · (τγ)^{r} as automorphisms of G.

Finally, (τγ)^{r} = τ · γ^{r} as r is odd, so γ^{θ} = conj(x) · τγ^{r}, contradicting (ii).

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Mathematisches Institut B, Universität Stuttgart, Pfaffenwaldring 57, D-70569 Stuttgart, Germany

E-mail address: hertweck@mathematik.uni-stuttgart.de