Outer group automorphisms may become inner in the integral group ring

Klaus W. Roggenkamp and Alexander Zimmermann

Abstract

In this note we shall construct a finite group G which has an automorphism α, which is not inner; however, the induced automorphism on SG is inner, where S is the ring of algebraic intgers in a suitably chosen algebraic number field. A consequence of our arguments is, that α is inner in KG for every field K.

1 Introduction

It follows from from a result of Coleman [1] that the natural map

Φ : Out(G) ⟶ Out(SG)

is injective for p–groups. Here S is the ring of integers in a global or a local number field K, in which p is not invertible. It is a question of Jackowski and Marciniak ([4, 3.7]) whether this happens for all finite groups G and coefficient ring ZZ = S.

However, there is not yet known an example of a finite group G, where this map is not injective for the algebraic integers of a global field K of characteristic 0. If one wants to construct α ∈ Aut(G) with Φ(α) inner on SG, then

1. α must be the identity on the conjugacy classes of G,

2. α must be inner in G on the Sylow p–subgroups, combining Sylow theorems and the result of Coleman (cf. [1] and [7, I.II § 2]).

To construct such an α is is a purely group theoretical problem. However, in order to show that α becomes inner on SG we have to use the following ingredients from integral representation theory:

1. We show that α is inner on ZZG semilocally. From this one cannot automatically conclude that α is inner on ZZG. The obstruction is an element in the class group of ZZG. We use class field theory – this is where K enters – to kill this obstruction.

2. The passage from the local to the semilocal situation becomes possible by interpreting automorphisms as invertible bimodules and using Fröhlich’s exact sequence of Picard groups [3].

3. Finally we are in the local respectively complete situation. Here we use Clifford theory ^{1} to show, that α acts as inner automorphism on the inertia groups, after having applied the theorem of Noether Deuring, to pass to a splitting field. The key point in our construction is to involve quaternion groups in order to keep the inertia groups small.

4. Let now K be a field, then there exists a rational prime p, such that KG ≃ K ⊗_{ZZ_{p}} ZZ_{p}G. Thus α induces an inner automorphism on KG.

2 A short review of Clifford theory

Since Clifford theory is essential to our arguments we give a brief summary of it as developed in [7] as far as it is needed here.

1. Let N be a normal complemented subgroup of a finite group G with (|G/N|, |N|) = 1.

2. Let, furthermore, R be a complete local Dedekind domain of characteristic 0 with residue field k of characteristic p > 0 such that the quotient field K of R is a splitting field for G and all of its subgroups.

3. We assume that p does not divide the order of N.

1The first author has learnt the way of using Clifford theory integrally from Leonard Scott.

4. We denote by I(M) the inertia group of an irreducible RN-lattice M in G.

5. We assume that I(M) is normal in G.

6. Let α be a group automorphism of RG that becomes inner in KG.

7. Let e_{M} be the central primitive idempotent of RN acting as the identity on M. For each t in I(M), the conjugate t · e_{M} · t^{−1} is again a central primitive idempotent of RN which acts as 1 on M, hence, e_{M} is centralized by I(M).

8. If α acts as the identity on N, we have α(e_{M}) = e_{M}.

9. If now I(M) < G, then let g_{1}, ..., g_{s} be a left transversal of I(M) in G and let e_{i} := g_{i} · e_{M} · g_{i}^{−1}; i = 1, ..., s be the conjugates of e_{M} in RG. Since the {g_{i}} form a transversal with respect to the inertia group of M, the conjugates e_{i} are pairwise orthogonal idempotents in RG. However, ∑_{i=1}^{s} e_{i} =: e is a central idempotent in RG. Now RG · e is isomorphic to an (s × s)-matrix ring, whose (i, j)-entry equals B_{i,j} := e_{i} · RG · e_{j} for 1 ≤ i, j ≤ s, the usual Pierce decomposition.

10. If, furthermore, x^{−1} · α(x) ∈ I(M) for all x ∈ G, then

α(e_{i}) = α(g_{i} · e_{M} · g_{i}^{−1}) = α(g_{i}) · α(e_{M}) · α(g_{i})^{−1} = e_{i}

For the rest of the section we assume that 1.– 10. hold.

We look at B_{i,j} more closely: An element g ∈ G can be written as n·h with n ∈ N and h ∈ G/N. Then,

h = g_{k} · t with t ∈ I(M) and a certain k. We denote by χ_{i} the character of N afforded by e_{i}. Using the normality of I(M) we calculate

e_{i} · n · h · e_{j} = χ_{i}(n) · e_{i} · g_{k} · t · e_{j}
= χ_{i}(n) · g_{i} · e_{M} · g_{i}^{−1} · g_{k} · g_{j} · e_{M} · g_{j}^{−1} · t.

If g_{k} · g_{j} ∉ g_{i} · I(M), which is equivalent to g_{k} ∉ g_{i} · g_{j}^{−1} · I(M), again by the normality of I(M), then the above equals 0. Hence,

B_{i,i} = g_{i} · e_{M}RI(M) · e_{M} · g_{i}^{−1}.

More precisely, the above calculations show

e_{i} · RG · e_{j} = g_{i} · e_{M} · RI(M) · e_{M} · g_{j}^{−1} .

This allows us to describe explicitely the action of α on B_{i,j}:

α : g_{i} · e_{M} · x · e_{M} · g_{j}^{−1} ⟶ α(g_{i}) · e_{M} · α(x) · e_{M} · α(g_{j}^{−1}) for x ∈ RI(M). (1)

A central^{2} automorphism α of RG fixes the B_{i,j} as set for all i, j if and only if it is conjugation by

( \begin{matrix}u_{1} & 0 & & & 0 \\ 0 & ... & ... \\ & ... & ... & ... \\ & & ... & ... & 0 \\ 0 & & & 0 & u_{s} \end{matrix} )

with elements {u_{i} ∈ (RI(M) · e_{M}) ∩ U(KI(M) · e_{M})|i = 1, .., s}.^{3}

If α acts as the identity on I(M) and if moreover α(g_{i}) ∈ g_{i} · Z(I(M)), then it acts as identity on g_{i} · e_{M} · RI(M) · e_{M} · g_{i}^{−1}, and furthermore u_{i} ∈ Z(RI(M) · e_{M}) for all i = 1, ..., s. Since R is local,

Outcent(RI(M) · e_{M}) ≃ Outcent(Mat_{n}(RI(M) · e_{M})).

Consequently, there is an inner automorphism γ of Mat_{n}(RI(M) · e_{M}) = RG · e such that

α = γ ∘ δ

with a central automorphism δ acting via the identification

B_{i,j} = e_{i} · RG · e_{j} = g_{i} · e_{M} · RI(M) · e_{M} · g_{j}^{−1}

on each of the entries B_{i,j}. Since δ is then just conjugation by

v ∈ (RI(M) · e_{M}) ∩ U(KI(M) · e_{M}),

2An automorphism of a ring acting as the identity on the centre of the ring is called central.

3Note that this can always be achieved by multiplication with a central element.

the unit γ has to have the diagonal form analogous to that of α. Hence, there is an element v ∈ (RI(M) · e_{M}) ∩ U(KI(M) · e_{M}) and elements γ_{i} ∈ U(RI(M) · e_{M}) such that

u_{i} = γ_{i} · v ∈ Z(RI(M) · e_{M}).

Therefore,

u_{i} · u_{j}^{−1} ∈ U(RI(M) · e_{M}) for all 1 ≤ i, j ≤ s.

We summarize these observations, which we shall apply in the next section, as

Proposition 1 With the above notation, assume that one of the elements u_{i} can be chosen to be a unit in RI(M) · e_{M}, then the automorphism α is inner.

3 The construction of the group and the automorphism

Let \overline{H} be the semidirect product of the cyclic group of order 8, generated by a, with its automorphism group C_{2} × C_{2}, generated by \overline{b} and \overline{c} such that \overline{b} inverts a and \overline{c} raises a to the third power. \overline{H} has a non inner central automorphism σ sending \overline{c} to a^{4}\overline{c} and fixing \overline{b} and a (cf. [4]). Let

H_{0} := (C_{8} ×(C_{2} × C_{2}))×< σ > .

Factoring the centre of a quaternion group Q_{8} of order 8, the group Q_{8} generated by b and c maps onto C_{2} × C_{2}, the image of b being \overline{b} and that of c being \overline{c}, the natural projection is called π_{Q}. We form the pullback

1 ⟶ C_{8} ⟶ H_{0} ⟶ C_{2} × C_{2}× < σ > ⟶ 1
∥  ↑  ↑ π_{Q} × id_{<σ>}
1 ⟶ C_{8} ⟶ H ⟶  Q_{8} × C_{2}  ⟶ 1.

For H we form two irreducible modules M_{1} and M_{2} such that for each i ∈ {1, 2} the element a acts as 1 on M_{i} and restricted to the quaternion group Q_{8} the modules M_{i} are the two dimensional irreducible representation of Q_{8} over a finite prime field IF_{i} of odd characteristic p_{i}. It is well known that there are irreducible two dimensional representations of the quaternion group of order 8 over every field of odd characteristic (see e.g. [2, (27.1) and (74.11)]). We choose p_{1} ≠ p_{2}. On M_{1} the element σ operates as 1 and on M_{2} it acts as −1. Let G be the semidirect product of M with H:

G := (M_{1} × M_{2}) ×H

Since a acts trivially on M := M_{1} ×M_{2} the automorphism α, conjugation by σ on H, can be extended to G as (id_{M}, α), also denoted by α. The extension is central since on H the elements h and α(h) are conjugate by either a or a^{2} or 1. All of these elements centralize M, and therefore, α as automorphism of G is central. In fact, for every integers i and j the elements a^{i}bcσ^{j} and a^{i}b^{3}cσ^{j} are centralized by aσ, the elements a^{i}cσ^{j} and a^{i}b^{2}cσ^{j} are centralized by a^{2}σ and the rest of H is centralized by σ itself.

4 The automorphism is semilocally inner

We claim that α is inner in RG with a suitable semilocal ring of integers R, containing a p_{1}·p_{2}·8-th root of unity, and neither p_{1} nor p_{2} nor 2 are invertible in R. Note that the Noether Deuring theorem ([6]) implies that there is no loss of generality if we assume that the quotient field of R is a splitting field for G. This can be applied since the outer central automorphism group of RG can be mapped into P icent(RG) (cf. [3]) by means of mapping the automorphism ϕ to the invertible bimodule _{ϕ}RG_{1} where the left module action is twisted by ϕ. For details see [3, Theorem 1].

Using Fröhlich’s result ([3, Theorem 6]) for semilocal domains R

Outcent(RG) ≃ ∏_{℘∈Spec(R)} Outcent(R_{℘}G)

and interpreting α as an invertible bimodule as above it is enough to show that α is inner for all completions of R at finite primes ℘: We extend α to a central automorphism α_{℘} of R_{℘}G and map the bimodule corresponding to α to the finite direct product of bimodules corresponding to α_{℘}.

For non semilocal Dedekind domains R the obstruction to globalize local automorphisms is a certain subgroup of the class group of the centre of the group ring. In fact, denoting by Cl(centre(Λ)) the locally free class group of the order Λ, the sequence

1 ⟶ Cl(centre(RG)) ⟶ P icent(RG) ⟶ ∏_{℘∈Spec(R)} P icent(R_{℘}G) ⟶ 1

Using class field theory (cf. [5, Satz 7]) this implies that there exists a ring of algebraic integers S in an algebraic number field L, L being finite over the quotient field of R, such that α becomes inner as an automorphism of SG.

4.1 The crucial calculation

We now come to the central point in the proof:

We shall show in the next subsection that the following three groups will occur as inertia groups at various primes ℘:

1. I_{1} :=< M, a, σ > is normal in G.

2. I_{2} :=< M, a, σ · b^{2} > is normal in G.

3. I_{3} :=< M, a >= I_{1} ∩ I_{2}.

We take this as guaranteed for the moment. At the prime ℘ the group ring has a ring direct factor of the form (g_{i} · e_{M} · R_{℘}I · e_{M} · g_{j}^{−1})_{1≤i,j≤s} with I ∈ {I_{1}, I_{2}, I_{3}}. Recall that e_{i} = ^{g_{i}}e_{M} for g_{i} ∈ H_{k} and H_{k} suitably chosen according to the inertia groups I_{k}.

We now consider the three cases I_{1} , I_{2} and I_{3}. In case I = I_{1} we can choose the transversal

H_{1} := {1, b, b^{2}, b^{3}, c, cb, cb^{2}, cb^{3}} = H_{1}^{0} ∪^{·} c · H_{1}^{0}.

In case I = I_{2} we can choose the transversal

H_{2} := {1, b, b^{2}, b^{3}, c, cb, cb^{2}, cb^{3}} = H_{2}^{0} ∪^{·} c · H_{2}^{0}.

In case I = I_{3} we can choose the transversal

H_{3} := {1, b, b^{2}, b^{3}, σ, bσ, b^{2}σ, b^{3}σ, c, cb, cb^{2}, cb^{3}, cσ, cbσ, cb^{2}σ, cb^{3}σ} = H_{3}^{0} ∪^{·} c·H_{3}^{0}.

We have seen that

e_{i} · R_{℘}G · e_{j} = g_{i} · e_{M} · R_{℘}I · e_{M} · g_{j}^{−1}

Thus, by Equation (1), in case I = I_{1} and in case I = I_{2} the action of α on (e_{i} · R_{℘}G · e_{j})_{1≤i,j≤s} is multiplication by a^{4} if 4 ≤ |i − j| and the identity otherwise, hence, conjugation by the matrix

( \begin{matrix}1_{4×4} & 0 \\ 0 & a^{4} · 1_{4×4} \end{matrix} ) .

Here we denote by 1_{m×m} the m by m unit matrix. Therefore, α acts as inner automorphism on the ring direct factor corresponding to e_{M}.

In case I = I_{3} a similar observation yields that α acts as conjugation by

( \begin{matrix}1_{8×8} & 0 \\ 0 & a^{4} · 1_{8×8} \end{matrix} )

Hence, by Proposition 1, α acts as inner automorphism on the ring direct factor corresponding to e_{M} as well.^{4}

4.2 Determination of the inertia groups

We choose a prime ℘ ∈ Spec(R) and discuss various cases for ℘.

\underline{Case 1:} 2 ∈ ℘. Then the group ring decomposes as

R_{℘}G = I_{R_{℘}}(M)G ∏ R_{℘}H.

On R_{℘}H the automorphism α is conjugation by σ . For I_{R_{℘}}(M)G we apply Clifford theory. Let χ be an irreducible nontrivial character of M as abelian groups (the quotient field of R is a splitting field for M) with kernel K ⊆ M_{1} × M_{2}.

For χ there are three cases which have to be considered separately:

1. If M_{2} ⊂ K then the inertia group I_{χ} contains M, σ and a. If h ∈ H lies in I_{χ} we observe that the sequence

1 ⟶< a, σ >⟶ H ⟶ Q_{8} ⟶ 1

4Note how sensitive this construction is with respect to the group G.

splits. Hence, we may assume that h ∈ Q_{8}, but there are fixed points on neither M_{1} nor M_{2}, a contradiction. Thus

I_{χ} =< M, a, σ >= I_{1},

and α is seen to be inner.

2. M_{1} ⊂ K then obviously I_{χ} ⊇< M, a, b^{2} · σ >. Similar arguments show

that

I_{χ} =< M, a, b^{2} · σ >= I_{2}.

Now, here α is also inner.

3. If neither M_{1} nor M_{2} belongs completely to K the inertia group

I_{χ} =< M, a >= I_{3}

is the intersection of the two inertia groups of the previous cases since M = M_{1} × M_{2} and (℘, |M|) = 1. Again we see that α acts as inner automorphism on the factor corresponding to that character of N.

Our automorphism now acts as inner automorphism on each of the factors and α_{℘} is inner in R_{℘}G provided 2 ∈ ℘.

\underline{Case 2:} p_{1} ∈ ℘. Then the group ring decomposes to

R_{℘}G = I_{R_{℘}}(M_{2})G ∏ R_{℘}G/M_{2}.

On the second factor the automorphism α is conjugation by σ and for the first factor we take an irreducible non trivial character χ of M_{2}. Then the inertia group is

I_{χ} =< M, a, b^{2} · σ >= I_{2}

as before. As above, α_{℘} is inner.

\underline{Case 3:} p_{2} ∈ ℘. Then the group ring decomposes to

R_{℘}G = I_{R_{℘}}(M_{1})G ∏ R_{℘}G/M_{1}.

On the second factor the automorphism α is conjugation by b^{2} · σ and for the first factor we take an irreducible non trivial character χ of M_{1}. Then the inertia group is

I_{χ} =< M, a, σ >= I_{1}

as before. Again, as above, Clifford theory tells us that α_{℘} is inner.

\underline{Case 4:} |G|R_{℘} = R_{℘} Then the group ring R_{℘}G is separable and every central automorphism is inner.

We have shown that α is an inner automorphism of RG.

Now we show that α is not inner as group automorphism. On G/M the automorphism is conjugation by σ. The conjugating elements on G/M are, therefore, contained in σ· < a^{4}, b^{2} >. Since a^{4} is central in G, we may assume that the conjugating element maps onto an element in σ· < b^{2} >. Therefore, the conjugating element in G is m · b^{2j} · σ for some m ∈ M and j ∈ ZZ. But on M_{1} this operates trivial if and only if j is even and on M_{2} it operates trivial if and only if j is odd. Thus it cannot exist and α is not inner in G.^{5}

Now, the invertible bimodule ^{6} _{α}(RG)_{1} is isomorphic to _{1}(RG)_{1}, and therefore, by [6], we have _{α}(ZZ_{π}G)_{1} ≃ _{1}(ZZ_{π}G)_{1} with π being any finite set of primes.

Proposition 2 For the group G of order 2^{7} · p_{1}^{2} · p_{2}^{2} as above the natural map

Outcent(G) ⟶ Outcent(ZZ_{π}G)

is not injective. For p_{1} = 5, p_{2} = 3 the group G has generators and relations^{7} G =< a, b, c, σ, x_{1}, x_{2}, y_{1}, y_{2} > subject to the relations {a^{8} = 1, b^{4} = 1, c^{4} = 1, (bc)^{2} = b^{2} = c^{2}, (σ, a) = 1, (σ, b) = 1, ^{σ}c = a^{4}c, ^{b}a = a^{7}, ^{c}a = a^{3}, x_{1}^{5} = 1, x_{2}^{5} = 1, y_{1}^{3} = 1, y_{2}^{3} = 1, (a, x_{1}) = 1, (a, x_{2}) = 1, (a, y_{1}) = 1, (a, y_{2}) = 1, (σ, x_{1}) = 1, (σ, x_{2}) = 1, ^{σ}y_{1} = y_{1}^{−1}, ^{σ}y_{2} = y_{2}^{−1}, ^{b}x_{1} = x_{2}, ^{b}x_{2} = x_{1}^{−1}, ^{c}x_{1} = x_{1}^{2}, ^{c}x_{2} = x_{2}^{3}, ^{b}y_{1} = y_{2},^{b} y_{2} = y_{1}^{2}, ^{c}y_{1} = y_{1}^{2} · y_{2}, ^{c}y_{2} = y_{1} · y_{2}, (x_{1}, x_{2}), (y_{1}, y_{2}), (x_{1}, y_{1}), (x_{2}, y_{2})} Remark 1 J. Krempa proved that there can at most be an elementary abelian two group in the kernel of the group homomorphism from Outcent(G) to Outcent(ZZG) (cf. [4, 3.2 Theorem]).

5These arguments are also checked by GAP.

6We adopt the notation of [3]

7The commutator of g and h is denoted by (g, h)

Claim 1 Outcent(G) is of order 2.

Proof. Let β be a central automorphism of G. The Schur-Zassenhaus theorem assures that we may vary β by conjugation with an element in M such that H :=< a, b, c, σ > is fixed under β. Therefore, we assume that β itself fixes H and we may calculate in H modulo M.

Let β be a central automorphism of H. Since the conjugacy class of a consists of the elements {a, a^{3}, a^{5}, a^{7}} the normal subgroup A :=< a > of G is fixed under β. Since β is central and elements are conjugate to their image we may assume that β(b) = b (if not we modify β by an inner automorphism). The conjugacy class of c consists of the elements {c, a^{2}c, a^{4}c, a^{6}c, c^{3}, a^{2}c^{3}, a^{4}c^{3}, a^{6}c^{3}}. If we vary by conjugation by σ and (or) b, centralizing b, we may assume that β(c) ∈ {c, a^{2}c}. The conjugacy class of bc is equal to {bc, a^{4}bc}, and therefore, c cannot be mapped to a^{2}c under β if b is fixed. We may assume that b and c are fixed. The conjugacy class of σ is equal to {σ, a^{4}σ}. The conjugacy class of aσ is equal to {aσ, a^{7}σ}. Therefore, from β(a) ∈ {a^{3}, a^{5}} it follows β(σ) = a^{4}σ and from β(a) ∈ {a, a^{7}} we conclude β(σ) = σ.

We, have only 4 candidates left: These are generated by the two automorphisms of order 2: λ_{1} and λ_{2} with λ_{1}(a) = a^{7} and the rest of the generators stay fixed and λ_{2}(a) = a^{3}, λ_{2}(σ) = a^{4}σ and the rest of the generators stay fixed. The conjugacy class of abc is equal to {abc, a^{5}bc, a^{7}b^{3}c, a^{3}b^{3}c}. Hence, neither λ_{1} nor λ_{2} are central automorphisms.

For the automorphism λ_{1} · λ_{2} we look at the conjugacy class of ay_{2} which has length 8. An automorphism has to map y_{2} to either y_{2} or to y_{2}^{−1} since the H–modules ^{λ_{1}λ_{2}}M and M are equal and an automorphism extending λ_{1}λ_{2} differs on M from the identity by an automorphism of M which preserves the orbits of H on M. But all of the elements in bc · C_{H}(a) moves y_{2} to neither y_{2} nor to y_{2}^{−1}. We see that also λ_{1}λ_{2} cannot be central ^{8}.

Assuming β to be the identity on G/M the automorphism is a module endomorphism of M as G/M-module. The endomorphism ring, however, is the prime field on each of M_{1} and M_{2} and only the element −1 is realizable by the operation of a group element. It may hence be realized by the operation of elements in the subgroup < b^{2}, σ > which causes to be α modulo conjugation by b^{2}.

8These statements are also checked by GAP.

We conclude: Up to inner automorphisms of G there is only one non trivial central automorphism of G inducing the identity on H.

Acknowledgement: We would like to thank the referee for pointing out a gap in the proof in an earlier version of the paper.

References

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[3] A. Fröhlich, The Picard groups of noncommutative rings, in particular of orders, Trans. Amer. Math. Soc. 180 (1973), 1–46.

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[5] H. Jackobinski, Über die Geschlechter von Gittern über Ordnungen, J. Reine Angew. Math. 230 (1968), 29–39.

[6] K. W. Roggenkamp, An extension of the Noether Deuring theorem, Proc. Amer. Math. Soc. 31 (1972), 423–426.

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Address of the authors: Mathematisches Institut B Universität Stuttgart Pfaffenwaldring 57 D–70550 Stuttgart Germany