On the holomorph of finite semisimple groups

Given a finite nonabelian semisimple group $G$, we describe those groups that have the same holomorph as $G$, that is, those regular subgroups $N\simeq G$ of $S(G)$, the group of permutations on the set $G$, such that $N_{S(G)}(N)=N_{S(G)}(\rho(G))$, where $\rho$ is the right regular representation of $G$.


Introduction
Let G be any finite group and let S(G) be the symmetric group on the set of elements of G.We denote by ρ : G → S(G) and λ : G → S(G) respectively the right and the left regular representations of G.The normalizer is the holomorph of G, and it is isomorphic to the natural extension of G by its automorphism group Aut (G).It is well-known that Hol (G) = N S(G) (λ(G)).
In [10] the multiple holomorph of G has been defined as NHol(G) = N S(G) (Hol (G)) and it is proved that the quotient group acts regularly by conjugation on the set of the regular subgroups N of S(G) that are isomorphic to G and have the same holomorph as G, that is, T (G) acts regularly on the set There has been some attention both in the distant past (see [11]) and quite recently (see [9], [5] [4] and [3]) on the problem of determining, for G in a given class of groups, those groups that have the same holomorph as G and, in particular, the set H(G) and the structure of the group T (G).Recently, in [4] the authors attack this problem when G is a perfect group, obtaining complete results for centerless groups (see Theorem 7.7 in [4]).However they leave open some interesting questions when the center is non-trivial.
The aim of this paper is to completely resolve the case when G is a finite semisimple group.
One of the main obstacles for describing the holomorph of a finite semisimple group (see [5,Remark 7.12] and also [2, ADV -4B]) was to completely classify those finite nonabelian simple groups that admit automorphisms acting like inversion on their Schur multiplier.In Proposition 2 we produce a complete analysis, whose proof depends on the Classification of Finite Simple Groups.It turns out that there is a small list L (see after Proposition 2 for its definition) of related quasisimple groups having automorphisms inverting their center.Our main result can be therefore stated as follows: Theorem.Let G be a finite nonabelian semisimple group and let be its unique central decomposition as a product of Aut (G)-indecomposable factors.Assume that the number of factors A i of G having components in L is exactly l, for some 0 ≤ l ≤ n.Then the set H(G) is completely known and it has cardinality Also, the group T (G) is an elementary abelian group of order 2 h .Moreover, if the centers of the factors A i are all amalgamated, then |H(G)| = 2 m and T (G) is elementary abelian of order 2 m , where m = min {n, n − l + 1}.

Semisimple groups
To establish the notation, note that we write permutations as exponents, and denote compositions of maps by juxtaposition.We compose maps left-to-right.We consider the right and the left regular "representations" of G, defined by Remark 1.Since composition of maps is left-to-right, with our definition the map λ is an antihomomorphism, not a homomorphism, from G to S(G).We have chosen this definition, over the standard one (where λ(g) maps x ∈ G to g −1 x) for the same reasons as in [5].
The first proposition recalls some basic facts (see [4,Proposition 2.4]).The proof is left to the reader.Proposition 1.Let G be any group and let inv be the inversion map on G defined by inv(g) = g −1 for every g ∈ G.The following hold: (3) ρ(g) inv = λ(g −1 ) and λ(g) inv = ρ(g −1 ) for every g ∈ G.In particular, inv conjugates ρ(G) to λ(G) (and vice versa), and it centralizes Aut (G); Recall that a quasisimple group is a perfect group X such that X/Z(X) is simple, and that a semisimple group is a central product of quasisimple groups, that is, a group X = X 1 X 2 . . .X t with each X i quasisimple and such that [X i , X j ] = 1 for every i = j.The quasisimple normal subgroups X i of X are called the components of X.Note in particular that semisimple groups are perfect.
Every finite semisimple group admits a unique decomposition as a central product of characteristic subgroups.
Lemma 1.Let G be a finite semisimple group.Then G is a central product of perfect centrally indecomposable Aut (G)-subgroups: Moreover, the integer n and the subgroups A i (for i = 1, 2, . . ., n) are uniquely determined (up to permutation).
Proof.Consider the Remak-Krull-Schmidt decomposition of Inn (G) ≃ G/Z(G) as an Aut (G)-group, and let this be , and thus M j induces by conjugation a central automorphism on M ′ i .Since perfect groups have no nontrivial central automorphisms (see [8], or [5, Lemma 7.1]), we have [M i , M j ] = 1 for every j = i.In particular, equation ( 1) and the fact that G is perfect imply the following central factorization of G: where A i = M ′ i , for each i = 1, 2, . . ., n.Note that the A i are perfect Aut (G)subgroups, which are indecomposable as Aut (G)-subgroups.Finally the uniquiness of the factorization (1) (see [12, 3.3.8])and, again, the fact that perfect groups have no nontrivial central automorphisms, imply that the central product decomposition (2) is also unique.
We make use of the same notation as [5].In particular, we define the following subsets of subgroups of S(G):

Note that H(G) ⊆ I(G) ⊆ J (G).
From now on we assume that G is a finite semisimple group and we write G as a central product of indecomposable Aut (G)-subgroups, in a unique way (by Lemma 1): Note that by Proposition 1 (1), we have that [ρ(A i ), λ(A j )] = 1 for every i = j.Fix I to be the set {1, 2, . . ., n}.For each subset J of I we denote the central product j∈J A j by A J .Then for each subset J of I we may define the subgroup where Moreover, since ρ(A J ) ≤ ρ(G) and λ(A J c ) ≤ λ(G), we have completing the proof of the Lemma.
Lemma 3. The inversion map inv conjugates G J to G J c , for every J ⊆ I.
Proof.The lemma is an immediate consequence of Proposition 1 (3).
Lemma 4. Assume that G is a finite semisimple group.Then, with the above notation, Proof.We first show that G J acts regularly on the set G. For an arbitrary g ∈ G , and therefore σ is the identity.
To show that each , it is enough to show that Aut (G) normalizes every G J .Fix J ⊆ I, let α ∈ Aut (G), and let ρ(x J )λ(x J c ) be an arbitrary element of G J , where x J ∈ A J and x J c ∈ A J c .Then we have that, for every g ∈ G, ), which lies in G J , since A J and A J c are characteristic subgroups of G.
Finally by [5,Theorem 7.8], |J (G)| = 2 n , and therefore to complete the proof it remains to show that G J = G K whenever J = K.If G J = G K then there exists an i ∈ I for which G J contains both ρ(A i ) and λ(A i ).But then the stabilizer of 1 in G J would contain ρ(x)λ(x −1 )|x ∈ A i , that is, all the conjugates of elements of A i .Since A i is not central in G, this contradicts the fact that G J is regular.

Automorphisms of finite quasisimple groups
In this section we classify all finite quasisimple groups that admit an automorphism acting like inversion on the center.This classification, which is used in the proof of our main result (Theorem A), is proved in Proposition 2 using the Classification of Finite Simple Groups.This result completely answers a question posed in [5,Remark 7.12] (see also [2, ADV -4B]), namely whether there are finite quasisimple groups L such that Z(L) is not elementary abelian, and such that Aut (L) does not induce inversion on Z(L), or acts trivially on it.
Before stating and proving Proposition 2, we introduce some notation and terminology related to automorphisms of finite nonabelian simple groups of Lie type.We refer the interested reader to the first two chapters of [7].
Let S be any finite group of Lie type.Then, by [13,Theorem 30], any automorphism of S is a product idfg, where i is an inner automorphism of S (an element of Inn (S)), d is a diagonal automorphism of S, f is a field automorphism of S and g is a graph automorphism of S. The following notation is quite standard.Inndiag(S), Φ S and Γ S denote, respectively, the group of inner-diagonal automorphisms of S, of field automorphisms of S, and of graph automorphisms of S. Further, Outdiag (S) is defined to be Inndiag(S)/Inn (S) (see [7,Definition 2.5.10]).Proposition 2. Let K be a finite quasisimple group.Then there exists an automorphism of K that inverts Z(K) if and only if K is not isomorphic to one of the following groups: (1) a covering of L 3 (4), with center containing (2) a covering of U 4 (3), with center containing Z 3 × Z 4 ; (3) U 6 (2), the universal covering of U 6 (2); Proof.As is well known (see, for example, [1]), any finite quasisimple group K is isomorphic to a quotient of the universal covering group of its simple quotient K/Z(K).Also, Aut (K) ≃ Aut (K/Z(K)) (see for instance [1, Section 33]).Therefore, for our purposes it is enough to consider the action of Aut (S) on the Schur multiplier M (S) when S varies among all finite nonabelian simple groups.In this situation, the outer automorphism group Out (S) acts on M (S), which, by [6, Section 5, 6-1], is isomorphic to the direct product of two factors of relatively prime orders: M c (S) and M e (S).The actions of Out (S) on both factors are completely described in [7, Theorem 6.3.1 and Theorem 2.5.12] for every finite nonabelian simple group S. In particular, Outdiag (S) centralises M c (S), and there is an isomorphism of Outdiag (S) on M c (S) preserving the action of Out (S).Note also that if one of the factors M c (S) or M e (S) has order at most 2, since they have coprime orders, to prove our statement it is enough to see if inversion is induced by Out (S) on the other factor.We may therefore consider the two cases: We prove that in this case there is always an automorphism of S inverting M (S).As noted above, it is enough to consider the action of Out (S) on Outdiag (S), which is Aut (S)-isomorphic to M c (S).But Outdiag (S) is always inverted by Out (S), since, by [7, Theorem 2.5.12],we have that: i) if S ∈ {A m (q), D 2m+1 (q), E 6 (q)}, then Outdiag (S) is inverted by a graph automorphism (by 2.5.12 (i)), ii) if S ∈ 2 A m (q), 2 D 2m+1 (q), 2 E 6 (q) , then Outdiag (S) is inverted by a field automorphism (by 2.5.12 (g)), iii) in all other cases, Outdiag (S) is either trivial or an elementary abelian 2-group and therefore it is inverted by the trivial automorphism.Case (2) |M e (S)| > 2. From [7, Table 6.3.1] and the knowledge of the corresponding factor M c (S) ([7, Theorem 2.5.12]),we can see that if S is not isomorphic to one of the following simple groups L 3 (4), U 6 (2), 2 E 6 (2), U 4 (3), then there exists an automorphism of S that inverts M (S) and, therefore, any quasisimple group K such that K/Z(K) ≃ S admits an automorphism inverting its center.We now consider separately the four special cases listed above.Let S = L 3 (4).Then M e (S) ≃ Z 4 × Z 4 and M c (S) ≃ Z 3 .Here Out (S) = Σ × u , with Σ = Outdiag(S)Γ S ≃ S 3 and u the image in Out (S) of a graph-field automorphism of order 2. By [7, Theorem 6.3.1]u is the only element of Out (S) that induces inversion of M e (S).Now, u is Aut (S)-conjugate to an element of the form φi, with φ a field automorphism and i a graph automorphism, where φ and i are commuting involutions (note that Φ S Γ S ≃ Z 2 × Z 2 ).The action of Φ S Γ S on M c (S) is the same as on Outdiag (S).Thus, by [7, Theorem 2.5.12(g) and (i)], both φ and i invert M c (S), and therefore u acts trivially on it.This argument shows that when K is the universal covering group of S, no inversion on Z(K) is induced by an automorphism of K. Assume now that K is a covering of S different from the universal one.If 3 ∤ |Z(K)| then u inverts Z(K).Assume therefore that 3| |Z(K)|.If Z(K) is cyclic of order 3 or 6, then φ inverts Z(K).Otherwise we may argue as follows.Since M c (S) and Outdiag (S) are Aut (S)-isomorphic, the elements of Out (S) that induce inversion on M c (S) are the six non central involutions, that is, the elements of the set Note that from Proposition 6.2.2 and the proof of Theorem 6.3.1 in [7], Outdiag (S) acts faithfully on the quotient group M e (S)/Φ(M e (S)), and hence on M e (S).Next, let t be an element of T .Since t inverts Outdiag (S), C Me(S)/Φ(Me(S)) (t) ≃ C Φ(Me(S)) (t) = Φ(M e (S)), so C Me(S) (t) = b ≃ Z 4 , and t inverts a unique cyclic subgroup of order 4.This in particular shows that when K is a covering extension with Here Out (S) = Outdiag (S) Φ S is isomorphic to a dihedral group of order 8 acting faithfully on M e (S).In particular, the nontrivial central element of Out (S) is the unique element inducing inversion on M e (S).Note that this element belongs to Outdiag (S) and therefore it centralizes M c (S).This implies that no inversion can be induced on M (S) by automorphisms of S. Therefore the universal covering group S has no automorphisms inverting its center.This argument can be extended to show that the same situation occours in any covering having center of order 12.However, for all other coverings S of S it can be easily checked that inversion on the center is induced either by the nontrivial central element of Outdiag (S) Φ S , when 3 divides Z( S) , or by a field automorphism of order two, when 3 ∤ Z( S) .
Let S = U 6 (2), or S = 2 E 6 (2).In both cases we have that M e (S) ≃ Z 2 × Z 2 , M c (S) ≃ Z 3 , and Out (S) ≃ S 3 acts faithfully on M e (S) (see [7,Proposition 6.2.2]).In particular, the trivial outer automorphism is the only one that inverts M e (S).Since it does not invert M c (S), the universal covering groups S have no automorphisms inverting their centers.On the contrary, every covering S different from the universal one possesses such automorphisms, which are either trivial if 3 ∤ Z( S) , or are field automorphisms.
For convenience, we write L for the set of quasisimple groups that appear as exceptions in Proposition 2; thus Remark 2. According to [7,Theorem 6.3.2],when S ≃ L 3 (4) (or when S ≃ U 4 (3)) there are precisely two non-isomorphic covering groups S such that S/Z( S) ≃ S and Z( S) ≃ Z 2 × Z 4 × Z 3 (respectively, Z( S) ≃ Z 3 × Z 4 ).In all other cases in the list L the covering group of the associated finite simple group is unique.Therefore, up to isomorphism, |L| = 9.
As an application of Proposition 2 we prove the following: Corollary 1. Assume that X is a finite semisimple group with all components not in L. Then there exists an automorphism of X that inverts the center Z(X).
Proof.Let X be a finite semisimple group.Then X is isomorphic to D/N , with . By Proposition 2, for each i ∈ {1, . . ., t} there exists an automorphism α i of K i that acts like the inversion on Z(K i ).We may therefore define α ∈ Aut (D) by x α = (x 1 , x 2 , . . ., x t ) α = (x α1 1 , x α2 2 , . . ., x αt t ) for x ∈ D (that is, where each x i ∈ K i ).Note that every element of Z(D) is inverted by α.In particular N α = N , and α induces an automorphism on D/N that inverts its center.

The holomorph of a semisimple group
We first consider the case in which G has no components belonging to L. Proposition 3. Assume that G is a finite nonabelian semisimple group and let G = A 1 A 2 . . .A n be its unique central decomposition as a product of Aut (G)indecomposable factors.Suppose that the components of G do not belong to L. Then H(G) = {G J |J ⊆ I} .Moreover, the group T (G) = NHol (G) /Hol (G) is elementary abelian of order 2 n .Proof.By Lemma 4, we have that J (G) = {G J |J ⊆ I}, and therefore H(G) ⊆ {G J |J ⊆ I}.We fix a subset J of I and, using Corollary 1, we choose an automorphism α J c of A J c that inverts Z(A J c ). Define the map ϕ , the map ϕ J is a well-defined bijection of G, that is, an element of S(G).We claim that the following hold: (1) ϕ J conjugates G I to G J , (2) ϕ J ∈ NHol (G) and, if J = I, then ϕ J ∈ Hol (G), (3) (ϕ J ) 2 ∈ Hol (G).Note that, by the arbitrary choice of J ⊆ I, once proved, (1) will imply that H(G) = J (G) = {G J |J ⊆ I}, while (2), (3) and the fact that T (G) acts regularly on H(G) will imply that T (G) is an elementary abelian 2-group of rank n.
(1) (G I ) ϕJ = G J .We claim that for x J ∈ A J and x ), or, equivalently, that: ).
Let g ∈ G and write g as g = y J y J c (with y J ∈ A J and y J c ∈ A J c ).Then , and Therefore (1) is proved.Note that, together with Lemma 4, we have proved that J (G) ⊆ H(G) and therefore H(G) = I(G) = J (G).
(2) ϕ J ∈ NHol (G).By (1) we have that Furthermore, G J = G I for J = I, thus we trivially have that ϕ J ∈ Hol (G) for J = I.
(3) (ϕ J ) 2 ∈ Hol (G).We claim that for every x J x J c ∈ G: J c ), or, equivalently, that: J c .This completes the proof of Proposition 3. Remark 3.For each fixed subset J of I we may define an operation • J on the set of elements of G, as follows: With this notation, it is straightforward to prove that (1) (G, • J ) is a group isomorphic to G J , for each J ⊆ I; (2) Aut (G) = Aut (G, • J ) for each J ⊆ I (see [5,Theorem 5.2 (d)]; (3) if G satisfies the assumptions of Proposition 3, each map ϕ J is an isomorphism between (G, • I ) and (G, • J ).
We consider now the general situation in which exactly l components of G do belong to L. If K ∈ L we call L-critical any subgroup U of Z(K) such that respectively: Proof.By Proposition 3 the result is clear for l = 0, so assume l > 0. Without loss of generality we may assume that A 1 , A 2 , . . ., A l are the central Aut (G)-indecomposable factors having components in L. We set L = {1, 2, . . ., l} and we claim that H(G) contains the set whose cardinality is 2 • |P(L c )| = 2 m .By the definition of H(G) and Lemmas 3 and 2, it is enough to show that G J c ≃ G I for each subset J of L c .By Corollary 1 there exists an automorphism α J of A J that inverts Z(A J ). Therefore the map ϕ J c defined as in Proposition 3 by ϕ J c (x J x J c ) = (x J ) −αJ x J c for each x J ∈ A J , x J c ∈ A J c , is a well-defined bijection of G that conjugates G I to G J c .Note that T (G) contains the elementary abelian 2-subgroup {ϕ J c |J ∩ L = ∅}.
Assume now that K ⊂ H(G).Note that G R ∈ H(G) \ K for some R ⊂ I if and only if G R c ∈ H(G) \ K.This shows in particular that |H(G)| is even.Moreover, by Remark 3, G R ∈ H(G) \ K if and only if G R is isomorphic to the group (G, • R Now, by Remark 3, any possible isomorphism α from G to (G, • R ) maps each A i to itself.In particular, if we take r ∈ R ∩ L and s ∈ R c ∩ L, the isomorphism α induces an automorphism on A r and an antihomomorphism on A s , as we have:  3) implies that the restriction of α to W = A r ∩ A s is a homomorphism, while condition (4) implies that the restriction of α • inv to W is a homomorphism.

Theorem 1 .
Let G be a finite nonabelian semisimple group and let G = A 1 A 2 . . .A n be its unique central decomposition as a product of Aut (G)-indecomposable factors.Assume that the number of factors A i of G having components in L is exactly l, for some 0 ≤ l ≤ n.Then the set H(G) is completely known and it has cardinality |H(G)| = 2 h , where m ≤ h ≤ n and m = min {n − l + 1, n}.Also, the group T (G) is an elementary abelian group of order 2 h .Moreover, if the centers of the factors A i are all amalgamated, then |H(G)| = 2 m and T (G) is elementary abelian of order 2 k m.

(
a r b r ) α = a α r b α r = a α r • J b α r for each a r , b r ∈ A r ,(3)(a s b s ) α = a α s b α s = b α s • J a α s for each a s , b s ∈ A s .(4)Condition(