On the holomorph of finite semisimple groups

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 Ai of G having components in L is exactly l for some 0≤l≤n. Then T⁢(G) is an elementary abelian group of order 2h for some h with min⁡{n-l+1,n}≤h≤n, acting regularly on H⁢(G). Moreover, if the centers of the factors Ai are all amalgamated, then |H⁢(G)|=2m, and T⁢(G) is elementary abelian of order 2m, where m=min⁡{n,n-l+1}, and therefore |H⁢(G)|=2m.

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].

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.

1. CS⁢(G)⁢(ρ⁢(G))=λ⁢(G) and CS⁢(G)⁢(λ⁢(G))=ρ⁢(G).

2. NS⁢(G)⁢(ρ⁢(G))=Aut⁢(G)⋉ρ⁢(G)=Aut⁢(G)⋉λ⁢(G)=NS⁢(G)⁢(λ⁢(G)).

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); therefore, inv normalizes NS⁢(G)⁢(ρ⁢(G)), that is, inv∈NHol⁢(G).

Lemma 1.

Let G be a finite semisimple group. Then G is a central product of perfect subgroups which are centrally indecomposable as Aut⁢(G)-subgroups,

Moreover, the integer n and the subgroups Ai (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

Since G is perfect, each Mi/Z⁢(G) is perfect. In particular, each Mi is equal to Mi′⁢Z⁢(G), where Mi′ is perfect. Now, for every j≠i, we have that

and thus Mj induces by conjugation a central automorphism on Mi′. Since perfect groups have no nontrivial central automorphisms (see [8] or [5, Lemma 7.1]), we have [Mi,Mj]=1 for every j≠i. In particular, equation (2.1) and the fact that G is perfect imply the following central factorization of G:

where Ai=Mi′ for each i=1,2,…,n. Note that the Ai are perfect Aut⁢(G)-subgroups, which are indecomposable as Aut⁢(G)-subgroups. Finally, the uniqueness of the factorization (2.1) (see [12, Theorem 3.3.8]) and, again, the fact that perfect groups have no nontrivial central automorphisms imply that the central product decomposition (2.2) is also unique. ∎

Lemma 2.

Each GJ is a subgroup of S⁢(G) that lies in Hol⁢(G).

Proof.

Since [ρ⁢(G),λ⁢(G)]=1, for every xJ,yJ∈AJ, xJc,yJc∈AJc, it follows that

which lies in GJ.

Moreover, since ρ⁢(AJ)≤ρ⁢(G) and λ⁢(AJc)≤λ⁢(G), we have

completing the proof of the lemma. ∎

Lemma 3.

The inversion map inv conjugates GJ to GJc 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, J⁢(G)={GJ∣J⊆I}.

Proof.

We first show that GJ acts regularly on the set G. For an arbitrary g∈G, write g=xJ⁢xJc, with xJ∈AJ and xJc∈AJc. Then the element ρ⁢(xJ)⁢λ⁢(xJc-1) of GJ sends 1 to g,

since [AJ,AJc]=1. Moreover, the stabilizer of 1 in GJ consists of the elements σ=ρ⁢(xJ)⁢λ⁢(xJc) such that xJ=xJc-1∈AJ∩AJc≤Z⁢(G), and therefore σ is the identity.

To show that each GJ is normal in Hol⁢(G), since [ρ⁢(G),λ⁢(G)]=1, it is enough to show that Aut⁢(G) normalizes every GJ. Fix J⊆I; let α∈Aut⁢(G), and let ρ⁢(xJ)⁢λ⁢(xJc) be an arbitrary element of GJ, where xJ∈AJ and xJc∈AJc. Then we have that, for every g∈G,

Therefore,

which lies in GJ, since AJ and AJc are characteristic subgroups of G.

Finally, by [5, Theorem 7.8], |𝒥⁢(G)|=2n, and therefore, to complete the proof, it remains to show that GJ≠GK whenever J≠K. If GJ=GK, then there exists an i∈I for which GJ contains both ρ⁢(Ai) and λ⁢(Ai). But then the stabilizer of 1 in GJ would contain {ρ⁢(x)⁢λ⁢(x-1)∣x∈Ai}, that is, all the conjugates of elements of Ai. Since Ai is not central in G, this contradicts the fact that GJ is regular. ∎

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 L3⁢(4), with center containing Z2×Z2×Z3,

2. a covering of U4⁢(3), with center containing Z3×Z4,

3. U6⁢(2)~, the universal covering of U6⁢(2),

4. E62⁢(2)~, the universal covering of E62⁢(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, Mc⁢(S) and Me⁢(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) centralizes Mc⁢(S), and there is an isomorphism of Outdiag⁢(S) on Mc⁢(S) preserving the action of Out⁢(S). Note also that if one of the factors Mc⁢(S) or Me⁢(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:

1. |Me⁢(S)|≤2,

2. |Me⁢(S)|>2.

Case (1) |Me⁢(S)|≤2. 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 Mc⁢(S). But Outdiag⁢(S) is always inverted by Out⁢(S) since, by [7, Theorem 2.5.12], we have that,

1. if S∈{Am⁢(q),D2⁢m+1⁢(q),E6⁢(q)}, then Outdiag⁢(S) is inverted by a graph automorphism (by [7, Theorem 2.5.12 (i)]),

2. if S∈{Am2⁢(q),D2⁢m+12⁢(q),E62⁢(q)}, Outdiag⁢(S) is inverted by a field automorphism (by [7, Theorem 2.5.12 (g)]),

3. 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) |Me⁢(S)|>2. From [7, Table 6.3.1] and the knowledge of the corresponding factor Mc⁢(S) ([7, Theorem 2.5.12]), we can see that if S is not isomorphic to one of the simple groups

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=L3⁢(4). Then Me⁢(S)≃Z4×Z4 and Mc⁢(S)≃Z3. Here

with Σ=Outdiag⁢(S)⁢ΓS≃S3 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 Me⁢(S). Now, u is Aut⁢(S)-conjugate to an element of the form ϕ⁢i, with ϕ a field automorphism and i a graph automorphism, where ϕ and i are commuting involutions (note that ΦS⁢ΓS≃Z2×Z2). The action of ΦS⁢ΓS on Mc⁢(S) is the same as on Outdiag⁢(S). Thus, by [7, Theorem 2.5.12 (g) and (i)], both ϕ and i invert Mc⁢(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 Mc⁢(S) and Outdiag⁢(S) are Aut⁢(S)-isomorphic, the elements of Out⁢(S) that induce inversion on Mc⁢(S) are the six non-central involutions, that is, the elements of the set

Note that, from [7, Proposition 6.2.2 and the proof of Theorem 6.3.1], Outdiag⁢(S) acts faithfully on the quotient group Me⁢(S)/Φ⁢(Me⁢(S)), and hence on Me⁢(S). Next, let t be an element of T. Since t inverts Outdiag⁢(S),

so CMe⁢(S)⁢(t)=〈b〉≃Z4, and t inverts a unique cyclic subgroup of order 4. This in particular shows that, when K is a covering extension with

no inversion is induced by automorphisms of K on Z⁢(K), while inversion is induced if Z⁢(K)≃Z4×Z3.

Let S=U4⁢(3). Then Me⁢(S)≃Z3×Z3 and Mc⁢(S)≃Z4. Here

is isomorphic to a dihedral group of order 8 acting faithfully on Me⁢(S). In particular, the nontrivial central element of Out⁢(S) is the unique element inducing inversion on Me⁢(S). Note that this element belongs to Outdiag⁢(S), and therefore it centralizes Mc⁢(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 occurs 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)⁢ΦS when 3 divides |Z⁢(S^)|, or by a field automorphism of order two when 3∤|Z⁢(S^)|.

Let S=U6⁢(2), or S=E62⁢(2). In both cases, we have that

and Out⁢(S)≃S3 acts faithfully on Me⁢(S) (see [7, Proposition 6.2.2]). In particular, the trivial outer automorphism is the only one that inverts Me⁢(S). Since it does not invert Mc⁢(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. ∎

Remark 2.

According to [7, Theorem 6.3.2], when S≃L3⁢(4) (or S≃U4⁢(3)), there are precisely two non-isomorphic covering groups S^ such that S^/Z⁢(S^)≃S and Z⁢(S^)≃Z2×Z4×Z3 (respectively, Z⁢(S^)≃Z3×Z4). 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.

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 D=K1×K2×⋯×Kt the direct product of nonabelian quasisimple groups Ki∉ℒ and N≤D such that N∩Ki=1 for each i∈{1,…,t}. Note that

By Proposition 2, for each i∈{1,…,t} there exists an automorphism αi of Ki that acts like the inversion on Z⁢(Ki). We may therefore define α∈Aut⁢(D) by xα=(x1,x2,…,xt)α=(x1α1,x2α2,…,xtαt) for x∈D (where each xi∈Ki). Note that every element of Z⁢(D) is inverted by α. In particular, Nα=N, and α induces an automorphism on D/N that inverts its center. ∎

Proposition 3.

Assume that G is a finite nonabelian semisimple group, and let G=A1⁢A2⁢…⁢An 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)={GJ∣J⊆I}. Moreover, the group T⁢(G)=NHol⁢(G)/Hol⁢(G) is elementary abelian of order 2n.

Proof.

By Lemma 4, we have 𝒥⁢(G)={GJ∣J⊆I}, so ℋ⁢(G)⊆{GJ∣J⊆I}. We fix a subset J of I and, using Corollary 1, choose an automorphism αJc of AJc that inverts Z⁢(AJc). Define the map φJ:G→G by φJ⁢(xJ⁢xJc)=xJ⁢(xJc)-αJc. Since αJc inverts AJ∩AJc≤Z⁢(G), the map φJ is a well-defined bijection of G, that is, an element of S⁢(G). We claim that the following hold:

1. φJ conjugates GI to GJ;

2. φJ∈NHol⁢(G) and, if J≠I, then φJ∉Hol⁢(G);

3. (φJ)2∈Hol⁢(G).

Note that, by the arbitrary choice of J⊆I, once proved, (1) will imply that ℋ⁢(G)=𝒥⁢(G)={GJ∣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) (GI)φJ=GJ. We claim that, for xJ∈AJ and xJc∈AJc,

or, equivalently, that

Let g∈G, and write g as g=yJ⁢yJc (with yJ∈AJ and yJc∈AJc). Then

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) φJ∈NHol⁢(G). By (1), we have that

since each GJ lies in ℋ⁢(G)=ℐ⁢(G), and therefore NS⁢(G)⁢(GJ)=NS⁢(G)⁢(GI); thus φJ∈NHol⁢(G).

Furthermore, GJ≠GI for J≠I; thus we trivially have that φJ∉Hol⁢(G) for J≠I.

(3) (φJ)2∈Hol⁢(G). We claim that, for every xJ⁢xJc∈G,

or, equivalently, that

Let g∈G, and write g=yJ⁢yJc, where yJ∈AJ, yJc∈AJc. Then

while

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:

for each g=gJ⁢gJc, h=hJ⁢hJc∈G, where gJ,hJ∈AJ and gJc,hJc∈AJc. Then (G,∘J) is a group. Note that ∘I coincides with the group operation of G, while ∘∅ with the opposite multiplication in G, that is, g1∘∅g2=g2⁢g1 for every g1,g2∈G. With this notation, it is straightforward to prove that

1. (G,∘J) is a group isomorphic to GJ 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 φJ is an isomorphism between (G,∘I) and (G,∘J).

Theorem 1.

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 Ai of G having components in L is exactly l for some 0≤l≤n. Then T⁢(G) is an elementary abelian group of order 2h for some h with min⁡{n-l+1,n}≤h≤n, acting regularly on H⁢(G). Moreover, if the centers of the factors Ai are all amalgamated, then |H⁢(G)|=2m, and T⁢(G) is elementary abelian of order 2m, where m=min⁡{n,n-l+1}, and therefore |H⁢(G)|=2m.

Proof.

By Proposition 3, the result is clear for l=0, so assume l>0.

Without loss of generality, we may assume that A1,A2,…,Al are the central Aut⁢(G)-indecomposable factors having components in ℒ. We set L={1,2,…,l} and claim that ℋ⁢(G) contains the set 𝒦={GJ,GJc∣J∩L=∅}, whose cardinality is 2⋅|𝒫⁢(Lc)|=2m. By the definition of ℋ⁢(G) and Lemmas 3 and 2, it is enough to show that GJc≃GI for each subset J of Lc. By Corollary 1, there exists an automorphism αJ of AJ that inverts Z⁢(AJ). Therefore, the map φJc defined as in Proposition 3 by φJc⁢(xJ⁢xJc)=(xJ)-αJ⁢xJc for each xJ∈AJ, xJc∈AJc is a well-defined bijection of G that conjugates GI to GJc. Note that T⁢(G) contains the elementary abelian 2-subgroup {φJc∣J∩L=∅}.