Groups that have the same holomorph as a finite perfect group
aa r X i v : . [ m a t h . G R ] O c t GROUPS THAT HAVE THE SAME HOLOMORPH ASA FINITE PERFECT GROUP
A. CARANTI AND F. DALLA VOLTA
Abstract.
We describe the groups that have the same holomorphas a finite perfect group. Our results are complete for centerlessgroups.When the center is non-trivial, some questions remain open.The peculiarities of the general case are illustrated by a couple ofexamples that might be of independent interest. Introduction
We are concerned with the question, when do two groups have thesame holomorph? Recall that the holomorph of a group G is the naturalsemidirect product Aut( G ) G of G by its automorphism group Aut( G ).To put this problem in proper context, recall that if ρ : G → S ( G ) isthe right regular representation of G , where S ( G ) is the group of per-mutations on the set G , then N S ( G ) ( ρ ( G )) = Aut( G ) ρ ( G ) is isomorphicto the holomorph of G . We will also refer to N S ( G ) ( ρ ( G )) as the holo-morph of G , and write it as Hol( G ). More generally, if N ≤ S ( G ) is aregular subgroup, then N S ( G ) ( N ) is isomorphic to the holomorph of N .We therefore begin to make the above question more precise by askingfor which regular subgroup N of S ( G ) one has N S ( G ) ( N ) = Hol( G ).W.H. Mills has noted [17] that such an N need not be isomorphicto G (see Example 3.1, but also Example 1.2 below, and the commentfollowing it). In this paper, we will be interested in determining thefollowing set, and some naturally related ones Definition 1.1. H ( G ) = n N ≤ S ( G ) : N is regular, N ∼ = G and N S ( G ) ( N ) = Hol( G ) o . G.A. Miller has shown [16] that the so-called multiple holomorph of G NHol( G ) = N S ( G ) (Hol( G )) Date : 13 October 2017, 11:45 CEST — Version 9.08.2010
Mathematics Subject Classification.
Key words and phrases. holomorph, multiple holomorph, regular subgroups, fi-nite perfect groups, central products, automorphisms.The authors are members of INdAM—GNSAGA. The first author gratefullyacknowledges support from the Department of Mathematics of the University ofTrento.
A. CARANTI AND F. DALLA VOLTA acts transitively on H ( G ), and thus the group T ( G ) = NHol( G ) / Hol( G )acts regularly on H ( G ).Recently T. Kohl has described [15] the set H ( G ) and the group T ( G )for G dihedral or generalized quaternion. In [5], we have redone, via acommutative ring connection, the work of Mills [17], which determined H ( G ) and T ( G ) for G a finitely generated abelian group.In this paper we consider the case when G is a finite, perfect group,that is, G equals its derived subgroup G ′ .If G has also trivial center, then one can show that if N E Hol( G )is a regular subgroup, then N ∈ H ( G ) (in particular, N ∼ = G ). Theelements of H ( G ) can be described in terms of a Krull-Remak-Schmidtdecomposition of G as a group with Aut( G ) as group of operators,that is, in terms of the unique decomposition of G as a direct productof non-trivial characteristic subgroups that are indecomposable as thedirect product of characteristic subgroups. The group T ( G ) turns outto be an elementary abelian 2-group.If G has non-trivial center, the regular subgroups N such that N E Hol( G ) can still be described in terms of the decomposition of G as thecentral product of non-trivial, perfect, characteristic subgroups, thatare indecomposable as a central product of characteristic subgroups.However, these N need not be isomorphic to G (see Example 1.2below, and the comment following it), and the structure of T ( G ) inthis case is not clear to us at the moment. The difficulties here areillustrated by the following examples, which might be of independentinterest. Example 1.2.
There is a group G which is the central product of twocharacteristic subgroups Q , Q , such that G is not isomorphic to thegroup ( G, ◦ ) obtained from G by replacing Q with its opposite.Recall that the opposite of a group Q is the group obtained by ex-changing the order of factors in the product of Q .We will see in Section 7 that the group ( G, ◦ ) is isomorphic to aregular subgroup N of S ( G ) such that N S ( G ) ( N ) = Hol( G ). Thereforein our context the latter condition does not imply N ∼ = G . Example 1.3.
There is a group G which is the central product ofthree characteristic subgroups Q , Q , Q such that Q and Q are notcharacteristic in the group obtained from G by replacing Q with itsopposite.As we will see in Section 7, this example shows that if N E Hol( G )is a regular subgroup, for G perfect, we may well have that N S ( G ) ( N )properly contains Hol( G ). However, if Aut( G ) and Aut( N ) have thesame order, then N S ( G ) ( N ) = Hol( G ) (see Lemma 2.6(2)). HE HOLOMORPH OF A FINITE PERFECT GROUP 3
Example 1.2 and 1.3 are given in Subsection 7.2 as Proposition 7.10and Proposition 7.9.The plan of the paper is the following. Sections 2 and 3 introducethe holomorph and the multiple holomorph. In these sections, and inthe following ones, we have chosen to repeat some elementary and well-known arguments, when we have deemed them handy for later usage.In Sections 4 and 5 we give a description of the regular subgroups N ofHol( G ), and of those that are normal in Hol( G ), in terms of a certainmap γ : G → Aut( G ). This leads to a group operation ◦ on G suchthat N is isomorphic to ( G, ◦ ). In Section 6 we show that the valuesof γ on commutators are inner automorphisms, and this leads us toconsider perfect groups.In Section 7 we study the case of finite, perfect groups. We firstobtain a description of the normal subgroups of Hol( G ) that are regular,in terms of certain central product decompositions of G (Theorem 7.5).We then discuss separately, as explained above, the centerless case,where we can give a full picture, and the general case, where somequestions remain open. Section 8 deals with a representation-theoreticmethod that is critical for the construction of the examples.We note, as in [15], that this work is related to the enumerationof Hopf-Galois structures on separable field extensions, as C. Greitherand B. Pareigis have shown [10] that these structures can be describedthrough the regular subgroups of a suitable symmetric group, which arenormalized by a given regular subgroup; this connection is exploited inthe work of L. Childs [8], N.P. Byott [3], and Byott and Childs [4].Our discussion of (normal) regular subgroups touches also on thesubject of skew braces [13], see Remark 5.3.We are very grateful to Robert Guralnick for several useful conver-sations. We are indebted to Derek Holt for kindly explaining to us incareful detail the example and the construction method which led toProposition 7.11. 2. The holomorph of a group
Notation 2.1.
We write permutations as exponents, and denote com-positions of maps by juxtaposition. We compose maps left-to-right.The holomorph of a group G is the natural semidirect productAut( G ) G of G by its automorphism group Aut( G ). Let S ( G ) be the group ofpermutations on the set G . Consider the right and the left regularrepresentations of G : ( ρ : G → S ( G ) g ( x xg ) ( λ : G → S ( G ) g ( x gx ) . A. CARANTI AND F. DALLA VOLTA
Notation 2.2.
We denote the inversion map g g − on a group byinv. Definition 2.3.
The opposite of the group G is the group obtained byexchanging the order of factors in the product of G .The map inv is an isomorphism between a group G and its opposite;compare with Proposition 2.4(4) below.The following well-known fact should be compared with Lemma 2.4.2of the paper [10] in which Greither and Pareigis set up the connection,already mentioned in the Introduction, between Hopf Galois extensionsand regular subgroups of symmetric groups. Proposition 2.4. (1) C S ( G ) ( ρ ( G )) = λ ( G ) and C S ( G ) ( λ ( G )) = ρ ( G ) .(2) The stabilizer of in N S ( G ) ( ρ ( G )) is Aut( G ) .(3) We have N S ( G ) ( ρ ( G )) = Aut( G ) ρ ( G ) = Aut( G ) λ ( G ) = N S ( G ) ( λ ( G )) , and this group is isomorphic to the holomorph Aut( G ) G of G .(4) Inversion on G normalizes N S ( G ) ( ρ ( G )) , centralizes Aut( G ) ,and conjugates ρ ( G ) to λ ( G ) , that is ρ ( G ) inv = λ ( G ) . Notation 2.5.
We write Hol( G ) = N S ( G ) ( ρ ( G )).We will refer to either of the isomorphic groups N S ( G ) ( ρ ( G )) andAut( G ) G as the holomorph of G .We now record another well-known fact. Lemma 2.6. (1) Let Ω be a set, and G a regular subgroup of S (Ω) . Then there isan isomorphism S (Ω) → S ( G ) that sends G to ρ ( G ) , and thus N S (Ω) ( G ) to Hol( G ) .(2) If N ≤ S ( G ) is a regular subgroup, then N S ( G ) ( N ) is isomorphicto the holomorph of N . It is because of Lemma 2.6(1) that we have done without a set Ω,and started directly with S ( G ) and its regular subgroup ρ ( G ). Proof.
We only treat (2), for further reference.Consider for such a regular subgroup N the bijection ϕ : N → Gn n . Then ψ : S ( G ) → S ( N ) σ ϕσϕ − HE HOLOMORPH OF A FINITE PERFECT GROUP 5 (recall that we compose left-to-right) is an isomorphism, which maps N onto ρ ( N ), as for x, n ∈ N we have x ϕnϕ − = (1 xn ) ϕ − = xn, that is, ϕnϕ − = ρ ( n ) . In particular, ψ ( N S ( G ) ( N )) = N S ( N ) ( ρ ( N )) = Hol( N ). (cid:3) Groups with the same holomorph
In view of Lemma 2.6(2), one may inquire, what are the regularsubgroups N ≤ S ( G ) for which(1) Hol( N ) ∼ = N S ( G ) ( N ) = N S ( G ) ( ρ ( G )) = Hol( G ) . W.H. Mills has noted in [17] that if (1) holds, then G and N neednot be isomorphic. Example 3.1.
The dihedral and the generalized quaternion groups oforder 2 n , for n ≥
4, have the same normalizer in S n , in suitable regularrepresentations. [14, 3.10], [15, 2.1].When we restrict our attention to the regular subgroups N of S ( G )for which N S ( G ) ( N ) = Hol( G ) and N ∼ = G , we can appeal to a resultof G.A. Miller [16]. Miller found a characterization of these subgroupsin terms of the multiple holomorph of G NHol( G ) = N S ( G ) (Hol( G )) . Consider the set H ( G ) = n N ≤ S ( G ) : N is regular, N ∼ = G and N S ( G ) ( N ) = Hol( G ) o . Using the well-known fact that two regular subgroups of S ( G ) are iso-morphic if and only if they are conjugate in S ( G ), Miller showed thatthe group NHol( G ) acts transitively on H ( G ) by conjugation. (SeeLemma 4.2 in the next Section for a comment on this.) Clearly thestabilizer in NHol( G ) of any element N ∈ H ( G ) is N S ( G ) ( N ) = Hol( G ).We obtain Theorem 3.2.
The group T ( G ) = NHol( G ) / Hol( G ) acts regularly on H ( G ) by conjugation. A. CARANTI AND F. DALLA VOLTA Regular subgroups of the holomorph
This section adapts to the nonabelian case the results of [6, Theo-rem 1] and [9, Proposition 2].Let G be a finite group, and N ≤ Hol( G ) a regular subgroup. Since N is regular, for each g ∈ G there is a unique element ν ( g ) ∈ N , suchthat 1 ν ( g ) = g . Now 1 ν ( g ) ρ ( g ) − = 1, so that ν ( g ) ρ ( g ) − ∈ Aut( G ) byProposition 2.4(2). Therefore for g ∈ G we can write uniquely(2) ν ( g ) = γ ( g ) ρ ( g ) , for a suitable map γ : G → Aut( G ). We have(3) ν ( g ) ν ( h ) = γ ( g ) ρ ( g ) γ ( h ) ρ ( h ) = γ ( g ) γ ( h ) ρ ( g γ ( h ) h ) . Since N is a subgroup of S ( G ), γ ( g ) γ ( h ) ∈ Aut( G ), and the expres-sion (2) is unique, we have γ ( g ) γ ( h ) ρ ( g γ ( h ) h ) = γ ( g γ ( h ) h ) ρ ( g γ ( h ) h ) , from which we obtain(4) γ ( g ) γ ( h ) = γ ( g γ ( h ) h ) . It is now immediate to obtain
Theorem 4.1.
Let G be a finite group. The following data are equiv-alent.(1) A regular subgroup N ≤ Hol( G ) .(2) A map γ : G → Aut( G ) such that (5) γ ( g ) γ ( h ) = γ ( g γ ( h ) h ) . Moreover, under these assumptions(a) the assignment g ◦ h = g γ ( h ) h. for g, h ∈ G , defines a group structure ( G, ◦ ) with the same unityas that of G .(b) There is an isomorphism ν : ( G, ◦ ) → N .(c) For g, h ∈ G , one has g ν ( h ) = g ◦ h. Proof.
Concerning the last statements, (5) implies that ◦ is associative.Then for each h ∈ G one has that 1 ◦ h = 1 γ ( h ) h = h , as γ ( h ) ∈ Aut( G ),and that ( h − ) γ ( h ) − is a left inverse of h with respect to ◦ . The bijection ν introduced above is a homomorphism ( G, ◦ ) → N by (3) and (4).Finally, g ν ( h ) = g γ ( h ) h = g ◦ h. (cid:3) HE HOLOMORPH OF A FINITE PERFECT GROUP 7
Note, for later usage, that (5) can be rephrased, setting k = g γ ( h ) , as(6) γ ( kh ) = γ ( k γ ( h ) − ) γ ( h ) . We record the following Lemma, which will be useful later. We usethe setup of Theorem 4.1.
Lemma 4.2.
Suppose N ∈ H ( G ) , and let ϑ ∈ NHol( G ) such that ρ ( G ) ϑ = N and ϑ = 1 . Then ϑ : G → ( G, ◦ ) is an isomorphism.Conversely, an isomorphism ϑ : G → ( G, ◦ ) conjugates ρ ( G ) to N .Proof. Note first that given any ϑ ∈ NHol( G ) such that ρ ( G ) ϑ = N ,we can modify ϑ by a suitable ρ ( g ), and assume 1 ϑ = 1.Suppose for y ∈ G one has ρ ( y ) ϑ = ν ( y σ ), for some σ ∈ S ( G ). Thus ρ ( y ) ϑ = ϑν ( y σ ), so that for x, y ∈ G one has( xy ) ϑ = x ρ ( y ) ϑ = x ϑν ( y σ ) = x ϑγ ( y σ ) y σ = x ϑ ◦ y σ . Setting x = 1 we see that ϑ = σ , and thus( xy ) ϑ = x ϑ ◦ y ϑ . For the converse, if the last equation holds then x ρ ( y ) ϑ = ( x ϑ − y ) ϑ = x ◦ y ϑ = x ν ( y ϑ ) . (cid:3) Normal regular subgroups of the holomorph
In this section, we adapt to the nonabelian case the results of [5,Theorem 3.1].Consider the sets I ( G ) = n N ≤ S ( G ) : N is regular, N S ( G ) ( N ) = Hol( G ) o and J ( G ) = { N ≤ S ( G ) : N is regular, N E Hol( G ) } . Clearly we have(7) H ( G ) ⊆ I ( G ) ⊆ J ( G ) . If N E Hol( G ), then Hol( G ) ≤ N S ( G ) ( N ). However the latter may wellbe properly bigger than the former, as shown by the following simpleexample. Example 5.1.
Let G = h (1 2 3 4) i ≤ S . Then N S ( G ) has order 8,but its regular subgroup N = h (1 3)(2 4) , (1 4)(2 3) i is normal in thewhole S . A. CARANTI AND F. DALLA VOLTA
Moreover, even when Hol( G ) = N S ( G ) ( N ), Example 3.1 shows that G and N are not necessarily isomorphic. Therefore all inclusions in (7)may well be proper.We will now give a characterization of the elements of J ( G ) in termsof the description of Theorem 4.1. Suppose N ∈ J ( G ). To ensure that N E Hol( G ), it is enough to make sure that N is normalized by Aut( G ),as if this holds, then the normalizer of N contains Aut( G ) N , which iscontained in Hol( G ), and has the same order as Hol( G ), as the regularsubgroup N intersects Aut( G ) trivially.In order for Aut( G ) to normalize N , we must have that for all β ∈ Aut( G ) and g ∈ G , the conjugate ν ( g ) β of ν ( g ) by β in S ( G ) lies in N .Since ν ( g ) β = ( γ ( g ) ρ ( g )) β = γ ( g ) β ρ ( g ) β = γ ( g ) β ρ ( g β )and γ ( g ) β ∈ Aut( G ), uniqueness of (2) implies γ ( g ) β ρ ( g β ) = γ ( g β ) ρ ( g β ) , so that(8) γ ( g β ) = γ ( g ) β for g ∈ G and β ∈ Aut( G ). Applying this to (6), we obtain that for h, k ∈ G (9) γ ( kh ) = γ ( k γ ( h ) − ) γ ( h ) = γ ( k ) γ ( h ) − γ ( h ) = γ ( h ) γ ( k ) , that is, γ : G → Aut( G ) is an antihomomorphism.Now note that (5) follows from (8) and (9), as γ ( g γ ( h ) h ) = γ ( h ) γ ( g ) γ ( h ) = γ ( g ) γ ( h ) . We have obtained
Theorem 5.2.
Let G be a finite group. The following data are equiv-alent.(1) A regular subgroup N E Hol( G ) , that is, an element of J ( G ) .(2) A map γ : G → Aut( G ) such that for g, h ∈ G and β ∈ Aut( G )(10) γ ( gh ) = γ ( h ) γ ( g ) γ ( g β ) = γ ( g ) β . Moreover, under these assumptions(a) the assignment g ◦ h = g γ ( h ) h. for g, h ∈ G , defines a group structure ( G, ◦ ) with the same unityas that of G .(b) There is an isomorphism ν : ( G, ◦ ) → N .(c) For g, h ∈ G , one has g ν ( h ) = g ◦ h. (d) Every automorphism of G is also an automorphism of ( G, ◦ ) . HE HOLOMORPH OF A FINITE PERFECT GROUP 9
Remark 5.3.
Note that under the hypotheses of Theorem 4.1, G be-comes a skew right brace (for which see [13] ) under the operations · and ◦ , that is, G is a group with respect to both operations, which areconnected by ( gh ) ◦ k = ( g ◦ k ) k − ( h ◦ k ) , for g, h, k ∈ G. The braces which correspond to the normal regular subgroups of
Hol( G ) satisfy the additional condition of Theorem 5.2 (d) . In the following, when dealing with N ∈ J ( G ), we will be using thenotation of Theorem 5.2 without further mention. Proof.
The last statement follows from( g ◦ h ) β = ( g γ ( h ) h ) β = ( g β ) γ ( h ) β h β = ( g β ) γ ( h β ) h β = g β ◦ h β , for g, h ∈ G and β ∈ Aut( G ). (cid:3) Let us exemplify the above for the case of the left regular represen-tation. Consider the morphism ι : G → Aut( G ) y ( x y − xy ) , that is, ι ( y ) ∈ Inn( G ) is conjugacy by y . If N = λ ( G ), then we havefor y ∈ G λ ( y ) = ι ( y − ) ρ ( y ) , as for z, y ∈ G we have z ι ( y − ) ρ ( y ) = yzy − y = yz = z λ ( y ) . Therefore γ ( y ) = ι ( y − ), and x ◦ y = x ι ( y − ) y = yx, that is, ( G, ◦ ) is the opposite group of G .Also, in [7] S. Carnahan and L. Childs prove that if G is a non-abelianfinite simple group, then H ( G ) = { ρ ( G ) , λ ( G ) } . In our context, thiscan be proved as follows. If G is a non-abelian finite simple group, and N ∈ H ( G ), then the normal subgroup ker( γ ) of G can only be either G or { } . In the first case we have x ◦ y = x γ ( y ) y = xy for x, y ∈ G ,so that x ν ( y ) = x ◦ y = xy = x ρ ( y ) , and N = ρ ( G ). In the second case, γ is injective. Since we have(11) γ ( x ◦ y ) = γ ( x ) γ ( y ) = γ ( yx ) , we obtain x ◦ y = yx , so N = λ ( G ) as we have just seen. Commutators
In this section we assume we are in the situation of Theorem 5.2.Let β ∈ Aut( G ), g ∈ G , and consider the commutator [ β, g − ] = g β g − taken in Aut( G ) G . Using (10), we get(12) γ ([ β, g − ]) = γ ( g β g − ) = γ ( g ) − γ ( g ) β = [ γ ( g ) , β ] . In the particular case when β = ι ( h ), for some h ∈ G , we obtain γ ([ h, g − ]) = γ ([ ι ( h ) , g − ]) = [ γ ( g ) , ι ( h )] = ι ([ γ ( g ) , h ]) , that is(13) γ ([ h, g − ]) = ι ([ γ ( g ) , h ]) . From this identity we obtain ι ([ γ ( g ) , h ]) = γ ([ h, g − ]) = γ ([ g − , h ]) − == ι ([ γ ( h − ) , g − ]) − = ι ([ g − , γ ( h − )]) , that is,(14) [ γ ( g ) , h ] ≡ [ g − , γ ( h − )] (mod Z ( G ))for all g, h ∈ G .In the rest of the paper we will deal with the case of finite perfectgroups, that is, those finite groups G such that G ′ = G . In this case,according to (13), we have γ ( G ) ≤ Inn( G ).7. Perfect groups
Let G be a non-trivial, finite, perfect group. We will determine J ( G ), and then discuss its relationship to H ( G ).Recall that an automorphism β of a group G is said to be central if[ x, β ] = x − x β ∈ Z ( G ) for all x ∈ G . In other words, an automorphismof G is central if it induces the identity on G/Z ( G ).We record for later usage a couple of elementary, well-known facts. Lemma 7.1.
Let G be a finite perfect group.(1) Z ( G ) = Z ( G ) .(2) A central automorphism of G is trivial.Proof. The first part is Grün’s Lemma [12].For the second part, if β is a central automorphism of G , then x [ x, β ]is a homomorphism from G to Z ( G ). Since G = G ′ , this homomor-phism maps G onto the identity. (cid:3) We now show that an element N ∈ J ( G ) yields a direct productdecomposition of Inn( G ). Proposition 7.2.
Let G be a finite, perfect group, and N ∈ J ( G ) . HE HOLOMORPH OF A FINITE PERFECT GROUP 11 (1) Z ( G ) ≤ ker( γ ) .(2) Inn( G ) = γ ( G ) × ι (ker( γ )) . Later we will lift the direct product decomposition (2) of Inn( G ) toa central product decomposition of G (Theorem 7.5(1)). Proof.
For the first part, let g ∈ Z ( G ). (14) yields [ γ ( g ) , h ] ∈ Z ( G )for all h ∈ G , that is, γ ( g ) is a central automorphism of G . ByLemma 7.1(2), γ ( g ) = 1.For the second part, we first show that γ ( G ) and ι (ker( γ )) commuteelementwise. Let g ∈ G and k ∈ ker( γ ). The results of Section 6 yield[ γ ( g ) , ι ( k )] = ι ([ γ ( g ) , k ]) = ι ([ g − , γ ( k − )]) = 1 . We now show that γ ( G ) ∩ ι (ker( γ )) = 1. Write an element of theperfect group G as x = n Y i =1 [ h i , g − i ] , for suitable g i , h i ∈ G .Using the first identity of (10) we get first(15) γ ( x ) = Y i = n γ ([ h i , g − i ]) = Y i = n [ γ ( g i ) , γ ( h − i )](note that the order of the product has been inverted by the applicationof γ ).Using (13) and the first identity of (10) we also get γ ( x ) = Y i = n γ ([ h i , g − i ]) = Y i = n ι ([ γ ( g i ) , h i ]) = ι Y i = n [ γ ( g i ) , h i ] ! . Now if γ ( x ) ∈ γ ( G ) ∩ ι (ker( γ )), part (1) yields Y i = n [ γ ( g i ) , h i ] ∈ ker( γ ) . We thus have, using (10) and (12)1 = γ Y i = n [ γ ( g i ) , h i ] ! = n Y i =1 γ ([ γ ( g i ) , h i ]) = n Y i =1 [ γ ( h − i ) , γ ( g i )] = γ ( x ) − , according to (15). Therefore γ ( x ) = 1, as claimed.Finally we have, keeping in mind part (1), | γ ( G ) × ι (ker( γ )) | = | γ ( G ) | · | ker( γ ) || Z ( G ) | = | G/Z ( G ) | = | Inn( G ) | , so that γ ( G ) × ι (ker( γ )) = Inn( G ). (cid:3) Regarding Inn( G ) and G as groups with operator group Aut( G ),we note that the second equation of (10) implies that both γ ( G ) and ι (ker( γ )) are Aut( G )-invariant, and so are H = ι − ( γ ( G )) and ker( γ ).(Clearly the latter statement is the same as saying that H and ker( γ ) are characteristic subgroups of G , but we prefer to use the same termi-nology of groups with Aut( G ) as a group of operators for both G andInn( G ).)We have G = H ker( γ ). We claim that [ H, ker( γ )] = 1, that is, G is the central product of H and ker( γ ), amalgamating Z ( G ). We willneed the following simple Lemma, which is hinted at by Joshua A.Grochow and Youming Qiao in [11, Remark 7.6]. Lemma 7.3.
Let G be a group, and H, K ≤ G such that G/Z ( G ) = HZ ( G ) /Z ( G ) × KZ ( G ) /Z ( G ) . Suppose KZ ( G ) /Z ( G ) is perfect.Then(1) K ′ is perfect, and(2) [ H, K ] = 1 .Proof.
Since KZ ( G ) /Z ( G ) is perfect, we have KZ ( G ) = K ′ Z ( G ),so that K ′ = K ′′ , and K ′ is perfect. As [ H, K ′ ] = [ H, K ′ Z ( G )] =[ H, KZ ( G )] = [ H, K ] ≤ Z ( G ), H induces by conjugation central auto-morphisms on K ′ , so that by Lemma 7.1(2) [ H, K ] = [
H, K ′ ] = 1. (cid:3) In our situation, take K = ker( γ ). We have that KZ ( G ) /Z ( G ) ∼ = ι ( K ) is perfect, as a direct factor of the perfect group G/Z ( G ). ThenLemma 7.3(2) implies that G is the central product of H and ker( γ ),amalgamating Z ( G ).We claim Lemma 7.4. γ ( y ) = ι ( y − ) for y ∈ H .Proof. Let y ∈ H . We claim that γ ( y ) ι ( y ) is a central automorphismof the perfect group G , so that by 7.1(2) γ ( y ) = ι ( y − ).If x ∈ K = ker( γ ), we have from (14)[ γ ( y ) , x ] ≡ [ y − , γ ( x − )] ≡ Z ( G )) , so that γ ( y ) induces an automorphism of the characteristic subgroup K which is the identity modulo Z ( G ), and so does γ ( y ) ι ( y ), as [ H, K ] = 1.Let now x ∈ H . Consider first the special case when G = H × K .Then γ is injective on H , so that (11) implies x ◦ y = yx , and thus x − x γ ( y ) ι ( y ) = x − y − ( x ◦ y ) y − y = x − y − ( yx ) y − y = 1 , that is, γ ( y ) ι ( y ) is the identity on H .In the general case, (11) implies that x ◦ y ≡ yx (mod K ), so thatas above x − x γ ( y ) ι ( y ) ≡ x − y − ( x ◦ y ) y − y ≡ K ) , that is, x − x γ ( y ) ι ( y ) ∈ K . Clearly x − x γ ( y ) ι ( y ) ∈ H , as H is characteristicin G . Therefore x − x γ ( y ) ι ( y ) ∈ H ∩ K ≤ Z ( G ), so that γ ( y ) ι ( y ) inducesan automorphism of H which is the identity modulo Z ( G ). HE HOLOMORPH OF A FINITE PERFECT GROUP 13
It follows that γ ( y ) ι ( y ) is a central automorphism of G = HK , asclaimed. (cid:3) For y ∈ H and x ∈ ker( γ ) we have x ◦ y = x γ ( y ) y = x ι ( y − ) y = yxy − y = yx = y γ ( x ) x = y ◦ x. Also, if x, y ∈ H we have x ◦ y = x γ ( y ) y = x y − y = yx .In the following we will be writing the elements of G as pairs in H × ker( γ ), understanding that a pair represents an equivalence classwith respect to the central product equivalence relation which identifies( xz, y ) with ( x, zy ), for z ∈ Z ( G ).We have obtained Theorem 7.5.
Let G be a finite perfect group.(1) If N ∈ J ( G ) , then G is a central product of its subgroups H = ι − ( γ ( G )) ′ and K = ker( γ ) . Both H and K are Aut( G ) -subgroups of G .(2) For x ∈ H we have γ ( x ) = ι ( x − ) .(3) ( G, ◦ ) is also a central product of the same subgroups. If werepresent the elements of G as (equivalence classes of) pairs in H × K , then (16) ( x, y ) ◦ ( a, b ) = ( ax, yb ) . (4) For ( a, b ) ∈ G , the action of ν ( a, b ) on ( x, y ) ∈ G is given by ( x, y ) ν ( a,b ) = ( x, y ) ◦ ( a, b ) = ( ax, yb ) , that is, N induces the right regular representation on K , andthe left regular representation on H . We note the following analogue of Proposition 2.4(1) and (4).
Proposition 7.6.
Let G be a finite perfect group, and let G = HK bea central decomposition, with Aut( G ) -invariant subgroups H, K . Con-sider the following two elements of J ( G ) .(1) N , for which ker( γ ) = KZ ( G ) and H = ι − ( γ ( G )) ′ , with γ ( x ) = ι ( x − ) for x ∈ H , and associated group operation ( x, y ) ◦ ( a, b ) = ( ax, yb ) .(2) N , for which ker( γ ) = HZ ( G ) and K = ι − ( γ ( G )) ′ , with γ ( x ) = ι ( x − ) for x ∈ K , and associated group operation ( x, y ) ◦ ( a, b ) = ( xa, by ) .Then(1) N inv1 = N .(2) inv : ( N , ◦ ) → ( N , ◦ ) is an isomorphism.(3) C S ( G ) ( N ) = N and C S ( G ) ( N ) = N . Proof.
The proof is straightforward. If N i = { ν i ( a, b ) : ( a, b ) ∈ G } asin Section 4,we have( x, y ) inv ν ( a,b ) inv = ( x − , y − ) ν ( a,b ) inv == ( ax − , y − b ) inv = ( xa − , b − y ) = ( x, y ) ν (( a,b ) inv ) , and then, as in Lemma 4.2, inv : ( N , ◦ ) → ( N , ◦ ) is an isomorphism. (cid:3) We now give a description of all possible central product decompo-sitions of the perfect group G as in Theorem 7.5.We deal first with the particular case when Z ( G ) = 1, where weare able to show that J ( G ) = I ( G ) = H ( G ) and determine this set,and the group T ( G ) = NHol( G ) / Hol( G ). When Z ( G ) is allowed to benon-trivial, we are able to determine J ( G ). However, examples showthat in this case H ( G ), I ( G ) and J ( G ) can be distinct, and we areunable at the moment to describe T ( G ).7.1. The centerless case.
Suppose Z ( G ) = 1, so that ι : G → Inn( G )is an isomorphism of Aut( G )-groups.Consider a Krull-Remak-Schmidt decomposition G = A × A × · · · × A n of G as an Aut( G )-group. Since Z ( G ) = 1, this is unique [18, 3.3.8,p. 83]. Therefore the only way to decompose G as the ordered directproduct of two characteristic subgroups H, K is by grouping togetherthe A i , so that there are 2 n ways of doing this. If G = H × K isone of these ordered decompositions, define an antihomomorphism γ : G → Aut( G ) by γ ( k ) = 1 for k ∈ K , and γ ( h ) = ι ( h − ), for h ∈ H .Then γ satisfies also the second identity of (10), and we have obtainedan element N ∈ J ( G ) as in Theorem (7.5)(3). The involution ϑ ∈ NHol( G ) given by ( h, k ) ϑ = ( h − , k ), for h ∈ H and k ∈ K is anisomorphism G → ( G, ◦ ). We have obtained Theorem 7.7.
Let G be a finite perfect group with Z ( G ) = 1 .(1) If N ∈ J ( G ) , that is, N E Hol( G ) is regular, then N ∈ H ( G ) ,that is, N ∼ = G .(2) If n is the length of a Krull-Remak-Schmidt decomposition of G as an Aut( G ) -group, then H ( G ) has n elements.(3) T ( G ) is an elementary abelian group of order n . Non-trivial center.
We now consider the situation when Z ( G )is (allowed to be) non-trivial.We describe the elements N of J ( G ), in analogy with the centerlesscase.As in the centerless case, we may consider the Krull-Remak-Schmidtdecomposition(17) Inn( G ) = A × A × · · · × A n HE HOLOMORPH OF A FINITE PERFECT GROUP 15 of Inn( G ) as an Aut( G )-group. This corresponds uniquely to the centralproduct decomposition of G (18) G = B B · · · B n , where B i = ι − ( A i ) ′ are perfect Aut( G )-subgroups, which are centrallyindecomposable as Aut( G )-subgroups. Therefore the central productdecomposition (18) is also unique. (Recall also that the Krull-Remak-Schmidt of G in terms of indecomposable Aut( G )-subgroups is unique,because of [18, 3.3.8, p. 83] and Lemma 7.1.(2).)As in the centerless case, we obtain that every decomposition G = H ker( γ ) as in Theorem 7.5 can be obtained by grouping together the B i in two subgroups H and K , and then defining an antihomomorphism γ : G → Aut( G ) by γ ( k ) = 1 for k ∈ KZ ( G ), and γ ( h ) = ι ( h − ), for h ∈ H , and then ◦ as in (16). As in the centerless case, this yields anelement N ∈ J ( G ). Moreover ( G, ◦ ) is still a central product of H and KZ ( G ), with ◦ as in Theorem 7.5(3).We have obtained the following weaker analogue of Theorem 7.7. Theorem 7.8.
Let G be a finite perfect group.If n is the length of a Krull-Remak-Schmidt decomposition of Inn( G ) as an Aut( G ) -group, then J ( G ) has n elements, that is, there are n regular subgroups N E Hol( G ) . Write Y = H ∩ KZ ( G ). As the subgroups H and KZ ( G ) are char-acteristic in G , we obtain that the elements of Aut( G ) can be describedvia the set of pairs { ( σ, τ ) : σ ∈ Aut( H ) , τ ∈ Aut( KZ ( G )) , σ | Y = τ | Y } . Theorem 5.2(d) states that Aut( G ) ≤ Aut( G, ◦ ). However, the lattergroup might well be bigger than the former. This is shown by thefollowing Proposition 7.9.
There exist perfect and centrally indecomposablegroups Q , Q , Q , and a central product G = Q Q Q , such that(1) each Q i is characteristic in G ,(2) in the group ( G, ◦ ) obtained by replacing Q with its opposite,the subgroups Q and Q are exchanged by an automorphism of ( G, ◦ ) , and thus are not characteristic in ( G, ◦ ) . Clearly the automorphism of the second condition lies in Aut( G, ◦ ) \ Aut( G ). This example shows that I ( G ) may well be a proper subsetof J ( G ).Clearly N ∈ I ( G ) if and only if Aut( G ) = Aut( G, ◦ ). However, inthis general situation H ( G ) might be a proper subset of I ( G ). This isshown by the following Proposition 7.10.
There exist perfect and centrally indecomposablegroups Q , Q , and a central product G = Q Q such that (1) Q , Q are characteristic in G ;(2) the group ( G, ◦ ) obtained from G by replacing Q with its oppo-site G is not isomorphic to G .Moreover, Q and Q are still characteristic in ( G, ◦ ) . Thus if N is the regular subgroup corresponding to ( G, ◦ ) of thisProposition, we have N ∈ I ( G ) \ H ( G ).So on the one hand not all central product decompositions lead toregular subgroups N which are isomorphic to G . And even when N is isomorphic to G , it is not clear whether ρ ( G ) and N are conjugateunder an involution in T ( G ), and therefore it is not clear to us at themoment whether T ( G ) is elementary abelian in this general case.To construct the groups of Proposition 7.9 and 7.10, we rely on thefollowing family of examples, based on a construction that we havelearned from Derek Holt. Proposition 7.11.
There exists a family of groups L p , for p ≡ a prime, with the following properties:(1) the groups L p /Z ( L p ) are pairwise non-isomorphic,(2) the groups L p are perfect, and centrally indecomposable,(3) Z ( L p ) is of order , and(4) Aut( L p ) acts trivially on Z ( L p ) . This is proved in Section 8.
Proof of Proposition 7.9.
Let Q , Q be two isomorphic copies of oneof the groups of Proposition 7.11, and Q another group as in Propo-sition 7.11, not isomorphic to Q , Q .Fix an isomorphism ζ : Q → Q . If Z ( Q ) = h a i , let a = a ζ . Let Z ( Q ) = h b i .Consider the central product G = Q Q Q , amalgamating a = a − = b .Consider the quotient G/Z ( G ). Since the groups of Proposition 7.11are centrally indecomposable, and have pairwise non-isomorphic cen-tral quotients, Krull-Remak-Schmidt implies that Q /Z ( G ) is charac-teristic in G/Z ( G ), and so is Q Q /Z ( G ). Therefore Q and Q Q arecharacteristic in G . Moreover, if there is an automorphism α of G thatdoes not map Q to itself, then applying to Q Q /Z ( G ) either Krull-Remak-Schmidt, or the results of [2, Theorem 3.1], and using eitherthe fact that Q /Z ( Q ) and Q /Z ( Q ) are perfect, or that they arecenterless, we obtain that α exchanges Q and Q .Therefore α | Q ζ − is an automorphism of Q , and thus maps a to a .Therefore α maps a to a ζ = a = a − . But α | Q is an automorphismof Q , and thus fixes b = a , a contradiction.Consider now the group ( G, ◦ ) obtained by replacing Q with itsopposite. Now the map which is the identity on Q , ζ inv on Q and HE HOLOMORPH OF A FINITE PERFECT GROUP 17 ζ − inv on Q induces an automorphism of ( G, ◦ ) which exchanges Q and Q . In fact we have for x, y ∈ Q ( x ◦ y ) ζ inv = ( yx ) ζ inv = ( y ζ x ζ ) inv = x ζ inv y ζ inv , and for x, y ∈ Q ( x ◦ y ) ζ − inv = ( xy ) ζ − inv = ( x ζ − y ζ − ) inv == y ζ − inv x ζ − inv = x ζ − inv ◦ y ζ − inv . Moreover a ζ inv1 = a inv2 = a − = a , and a ζ − inv2 = a inv1 = a − = a ,which is compatible with the identity on Q . (cid:3) Proof of Proposition 7.10.
Let Q , Q to be two non-isomorphic groupsas in Proposition 7.11, and let G = Q Q , amalgamating the centers.Consider the group ( G, ◦ ) obtained by replacing Q with its opposite.If the map ϑ : G → G is an isomorphism of G onto ( G, ◦ ), by thearguments of the previous proofs it has to map each Q i to itself. Then ϑ induces an anti-automorphism on Q , thus inverting Z ( G ), and anautomorphism on Q , thus fixing Z ( G ) elementwise, a contradiction. (cid:3) Proof of Proposition 7.11
Consider the groups T p = PSL(3 , p ), where p ≡ F p be the field with p elements.It is well known ([20, Theorem 3.2]) that the outer automorphismgroup of T p is isomorphic to S , where an automorphism ∆ of order3 is diagonal, obtained via conjugation with a suitable δ ∈ PGL(3 , p ),and one of the involutions is the transpose inverse automorphism ⊤ .Moreover, the Schur multiplier M ( T p ) of T p has order 3 [20, 3.3.6],it is inverted by ⊤ , and clearly centralized by ∆.Let P be the natural F p -permutation module of T p in its actionon the points of the projective plane. P is the direct sum of a copyof the trivial module, and of a module N . The structure of N isinvestigated in [21], [1]. In particular, it is shown in [1] that N has aunique composition series { } ⊂ N ⊂ N , such that N and N = N/N are dual to each other, exchanged by ⊤ , but not isomorphic to eachother. Note that ∆ still acts on P , as it comes from conjugation withan element δ ∈ PGL(3 , p ), and thus also acts on N and N .Consider the natural semidirect product S p of N by T p . It has beenshown by K.I. Tahara [19] that the Schur multiplier of T p is a directsummand of the Schur multiplier of S p , that is, M ( S p ) = M ( T p ) ⊕ K for some K . We may thus consider the central extension L p of M ( T p )by S p , which is the quotient of the covering group of S p by K . Thus Z ( L p ) = M ( T p ) (Derek Holt has shown to us calculations for smallprimes, based on the description of [19], which appear to indicate thatactually K = { } here.) An automorphism σ of L p induces automorphisms α of N and β of T p . Abusing notation slightly, we have, for n ∈ N and h ∈ T p ,( n h ) α = ( n h ) σ = ( n σ ) h σ = ( n α ) h β . If β is an involution in Out( T p ), say β = ∆ − ⊤ ∆, then we have h β = δ − h ⊤ δ , so that ( n h ) α ∆ − = ( n α ∆ − ) h ⊤ , that is, α ∆ − is an isomorphism of N with its dual, a contradiction.Then Aut( L p ) induces on T p only inner automorphisms and at mostouter automorphisms of order 3, all of which centralize M ( T p ) = Z ( L p ). References
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E-mail address : [email protected] URL : http://science.unitn.it/ ∼ caranti/ (F. Dalla Volta) Dipartimento di Matematica e Applicazioni, EdificioU5, Università degli Studi di Milano–Bicocca, Via R. Cozzi, 53, I-20126Milano, Italy
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