THE AUTOMORPHISM GROUP FOR p-central p-groups. Communicated by Alireza Abdollahi. 1. Introduction

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1 International Journal of Grou Theory ISSN (rint): , ISSN (on-line): Vol 01 No2 (2012), c 2012 University of Isfahan wwwtheoryofgrousir wwwuiacir THE AUTOMORPHISM GROUP FOR -CENTRAL -GROUPS ANITHA THILLAISUNDARAM Communicated by Alireza Abdollahi Abstract A -grou G is -central if G (G), and G is 2 -abelian if (xy) 2 = x 2 y 2 for all x, y G We rove that for G a finite 2 -abelian -central -grou, excluding certain cases, the order of G divides the order of Aut(G) 1 Introduction The following conjecture has been the subject of much debate over the ast forty years or so For G a grou, we denote the grou of automorhisms of G by Aut(G) Here G denotes the order of the grou G Well-known Conjecture For G a non-cyclic -grou of order n with n 3, G divides Aut(G) Results in favour of the conjecture have been made by Buckley [1]; Davitt [2, 3, 4, 5]; Exarchakos [6]; Faudree [7]; Fouladi, Jamali & Orfi [8]; Gaschütz [9]; Gavioli [10]; Hummel [12]; Otto [4, 5, 14]; and Yadav [18] See Result A below for further details I aologise if I have unknowingly omitted other references Notice that each non-central element g of G induces a non-trivial automorhism of G via conjugation This defines an inner automorhism of G Inn(G) denotes the subgrou of inner automorhisms of G, which is normal in Aut(G) The non-inner automorhisms are called outer automorhisms They MSC(2010): Primary: 20D15; Secondary: 20D45 Keywords: automorhism grou, -central, 2 -abelian Received: 09 December 2011, Acceted: 15 March

2 60 A Thillaisundaram are elements in Aut(G) Inn(G) of G by Out(G) and so are defined modulo Inn(G) We denote the grou of outer automorhisms Certainly as Inn(G) = G (G), we can rehrase the question to whether or not (G) divides Out(G) The conjecture has been established to be true for several classes of -grous, as listed below in Result A We note that a grou G is the central roduct of two subgrous H and K if (a) [H, K] = 1, (b) G = HK, and (c) H K = (G); and a grou is modular if each subgrou commutes with every other subgrou, ie for H, K G, we have HK = KH Result A The conjecture holds for the following finite -grous: -abelian -grous [2]; -grous of class 2 [7]; -grous of maximal class (or coclass 1) [14]; -grous of coclass 2 [8]; -grous with centre of order [9]; -grous of order at most 7 [3, 6, 10]; modular -grous [4]; -grous with G (G) metacyclic [5]; -grous with G (G) 4 [3]; G = A B where A is abelian and B divides Aut(B) [14]; G a central roduct of non-trivial subgrous H and A, where A is abelian and H divides Aut(H) [12]; G with a non-trivial normal subgrou N such that N [G, G] = 1, and G N divides Aut( G N ) [1]; G such that x(g) x G for all x G\(G), where x G denotes the conjugacy class of x in G [18] For n N, we have G {n} = {x n x G} and G n = G {n} We define the centre of G as (G) = { x G x 1 gx = g for all g G } We say that a -grou G is -central if G (G) We define G to be 2 -abelian if for all x, y G, we have (xy) 2 = x 2 y 2 In this aer, we rove the conjecture for 2 -abelian -central -grous Theorem 11 For an odd rime, let G be a non-abelian 2 -abelian -central -grou, with G 3 Suose that the centre (G) of G is of the form (G) = e 1 en where 3 e 1 e n and n 3 Then G divides Aut(G) As -abelian grous are 2 -abelian, Theorem 11 artially generalizes the fact that -abelian -grous satisfy the conjecture An examle of a 2 -abelian -central -grou, that is not itself -abelian, is the

3 Automorhisms of -Central -Grous 61 wreath roduct W = C C, which is of exonent 2 From ([13], Examles 315(iii)), we know that W is of order +1 with maximal class and is not regular A result from [17] says that -grous of maximal class with order n, where 4 n + 1, are -central Also, we have an equivalence from [17] that a -grou is -abelian if and only if it is -central and regular Using these two facts, we confirm that W is 2 -abelian -central and not -abelian In the following, we rove Theorem 11, using results on extending automorhisms of subgrous (by Passi, Singh & Yadav [15]) and on counting automorhisms of abelian -grous (by Hillar & Rhea [11]) This aer is an extract from my PhD thesis under the suervision of Rachel Camina 2 Proof of Theorem 11 Let G be a 2 -abelian -central -grou and denote (G) by simly Let Out(G) denote the largest ower of that divides Out(G) This corresonds to the order of a Sylow -subgrou of Out(G) To rove the theorem, as Inn(G) = G, it suffices to show that Out(G) Let E : 1 N G Q 1 be an extension of the grou N by the grou Q Note: if N, then E is termed a central extension Here Aut G (G) is the subgrou of automorhisms of G that induce the identity on G For N normal in G, we define Aut N (G) to be the subgrou of automorhisms of G that normalize N Our lan is to comute a lower bound for the size of a Sylow -subgrou of Out(G) To do this, we choose to consider elements of Aut() that extend to elements of Aut G (G) Naturally any such extension of a non-identity automorhism of is non-inner Such extendable elements of Aut() are determined by the following result t : Q G is a left transversal, and µ : Q Q N is defined by t(xy)µ(x, y) = t(x)t(y) Also, N Q denotes the grou of all mas ψ from Q to N such that ψ(1) = 1 In the following, Lemma 21 [15] Let 1 N G Q 1 be a central extension Using the notation above, if γ Aut N (G), then there exists a trilet (θ, φ, χ) Aut(N) Aut(Q) N Q such that for all x, y Q and n N the following conditions are satisfied: (1) γ(t(x)n) = t(φ(x))χ(x)θ(n),

4 62 A Thillaisundaram (2) µ(φ(x), φ(y))θ(µ(x, y) 1 ) = χ(x) 1 χ(y) 1 χ(xy) Conversely, if (θ, φ, χ) Aut(N) Aut(Q) N Q is a trilet satisfying equation (2), then γ defined by (1) is an automorhism of G normalizing N We take N =, Q = G in the lemma above It is clear from equation (1) of Lemma 21 that when φ = 1, the automorhism γ induces the identity on G So we set φ = 1 Given a suitable θ Aut(), we construct the required χ to satisfy: (21) µ(x, y)θ(µ(x, y) 1 ) = χ(x) 1 χ(y) 1 χ(xy) In [11], Hillar and Rhea give a useful descrition of the automorhism grou of an arbitrary abelian -grou, and they comute the size of this automorhism grou We sketch their results here The first comlete characterization of the automorhism grou of an abelian grou was, however, given by Ranum [16] We will use Hillar and Rhea s account to characterize Aut() First, we set u the relevant notation and results leading to our desired descrition We begin with an arbitrary abelian -grou H, where H = and 1 e 1 e n are ositive integers e 1 en Hillar and Rhea first describe End(H ), the endomorhism ring of H, as a quotient of a matrix subring of n n Then, as we will see below, the units Aut(H ) End(H ) are characterized from this descrition An element of H is reresented by a column vector (α 1,, α n ) T where α i Definition 22 ([11], Definition 31) by e i R = {(a ij ) n n : e i e j a ij for all i and j satisfying 1 j i n} From [11], we have that R forms a ring Let π i : e i be defined by x x mod e i Let π : n H be the homomorhism given π(x 1,, x n ) T = (π 1 (x 1 ),, π n (x n )) T Here is the descrition of End(H ) as a quotient of the matrix ring R Theorem 23 [11] The ma ψ : R End(H ) given by is a surjective ring homomorhism ψ(a)(α 1,, α n ) T = π(a(α 1,, α n ) T ) Let K be the set of matrices A = (a ij ) R such that e i a ij for all i, j This forms an ideal

5 Automorhisms of -Central -Grous 63 Lemma 24 [11] The ideal K, as defined above, is the kernel of ψ Theorem 23 and Lemma 24 give that End(H ) is isomorhic to R K For more details, the reader is referred to [11] The following is a comlete descrition of Aut(H ) Theorem 25 ([11], Theorem 36) An endomorhism M = ψ(a) is an automorhism if and only if A(mod ) GL n (F ) Hillar and Rhea illustrate how to calculate Aut(H ), which is resented in the theorem below First, the following numbers are defined: d k = max{m : e m = e k }, c k = min{m : e m = e k } Since e m = e k for m = k, we have the two inequalities d k k and c k k and Note that etc So we have c 1 = c 2 = = c d1, c d1 +1 = = c dd1 +1, c 1 = = c d1 < c d1 +1 = = c dd1 +1 < c d d = We introduce the numbers e i, C i, D i as follows Define the set of distinct numbers {e i } such that {e i} = {e j } and e 1 < e 2 < Let l N be the size of {e i } So e 1 = e 1, e 2 = e d 1 +1,, e l = e n Now define D i = max{m : e m = e i} for 1 i l and C i = min{m : e m = e i} for 1 i l Note that C 1 = 1 and D l = n For convenience, we also define C l+1 = n + 1 Theorem 26 ([11], Theorem 41) The abelian grou H = / e 1 / en has n n n Aut(H ) = ( d k k 1 ) ( e j ) n d j ( ei 1 ) n ci+1 k=1 j=1 Proof Their calculation involves finding all elements of R that are invertible modulo, and comuting the distinct ways of extending such elements to automorhisms of the grou

6 64 A Thillaisundaram So, we need to count all matrices M R that are invertible modulo These M are uer block triangular matrices which may be exressed in the following three forms M = m 11 m D1 1 m D1 D 1 m C2C2 m D2 C 2 2 m Cl C l or M = 0 m Dl C l m Dl D l m 11 m 12 m 1n m d1 1 m d2 2 = m 2c2 0 m ncn m nn m 1c1 The number of such M is 0 m dnn n ( d k k 1 ), k=1 since we require linearly indeendent columns So the first ste to calculating Aut(H ) is done The second half of the comutation is to count the number of extensions of M to Aut(H ) To extend each entry m ij from m ij to a ij e i ej (if e i > e j ), or a ij e i (if e i e j ), such that a ij m ij (mod ), we have e j ways to do so for the necessary zeros (that is, when e i > e j ), as any element of e i ej e i works Similarly, there are ei 1 ways for the not necessarily zero entries (that is, when e i e j ), as any element of e i will do e i We aly Hillar and Rhea s method to M = I n n We consider all extensions of I n n to Aut() Using Lemma 21, we identify which of these elements of Aut() can be extended to Aut G (G) To this end, we rove the following Proosition 27 Let G be a finite non-abelian 2 -abelian -central -grou Suose = (G) = e 1 e 2 en where 2 e 1 e 2 e n and n N Let θ Aut() be such that:

7 Automorhisms of -Central -Grous 65 (a) θ is reresented as a matrix A R ; (b) A(mod ) I n n ; (c) a ij 0 mod e i e j +2 for i j with e i e j, and a ii 1 mod 2 Then θ can be extended to θ Aut G (G) Proof By Lemma 21, we know that θ Aut() can be extended to Aut(G) if there exist φ and χ such that condition (2) of the lemma holds Our strategy is to take φ = 1 and to construct a suitable χ Recall equation (21): µ(x, y)θ(µ(x, y) 1 ) = χ(x) 1 χ(y) 1 χ(xy) We aim to construct χ such that equation (21) is satisfied We exress as where {z 1,, z n } generates z 1 z 2 z n = e 1 e 2 en Before we rove that a general θ Aut() which satisfies (a) to (c) can be extended to θ in Aut(G), we illustrate our method by considering the following automorhism θ 0 (in its matrix reresentation): ϕ(θ 0 ) = A 0 = The automorhism θ 0 clearly satisfies our conditions (a) to (c) Writing µ(x, y) as z α 1 1 zα 2 2 zαn n for α i e i, we have that θ 0(µ(x, y)) is given by α α 1 α α 2 A 0 = which translates to α n The left-hand side of (21) is then (1 + 2 )α 1 (1 + 2 )α 2 =, (1 + 2 )α n z (1+2 )α 1 1 z (1+2 )α n n (z α 1 1 z α 2 2 z αn n ) 2 = (µ(x, y) 1 ) 2 α n

8 66 A Thillaisundaram As µ(x, y) = t(xy) 1 t(x)t(y), we see that by the 2 -abelian and the -central roerties, (22) µ(x, y) 2 = t(x) 2 t(y) 2 t(xy) 2 So setting χ 0 (x) = t(x) 2 (and θ 0 (z) = z 1+2 ) works as (21) is fulfilled Note that χ 0 is defined on Q as required Thus θ 0 can be extended to θ 0 Aut G (G) We now consider a general element θ satisfying conditions (a) to (c) We may exress θ as the matrix A below: where r i 2 and s i 1 + s 1 r 1 a 12 a 1n a s 2 r 2 A =, an 1,n a n1 a n,n s n rn e i r i, and for i j, { } 0 mod if ei < e j a ij 0 mod e i e j +2 if e i e j Recall that µ(x, y) = z α 1 1 zα 2 2 zαn n, and the left-hand side of (21) is µ(x, y)θ(µ(x, y) 1 ) The left-hand side of (21) is now (z α 1s 1 r 1 1 z αnsnrn n ) [ z (a 12α 2 ++a 1n α n) 1 ] [ z (a n1α 1 ++a n,n 1 α n 1 ) n ] Now we set u the reliminaries for constructing χ We note that t(x), as G is -central So we may write for some β i for some γ i t(x) = z β 1 1 zβn n e i Similarly we have t(y) = z γ 1 1 zγn n e i, and t(xy) = z δ 1 1 zδn n for some δ i e i Again we consider equation (22) In terms of z 1,, z n, we have We note that for each i = 1,, n, z α zn αn2 = (z β 1 1 zn βn )(z γ 1 1 zn γn = z (β 1+γ 1 +δ 1 ) 1 z (βn+γn+δn) n (23) α i 2 = (β i + γ i + δ i ) + k i e i for some k i )(z δ 1 n ) 1 z δn We construct χ, which is deendent on β i, γ i, δ i, as the comosition of the two mas below:

9 Automorhisms of -Central -Grous 67 and x t(x) = z β 1 1 zβn n, side [ (z β 1s 1 r zn βnsnrn 1 ) Note again that χ is defined on Q z β 1 1 zβn n z ( a 12 1 β 2++ a 1n β n) ] [ z ( a n1 β 1++ a n,n 1 β n 1 ) n Using (23), we check that the right-hand side of (21) matches the reviously comuted left-hand The right-hand side of (21) is χ(x) 1 χ(y) 1 χ(xy) ] ( a12 β 2++ a 1n β n) ( an1 β 1++ a n,n 1 β n 1 ) = (z β 1s 1 r zn βnsnrn 1 ) [z1 ] [zn ] ( a12 γ 2++ a 1n γ n) ( an1 γ 1++ a n,n 1 γ n 1 ) (z γ 1s 1 r zn γnsnrn 1 ) [z1 ] [zn ] ( a12 δ 2++ a 1n δ n) (z δ 1s 1 r zn δnsnrn 1 ) [z1 ] [z an1 ( δ 1++ a n,n 1 δ n 1 ) n ] [z ( a 12 (β 2+γ 2 +δ 2 )++ a 1n (β n+γ n+δ n) = (z (β 1+γ 1 +δ 1 )s 1 r z (βn+γn+δn)snrn 1 n ) ( an1 (β 1+γ 1 +δ 1 )++ a n,n 1 (β n 1 +γ n 1 +δ n 1 ) 1 ] [zn ] Substituting (23) β i + γ i + δ i = α i k i e i 1 into the above gives the following (z (α 1 k 1 e 1 1 )s 1 r z (αn knen 1 )s n rn 1 n ) [z a 12 (α 2 k 2 e 2 1 )++ a 1n (α n k n en 1 ) 1 ] [z an1 (α 1 k 1 e 1 1 )++ a n,n 1 (α n 1 k n 1 e n 1 1 ) n ] We simlify the above using the following facts: (i) o(z i ) = e i and r i 2; (ii) a ij e j 2 0 mod e i for i j So the right-hand side of (21) is now [ (z α 1s 1 r 1 1 zn αnsnrn ) z (a 12α 2 ++a 1n α n) 1 ] [ z (a n1α 1 ++a n,n 1 α n 1 ) n ] and this matches the left-hand side of (21), as required

10 68 A Thillaisundaram Now, we calculate all such matrices in R satisfying conditions (a) to (c) in Proosition 27, as these extend to distinct elements in Aut G (G) We recall the general form of a matrix in R Note the l vertical and l horizontal blocks in the matrix, corresonding to the set {e i : 1 i l} m 11 m D1 1 m D1 D 1 m C2C2 m D2 C 2 2 m Cl C l 0 m Dl C l m Dl D l For the diagonal entries we have 2 e 1 2 en = choices When e i < e j, we have ei 1 choices as any element of e i works When e i = e j and i j, we have ei 2 choices as any element of 2 e i works For each row i, we have n C i off-diagonal entries which corresond to e i e j Of these C i+1 C i 1 corresond to e i = e j and n C i corresond to e i < e j So the number of choices (groued according to the l blocks) for these entries is ( e i 1 ) (n C i+1+1)(c i+1 C i ) ( e i 2 ) (C i+1 C i 1)(C i+1 C i ) When e i > e j, there are ej 2 choices as any element of e i e j +2 e i works For each column j, there are n D j entries corresonding to e i > e j So the number of choices for these entries is ( e j 2 ) (n D j)(c j+1 C j ) = j=1 This enables us to rove the following lemma ( e j 2 ) (n C j+1+1)(c j+1 C j ) j=1 Lemma 28 Using the notation from before, for e 1 2, Aut G (G) ( e i 1 ) (n C i+1+1)(c i+1 C i ) ( e i 2 ) (n C i)(c i+1 C i ) Furthermore, the non-trivial automorhisms calculated above are all non-inner automorhisms

11 Automorhisms of -Central -Grous 69 Proof The number of extensions θ Aut G (G) from Proosition 27 is ( e i 1 ) (n C i+1+1)(c i+1 C i ) which simlifies to Therefore ( e i 2 ) (C i+1 C i 1)(C i+1 C i ) ( e i 1 ) (n C i+1+1)(c i+1 C i ) Aut G (G) ( e i 2 ) (n C i+1+1)(c i+1 C i ), ( e i 2 ) (n C i)(c i+1 C i ) ( e i 1 ) (n C i+1+1)(c i+1 C i ) ( e i 2 ) (n C i)(c i+1 C i ) It is clear that all the automorhisms θ as in Proosition 27 are non-inner, since θ acts non-trivially on It remains to show that these non-inner automorhisms have order a ower of Denote by P this finite set of non-inner automorhisms; more recisely, P = { θ Aut G (G) θ := θ satisfies (a) to (c) of Proosition 27} It is sufficient to show that P is a subgrou, as then it follows that every element of P has th ower order since P is a -grou To rove that we have a subgrou, we need to show that P is multilicatively closed That is, for θ 1, θ 2 P, the comosite θ 1 θ 2 P It is enough to consider the restriction to Aut() since P is characterized by conditions (a) to (c) on Aut() We have θ 1, θ 2 Aut() such that θ 1 = θ 1 and θ 2 = θ 2 Working with the matrix reresentations, we note that θ 1, θ 2 satisfy conditions (a) to (c) of Proosition 27 Let ϕ(θ 1 ) = (a ij ) and ϕ(θ 2 ) = (b ij ) Then ϕ(θ 1 θ 2 ) = ϕ(θ 1 )ϕ(θ 2 ) = (c ij ) where c ij = k a ikb kj It is immediate that ϕ(θ 1 ) ϕ(θ 2 ) R since R is a ring So (a) is satisfied for θ 1 θ 2 Using the exression for c ij, it is clear that (b) is satisfied for θ 1 θ 2 For (c), we consider c ij for three cases: (1) i < j, (2) i = j and (3) i > j Case (1) We have i < j and so e i e j We write c ij = a ik b kj + a ik b kj + a ik b kj k i i<k<j k j If e i < e j, we need to show that c ij 0 mod As a ij and b ij for i j, it is straightforward that c ij 0 mod If e i = e j, we need to show that c ij 0 mod 2 Again as a ij and b ij for i j, we have that c ij a ii b ij + a ij b jj mod 2 We further have that 2 a ij and 2 b ij since e i = e j So c ij 0 mod 2, as required

12 70 A Thillaisundaram Case (2) We have i = j, and so c ii = a ik b ki + a ii b ii + a ik b ki k<i k>i We need to show that c ii 1 mod 2 As before, c ii a ii b ii mod 2 Since a ii 1 mod 2 and b ii 1 mod 2, we have that c ii 1 mod 2, as required Case (3) Here i > j and hence e i e j We write c ij = a ik b kj + a ij b jj + a ik b kj + a ii b ij + a ik b kj k<j j<k<i k>i For k < j, we have e i e j +2 divides e i e k +2, which in turn divides a ik Similarly for k > i, we have e i e j +2 divides e k e j +2, which divides b kj For k = j, we have e i e j +2 divides a ij Similarly for k = i, we have e i e j +2 divides b ij For j < k < i, we have e i e k +2 divides a ik and e k e j +2 divides b kj So e i e j +4 divides a ik b kj Therefore c ij 0 mod e i e j +2 as required So (c) is satisfied for θ 1 θ 2 Thus θ 1 θ 2 P Therefore, P is a subgrou as required PROOF OF THEOREM 11 We recall that we need Out(G) to rove the theorem, we now analyse our lower bound for Out(G), as given in Lemma 28 As e 1 > 2, we have Out(G) = ( 2 ) (n C i+1+1)(c i+1 C i ) (n C i+1+1)(c i+1 C i ) = (n C i+1+1)(c i+1 C i ) (n C i+1+1)(c i+1 C i ) (C i+1 C i )(2n+1) (C 2 i+1 C2 i ) = (2n+1)(C l+1 C 1 ) (C 2 l+1 C2 1 ) = n2 3n As n 3, we have that Out(G) as required Acknowledgments (n C i)(c i+1 C i ) (n C i)(c i+1 C i ) (C i+1 C i )[2n+1 (C i+1 +C i )] I am very grateful to Rachel Camina for her time and helful comments Also thank you to Chris Brookes and to Gavin Armstrong for a thorough reading of this work This research has been made ossible through the generous suort from the Cambridge Commonwealth Trust, the Cambridge Overseas Research Scholarshi, and the Leslie Wilson Scholarshi (from Magdalene College, Cambridge)

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