Groups of Prime Power Order with Derived Subgroup of Prime Order

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1 Journal of Algebra 219, (1999) Article ID jabr , available online at on Groups of Prime Power Order with Derived Subgroup of Prime Order Simon R. Blackburn* Department of Mathematics, Royal Holloway, University of London Egham, Surrey TW20 0EX, United Kingdom Communicated by Jan Saxl Received January 18, 1999 The paper classifies (up to isomorphism) those groups of prime power order whose derived subgroups have prime order Academic Press Key Words: p-groups; small derived subgroup. 1. INTRODUCTION Let p be a fixed prime and let n be a non-negative integer. It is well known that there is a bijection between the isomorphism classes of abelian groups of order p n and the partitions of n; under this bijection, the sequence k 1 k 2 k r of integers such that r i=1 k i = n and such that 0 <k 1 k 2 k r corresponds to the isomorphism class containing the group P defined by P = C p k 1 C p k 2 C p kr This bijection classifies abelian groups of order p n in terms of a vector of integers having certain properties. Furthermore, this bijection shows that the number of isomorphism classes of abelian groups of order p n is a function of n only (so does not depend on p). The main aim of this paper is to prove similar results for a different class of groups, namely, the class of all (finite) p-groups P having derived subgroup P of order p. This class of groups arises in the study of the * The author is supported by an E.P.S.R.C. Advanced Fellowship /99 $30.00 Copyright 1999 by Academic Press All rights of reproduction in any form reserved.

2 626 simon r. blackburn breadth of p-groups (a p-group P lies in if and only if P has breadth 1; see Knoche [4]). The groups in may also be characterised as those non-abelian p-groups P such that the probability that two elements chosen uniformly at random from P commute is at least 1 ; this result follows from p the work of Rusin [7]. We classify the groups in in the sense that we provide a bijection between the set of isomorphism classes of groups of order p n in and a set n of integer vectors having certain properties. If x n, we provide an explicit presentation for the p-group P corresponding to x. In particular, this result shows that the number of groups in of order p n depends only on n (and not on p). On our way to proving this result, we classify the flags in a vector space of even dimension over any field, up to the natural action by the general symplectic group. The remainder of this paper is organised as follows. Section 2 summarises the basic results and definitions from the theory of alternating forms that we use, and proves two technical results that we require. Section 3 classifies flags up to the action by the general symplectic group. Section 4 classifies p-groups P such that P =p in the case when p is odd. Finally, Section 5 sketches the changes that need to be made to the proof when p = PRELIMINARIES: ALTERNATING FORMS This section reviews the definitions in the theory of alternating forms that we will use and proves the basic results that we need in the next section. See Cohn [1, Sect. 8.6], for example, for an introduction to the theory of alternating forms. Let be a field and let V be a vector space over. A function φ V V is an alternating form if φ is bilinear and has the property that φ x x =0 for all x V. (Note that an alternating form has the property that φ x y = φ y x for all x y V ). From now on, we regard V as having been equipped with a fixed alternating form φ. The radical of φ is the subspace R of V defined by R = v V φ v x =0 for all x V We say that φ is non-degenerate if the radical is the zero subspace. Let U be a subspace of V. We define U to be the subspace U = v V φ v u =0 for all u U We say that U is totally isotropic if U U.

3 derived subgroup of prime order 627 Let e 1 f 1 e 2 f 2 e r f r be a basis of V. We say that this basis is a symplectic basis of V if the subspaces e 1 e 2 e r and f 1 f 2 f r are both totally isotropic and if for all i j 1 2 r we have that { 0 if i j φ e i f j = 1 if i = j Every vector space V equipped with a non-degenerate form φ has a symplectic basis; in particular, the dimension of V is always even in this case. Lemma 1. Let V be a vector space of dimension 2r over a field and let φ be a non-degenerate form on V. Then every maximal totally isotropic subspace of V has dimension r. Proof. Let U be a maximal totally isotropic subspace of V. Let λ V U be the linear function such that λ v is the mapping defined by x φ x v for all x U. Suppose that λ is not surjective. Then there exists u U \ 0 that is in the kernel of λ v for all v V. By the definition of λ, this implies that φ x v =0 for all v V, and hence that u is in the radical of φ. Hence u = 0 since φ is non-degenerate. This contradiction implies that λ is surjective. Since U is totally isotropic, U ker λ. If there exists v ker λ \ U, then U v would be totally isotropic and would strictly contain U, contradicting the maximality of U. Hence ker λ = U. Thus, dim U = dim U = dim im λ =2r dim ker λ =2r dim U and hence dim U = r as required. Lemma 2. Let V be a vector space of dimension 2r over a field and let φ be a non-degenerate form on V. Let U be a maximal totally isotropic subspace of V. Then there exists a maximal isotropic subspace W such that V = U W. Proof. Let U be a maximal totally isotropic subspace of V ; then dim U = r, by Lemma 1. Let W be a subspace of V that is maximal with respect to being totally isotropic and having trivial intersection with U. To prove the lemma, it suffices to show that dim W = r. Let λ V U be defined as in Lemma 1. Then λ is surjective and ker λ = U. An element u U is in W if and only if u is in the annihilator of λ W. Since U W = 0, we have that dim λ W =dim W, thus W U has codimension dim W in U. But W has codimension at most dim W in V ; hence U + W = V. Suppose, for a contradiction, that dim W<r. Hence there exists an element v V \ U W. Since V = U + W, we may write v = u + w for

4 628 simon r. blackburn some u U and w W. Note that, by our choice of v, we have that w W. Define W = W w. Since w W, we have that W is totally isotropic. Suppose u 1 U W. Then u 1 = γw + w for some γ and some w W. If γ 0, we have a contradiction, for then w = 1 γ u 1 w U W, which contradicts our choice of v above. Hence u 1 = w, and so u 1 U W = 0. Hence u 1 = 0. Thus U W = 0. By the maximality of W, we have that W = W. But W strictly contains W, since w W. This contradiction establishes that dim W = r, as required. Lemma 3. Let V be a vector space of dimension 2r over a field and let φ be a non-degenerate form on V. Suppose that V = U W, where U and W are totally isotropic subspaces of V (of dimension r). Let e 1 e 2 e r be a basis of U. Then there exists a basis f 1 f 2 f r of W such that e 1 f 1 e 2 f 2 e r f r is a symplectic basis of V. Proof. Let ψ 1 ψ r U be the dual basis to e 1 e 2 e r, so for all i j 1 2 r, { 0 if i j e i ψ j = (1) 1 if i = j Let λ V U be the function defined in Lemma 1. We have that λ is bijective when restricted to W, since λ is surjective and W is a complement to the kernel, U, ofλ. Define the elements f 1 f 2 f r to be the unique elements of W such that λ f i =ψ i for all i 1 2 r. We have that the basis e 1 f 1 e 2 f 2 e r f r is a symplectic basis of V by (1) and the fact that U and W are totally isotropic. Corollary 4. Let V be a vector space over a field and let φ be an alternating form on V. Let R be the radical of φ. Let E be a totally isotropic subspace of V such that E R = 0 and suppose that E = k 1 i=1 E i for some subspaces E 1 E 2 E k 1 of V. Then there exist subspaces F 1 F 2 F k 1 F k of V and a subspace E k of V such that all the following properties hold. (i) V = R E E k F 1 F 2 F k. (ii) The subspaces E E k and F 1 F 2 F k are totally isotropic. (iii) For all i 1 2 k, the form φ is non-degenerate when restricted to E i F i. (iv) For all i j 1 2 k we have that E i Fj whenever i j. Proof. By passing to a complement of R containing E if necessary, we may assume that R = 0 and hence that φ is non-degenerate. Let r be defined by dim V = 2r. LetU be a maximal totally isotropic subspace containing E; then U has dimension r by Lemma 1. Let W be a

5 derived subgroup of prime order 629 maximal totally isotropic subspace of V which is a complement to U; such a subspace exists by Lemma 2. Define integers d 0 d 1 d 2 d k 1 d k by setting d i = dim E 1 E 2 E i for all i 0 1 k 1 and by setting d k = r. Choose a basis e 1 e 2 e r for U such that for all i 1 2 k 1, we have that e di 1 +1 e di is a basis for E i. By Lemma 3, there exist vectors f 1 f 2 f r W with the property that e 1 f 1 e 2 f 2 e r f r form a symplectic basis for V. For all i 1 2 k, define F i = f di 1 +1 f di. Set E k = e dk 1 +1 e dk 1 +2 e dr. It is not difficult to check, using the fact that e 1 f 1 e r f r form a symplectic basis, that the subspaces F i and E k defined in this way satisfy the conditions of the corollary. 3. FLAGS UNDER THE SYMPLECTIC GROUP Let V be a vector space over a field. Define the set F t of t-flags of V to be the set whose elements are the flags of V of the form 0 = V 0 V 1 V t V t+1 = V (2) (Note that the containments in these chains of subspaces are not necessarily proper.) Let φ be a non-degenerate alternating form on V.LetG be the general symplectic group on V (so G may be identified with the set of linear transformations of V that preserve φ up to scalar multiplication). Now, G acts naturally on V and this action induces an action on the set F t of t-flags of the form (2). The aim of this section is to specify a complete set of invariants for a t-flag (2), so that two t-flags are associated with the same set of invariants if and only if they are in the same orbit under the action of G. Let V 0 V 1 V t+1 be a t-flag of the form (2). Define the set J t by J t = i j 2 1 i j t + 1 (3) We define the type of the flag V 0 V 1 V t+1 to be a set A = α i j i j J t of real numbers indexed by the elements of J t ; we define A in the following way. For i 1 2 t + 1, define R i to be the radical of the form φ when restricted to the subspace V i.so R i = v V i φ v x =0 for all x V i For i j J t, define dim R i R j 1 +V i 1 dim R i R j +V i 1 ( α i j = dim V i dim V i 1 α k i ) α i k k i 1 i+1 k t+1 if i<j if i = j

6 630 simon r. blackburn It is clear that the action of G preserves the type A of a flag. We will show later in this section that any two t-flags of the same type are in the same orbit under G. We begin by stating and proving a proposition which implies that the elements α i j of A are always non-negative integers that sum to dim V /2. Proposition 5. Let V be a vector space over and suppose that φ is a non-degenerate antisymmetric form on V. Let V 0 V 1 V t+1 be a t-flag in V and let A = α i j i j J t be fixed. Then V 0 V 1 V t+1 is of type A if and only if there exist subspaces E i j and F i j of V for all i j J t having the following properties: (i) For all i j J t, (ii) dim E i j = dim F i j = α i j For all l 1 2 t+ 1, ( ) ( V l = E i j i j J t i l i j J t j l F i j ) (iii) For all i j J t, the form φ is non-degenerate when restricted to E i j F i j. (iv) For all i j i j J t such that i j i j, E i j Fi j. (v) The subspaces and i j J t E i j i j J t F i j are totally isotropic. (vi) Let R i be the radical of the form φ when restricted to V i. For all l l 1 2 t+ 1 such that l l, we have that l t+1 R l R l = E i j i=1 j=l +1 Proof. Suppose subspaces E i j and F i j of V exist that satisfy conditions (i) to (vi) above. Let r s J t.ifr<s, then α r s = dim E r s = dim (by (i)) (( r 1 t+1 i=1 j=i (( r 1 t+1 i=1 j=i dim E i j ) ( t+1 ) E r j j=s ) ( t+1 E i j j=s+1 ( t+1 r 1 i=1 j=i ( t+1 r 1 i=1 j=i E r j ) F i j )) F i j )) (since the subspaces E i j, F i j form a direct sum) = dim R r R s 1 +V r 1 dim R r R s +V r 1 by (ii), (vi)

7 derived subgroup of prime order 631 If r = s, then 2α r s = dim E r r F r r (( r t+1 ) ( r )) r = dim E i j F i j dim t+1 j=r+1 i=1 j=i (( r 1 t+1 i=1 j=i E i j ) i=1 j=i ( r 1 r 1 i=1 j=i r 1 dim E r j dim F i r = dim V r dim V r 1 i=1 t+1 j=r+1 F i j )) r 1 α r j α i r Thus V 0 V 1 V t+1 is a flag of type A. Conversely, let V 0 V 1 V t+1 be a flag of type A. We will choose subspaces E i j by induction on i, and the subspaces F i j by induction on j. Our inductive hypothesis will take the following form. Let k t + 1. Define H k to be the assertion that the following statements all hold. (a) For all integers i and j such that 1 i j k, For all i j J t such that i k, (b) dim E i j = dim F i j dim E i j = α i j For all l 1 2 k, ( ) ( V l = E i j i j J t i l i j J t j l i=1 F i j ) (c) For all integers i and j such that 1 i j k, the form φ is non-degenerate when restricted to E i j F i j. (d) For all i j i j J t such that i j = i j and such that i j k, E i j Fi j. (e) The subspaces i j J t i k E i j and i j J t j k F i j are both totally isotropic. (f) Let R i be the radical of the form φ when restricted to V i. For all l l 1 2 t+ 1 such that l l and such that l k, we have that R l R l = l t+1 i=1 j=l +1 E i j

8 632 simon r. blackburn Note that the assertion H 0 is trivial, and that the assertion H t+1 implies conditions (i) to (vi) of the proposition. Let k 1 2 t + 1. Assume that we have chosen subspaces E i j, where i k 1 and F i j where j k 1 such that H k 1 holds. We show that we may choose subspaces E k k E k k+1 E k t+1 and F 1 k F 2 k F k k such that H k holds. For all j k + 1 k+ 2 t+ 1, define E k j to be a complement to R k R j +V k 1 R k R j 1 in R k R j 1. The natural map from R k R j 1 onto the space R k R j 1 + V k 1 / R k R j +V k 1 is bijective when restricted to E k j (since E k j is a complement to its kernel), hence dim E k j = α k j for all j k + 1 t+ 1 (4) Since E k j R k for all j k + 1 k+ 2 t+ 1, ( ) E k k+1 + E k k+2 + +E k t+1 + E i j is totally isotropic (5) Furthermore, i j J t i k 1 E k j F r s for k + 1 j t r s k 1 (6) We claim that ( the sum i j J t \ k k i k ) ( E i j + i j J t j k 1 By H k 1 part (b), it is enough to show that the sum ( t+1 ) E k j + V k 1 j=k+1 ) F i j is direct. (7) is direct. For all j k + 1 t+ 1, let u k j E k j and suppose that not all of these elements are zero. Suppose that there exists u V k 1 such that ( t+1 ) u k j + u = 0 j=k+1 Let l be the smallest value of j such that u k l 0. By our choice of E k l,we have that u k l R k R l +V k 1. Now, for all j such that l + 1 j t + 1 we have that u k j E k j R k R j 1 R k R l. Hence, since u V k 1, u k l = u t+1 j=l+1 This contradiction establishes the claim (7). u k j R k R l +V k 1

9 derived subgroup of prime order 633 We are now in a position to establish part (f) of the assertion H k.by H k 1, it is sufficient to prove that for all l k k + 1 t+ 1, Now, R k R l = k t+1 i=1 j=l +1 E i j R k R l V k 1 = R k V k 1 R l = R k R k 1 R l = R k 1 R k R k 1 R l = R k 1 R l = k 1 t+1 i=1 l=l +1 E i j Hence, since E k j R k R j 1 R k R l for all j such that l + 1 j t + 1, to establish part (f) of H k it suffices to show that t+1 j=l +1 E k j generates R k R l modulo V k 1. This statement is trivial if l = t + 1, since R t+1 = 0. But, if t+1 j=l +2 E k j generates R k R l +1 modulo V k 1 then t+1 j=l +1 E k j generates R k R l modulo V k 1, by our choice of E k l +1. Hence, by induction on t + 1 l, the statement, and hence part (f) of H k, follows. We now choose the subspaces F i k, where 1 i k, and the subspace E k k. Define the subspace M of V k by M = E i j F i j (8) i j J t j<k Let N be the subspace of V k defined by N = M V k Note that the form φ is non-degenerate when restricted to M, byh k 1. Thus V k = M N. Note also that k t+1 i=1 j=k+1 E i j = R k N and (since M N ) the radical of the form φ restricted to N is R k. Furthermore, k 1 E i k N i=1

10 634 simon r. blackburn this direct sum is clearly disjoint from R k, by (7) and by H k part (f); moreover, H k 1 part (e) implies that this direct sum is a totally isotropic subspace. By Corollary 4 we may choose subspaces E k k and F i k (where 1 i k) ofn having the following properties: (A) N = R k k i=1 E i k k i=1 F i k. (B) The subspaces k i=1 E i k and k i=1 F i k are totally isotropic. (C) For all i 1 2 k, the form φ is non-degenerate when restricted to E i k F i k. (D) For all i i 1 2 k such that i i, we have that E i k Fi k. We now check that H k is satisfied. We have that V k = M N ( = i j J t j<k ( = i j J t i k ) ( k E i j F i j R k ) ( E i j i j J t j k F i j ) i=1 ) ( k E i k i=1 by H k part (f). Hence H k part (b) holds. In order to establish H k part (a), it is sufficient to show that and F i k ) by (8) and (A) dim E i k = dim F i k for all i 1 2 k (9) dim E k j = α k j for all j k k + 1 t+ 1 (10) Now, (B) implies that the subspaces E i k and F i k are totally isotropic, and then (C) implies that (9) holds. Furthermore, (4) implies that (10) holds whenever j k. This in turn implies, when taken together with (9) and H k part (b), that dim E k k = α k k. Hence H k part (a) holds. We have that H k part (c) follows from (C) and H k 1 part (c). To show that H k part (d) holds, it is sufficient to establish the statement that E i j F i j for all i j i j J t such that i j i j such that i j k and such that at least one of i and j is equal to k. Ifj = k, then the statement follows from (D) and the fact that N M ; thus we may assume that j k 1. The statement holds when i = k and j = k, again by (D) and the fact that N = M. When i = k and j k + 1, the statement is implied by (6). Thus H k part (d) holds.

11 derived subgroup of prime order 635 Finally, H k part (e) follows from H k 1, from (B) and from the fact that N M. Hence H k follows from H k 1, and so the proposition follows by induction on k. Corollary 6. Let V be a vector space over and let φ be a nondegenerate alternating form on V. Let V 0 V 1 V t+1 be a t-flag in V and let A = α i j i j J t be a set of integers whose sum is dim V /2. Let I A be the set of triples of integers defined by i j k I A if and only if i j J t and 1 k α i j Then the flag V 0 V t+1 is of type A if and only if there exist sets e a V a I A and f a V a I A such that (i) For all l 1 2 t+ 1, the set e i j k i j k I A i l f i j k i j k I A j l is a basis for V l. (ii) (iii) For all a a I A we have that { 1 if a = a φ e a f a = 0 otherwise. For all a a I A we have that φ e a e a =φ f a f a =0 We say that a basis e a a I A f a a I A is a standard basis for the flag V 0 V 1 V t+1 if this basis satisfies the conditions (i), (ii), and (iii) of Corollary 6. Proof. Suppose that there exist sets e a V a I A and f a V a I A having properties (i) to (iii) above. For all i j J t, we define subspaces E i j and F i j by E i j = e i j k 1 k α i j and F i j = f i j k 1 k α i j It is not difficult to check that the subspaces E i j and F i j satisfy conditions (i) to (vi) of Proposition 5, and hence V 0 V 1 V t+1 is of type A. Conversely, suppose that V 0 V 1 V t+1 is a flag of type A. Let E i j and F i j for all i j J t be subspaces which satisfy the conditions of Proposition 5. For all i j J t, let e i j 1 e i j αi j be a basis of E i j. Let f i j 1 f i j αi j be a basis for F i j such that

12 636 simon r. blackburn e i j 1 f i j 1 e i j αi j f i j αi j is a symplectic basis for E i j F i j ; this is possible by Lemma 3. It is not difficult to check that the elements in e a a I A and f a a I A satisfy conditions (i), (ii), and (iii) above, and so the corollary follows. We are now in a position to prove the main theorem of this section. Theorem 7. Let V be a vector space over a field and let φ be a nondegenerate alternating form on V. Let G be the general symplectic group acting on V. Let be the mapping between the set of orbits of t-flags of V under the action of G and the set of sets A of non-negative integers indexed by J t and summing to dim V /2 which is defined by mapping an orbit of flags to the type of any one of its representatives. Then is a bijection. Proof. From the definition of the type of a flag, it is clear that the type of a flag is preserved under the action of G. Hence the mapping is well defined. We first show that is surjective. Let A = α i j i j J t be a set of integers whose entries sum to dim V /2. Define the set I A as in the statement of Corollary 6. By indexing the elements of a symplectic basis in a suitable way, we may produce sets e a a I A and f a a I A that satisfy properties (ii) and (iii) of Corollary 6. For all l t+ 1, define the subspace V l by V l = e i j k i j k I A and i l f i j k i j k I A and j l It is clear that V 0 V 1 V t+1 is a t-flag in V. Furthermore, by Corollary 6, this flag has type A. Hence is surjective. We now show that is injective. To show this, we prove that any two t-flags in V of the same type are in the same orbit under the action of G on the t-flags of V. Let V 0 V 1 V t+1 and V 0 V 1 V t+1 be t-flags in V of type A, where A = α i j i j J t. Define the set I A as in the statement of Corollary 6. Let e a a I A f a a I A be a standard basis for the flag V 0 V 1 V t+1.let e a a I A f a a I A be a standard basis for the flag V 0 V 1 V t+1.letg be the linear transformation which maps e a to e a for all a I A and maps f a to f a for all a I A. Now, g G, by conditions (ii) and (iii) of Corollary 6. But g maps V i to V i for all i 0 1 t + 1, by condition (i) of Corollary 6. Hence the flags V 0 V 1 V t+1 and V 0 V 1 V t+1 are in the same orbit under G, asrequired. This shows that is injective, so is a bijection. Hence the theorem follows.

13 derived subgroup of prime order CLASSIFYING THE GROUPS IN WHEN p IS ODD Let p be an odd prime. This section contains the classification of groups P of order p n in the class of groups having a derived subgroup of order p. Before stating the main theorem of this section, we make the following definition. Let n be a positive integer. Define a set n to be the set of triples ρ e A which satisfy the following properties. (i) ρ = k 1 k 2 k r is a partition of some integer c where c<n and where n c is even. (i) e is an integer such that ρ has at least one part of size e. (iii) Let J t be defined as in (3). Then A = α i j i j J c+1 where the elements α i j are non-negative integers such that (a) i j J c+1 α i j = n c /2. (b) For all integers i, define m i to be the number of parts of ρ of size i. Then for all k 2 3 c+ 2, m k 1 if k e α i k + 1 if k = e + 1 α k j m 1 i k k j c+2 e 1 if k = e + 2 if k>e+ 2 m k 2 Theorem 8. Let p be a prime and let n be a positive integer. Then there is a bijection between the set of isomorphism classes of groups P of order p n and the set n defined above. We defer the proof of Theorem 8 when p = 2 to Section 5. This section aims to prove Theorem 8 when p is odd. The rest of the section is organised as follows. First, we state some basic facts about groups P. We will then explicitly define the bijection of Theorem 8 in the case when p is odd. Finally, we prove Theorem 8 in the case when p is odd. Any standard group theory that we use may be found, for example, in Gorenstein [2]. Let P have order p n. Since P is nilpotent, P P is strictly contained in P. Since P has order p, we have that P P =1. Thus P has nilpotency class 2 and P Z P. For all x y P we have that x p y = x y p = 1 (the first equality follows since P has nilpotency class 2, the second equality follows since P has order p). Hence P =P P p Z P. (Here P is the Frattini subgroup of P.) Let V = P/Z P. Then, since P Z P, V is an elementary abelian p-group, and hence we may regard V as a vector space over p.leth P be a non-identity element of P. We identify P with p by identifying the

14 638 simon r. blackburn element h i P with the element i p. Using this identification, we may define a mapping φ V V p by defining φ xz P yz P = x y for all x y P It is not difficult to check that φ is well defined and that φ is a nondegenerate alternating form on V. Note that different choices for the element h P give rise to forms φ which are scalar multiples of each other, hence φ is canonical up to scalar multiplication. Let π V Z P /Z P p be the map induced by the p-power map x x p in P. Now xy p = x p y p y x 1/2 p p 1 for all x y P. When p is odd, p divides 1 2 p p 1 and so x y 1/2 p p 1 = 1. Thus π is a homomorphism (and so may be regarded as a linear mapping) whenever p is odd. We now define the bijection of Theorem 8 when p is odd. Let P have order p n. We define P to be be a certain triple ρ e A n ;we call this triple the type of P. We define ρ e A as follows. Let Z P = C p k 1 C p k 2 C p kr where 0 <k 1 k 2 k r. Define ρ = k 1 k 2 k r So ρ is a partition of c, where Z P = p c. Note that c<n, since P is not abelian. Furthermore, n c is even, since V = P/Z P is a vector space of dimension n c which is equipped with a non-degenerate alternating form. Thus ρ satisfies property (i) of the definition of n. We define e to be the largest integer such that P Z P pe 1 We claim that Z P must have a direct factor isomorphic to C p e. For otherwise, defining Z = z Z P z p = 1, we have that P Z Z P pe 1 = Z Z P pe Z P pe which contradicts the definition of e. So the claim follows. Hence e satisfies property (ii) in the definition of n. For i 0 1 c+ 1, define Z i Z P by z Z P z pi = 1 if i<e Z i = z Z P z pe 1 P if i = e z Z P z pi 1 = 1 if i>e Note that Z e Z e+1, since for all z Z e we have that z pe = z pe 1 p P p = 1. Note also that Z e 1 Z e ; indeed, Z e 1 is the kernel of the

15 derived subgroup of prime order 639 homomorphism from Z e to P given by x x pe 1. This homomorphism is onto, since P Z P pe 1,soZ e 1 has index p in Z e. Thus we have that Z 0 Z 1 Z c+1 Define subspaces Y 0 Y 1 Y c+1 of Z P /Z P p by Y i = Z i Z P p /Z P p We claim that the subspace Y e /Y e 1 has dimension 1. Clearly, Y e /Y e 1 has dimension at most 1, since Z e 1 has index p in Z e. Suppose, for a contradiction, that Y e = Y e 1.Letz Z e \ Z e 1. Since Z e Z P p = Z e 1 Z P p we have that z = xy p for some x Z e 1 and y Z P. But then y pe = z pe 1 P \ 1 Hence P Z P pe, contradicting our choice of e. Thus Y e /Y e 1 has dimension 1, as required. For i 0 1 c+ 2, define V i V by { 0 if i = 0 V i = π 1 Y i 1 = v V π v Y i 1 if i>0 Note that, since p is odd, π is a homomorphism and hence V 0 V c+2 are subspaces of V. We define A = α i j i j J c+1 to be the type of the flag V 0 V 1 V c+2 (with respect to the form φ V V p defined above). Note that, since φ is canonical up to scalar multiplication, the type A of the flag depends only on the isomorphism class of P (and not on the choice of h P \ 1 ). Since V =p n c, we have that i j J c+1 α i j = n c /2. Thus the triple ρ e A satisfies property (iii)(a) in the definition of n. Since for all i 2 3 c+ 2, 1 j i α j i + i j c+2 α i j = dim V i dim V i 1 dim Y i 1 dim Y i 2 the triple ρ e A satisfies property (iii)(b) of the definition of n. Hence ρ e A n. Since our definition of the type ρ e A of P depends only on the isomorphism class of P, the map taking a group P to the type of P induces a well defined map from the set of all isomorphism classes of groups P of order p n to the set n (in the case when p is odd). We show that is a bijection. The next proposition implies that is injective. Before stating the proposition, we define some notation. Let ρ e A n. Suppose that ρ is a partition of c and define (for all integers i) m i to be the number of parts of

16 640 simon r. blackburn ρ of size i. LetA = α i j i j J c+1. We define, for i 2 c+ 2, the integer r i by m i 1 if i e 1 if i = e + 1 r i = m e 1 if i = e + 2 if i>e+ 2 m i 2 For all i 2 3 c+ 2, we define the non-negative integer d i by d i = r i α j i α i j 1 j i i j c+2 Finally, for all i 1 2 c+ 2, we define { i 1 if i e + 1 s i = i 2 if i e + 2 So we have that c+2 r i s i = c i=2 Suppose that P has type ρ e A. Then for all i 2 3 c+ 2, every element z of Z i 1 has order at most p s i, and if z has order strictly less than p s i, then z Zi 2. Note also that and that r i = dim Y i 1 dim Y i 2 d i = dim Y i 1 dim Y i 2 + im π Y i 1 Proposition 9. Let p be an odd prime and let n be a positive integer. Let P have order p n and let ρ e A n. Define the integers c, α i j, d i and s i as above and define the set I A as in Corollary 6. Then P has type ρ e A if and only if there exists a generating set X for P, where X is the set x a a I A y a a I A z i k 2 i c k d i with the property that X together with the following relations forms a presentation for P, x ps i +1 i j k = 1 for all i j k I A (11) y ps j +1 i j k = 1 for all i j k I A (12) z ps i i k = 1 for 2 i c k d i (13)

17 derived subgroup of prime order 641 x p a x =1 for all x X and for all a I A (14) y p a x =1 for all x X and for all a I A (15) z i k x =1 for all x X where 2 i c i d i (16) x a x a =1 for all a a I A (17) y a y a =1 for all a a I A (18) x a y a =1 for all a a I A where a a (19) x a y a =b pe 1 for all a I A (20) where b = z e+1 1 x p e+1 j 1 y p j e+1 1 (We remark that since e+1 d e+1 + α j e+1 + j=1 if d e+1 = 1 if α e+1 j = 1 where j>e+ 1 if α j e+1 = 1 where j<e+ 1 c+2 j=e+1 α e+1 j = r e+1 = 1 precisely one of the three alternatives occurs in the definition of b, and the value of j occurring in the last two of these alternatives is uniquely defined.) Proof. Suppose that P has type ρ e A. We first show that the group P has a collection of generators that satisfies the relations (11) to (20) above. We then show that this collection of generators and relations is a presentation for P. Let h P \ 1. Define the subgroups Z 0 Z 1 Z c+1 of Z P, the subspaces Y 0 Y 1 Y c+1 of Z P /Z P p, the quotient V, the alternating form φ, and the subspaces V 0 V 1 V c+2 of V as in the definition of the type of P, above. Let e a a I A f a a I A be a standard basis for the flag V 0 V 1 V c+2. For all i 2 c + 2, we have that Y i 1 / Y i 2 + im π Y i 1 has dimension d i.letg i 1 g i di be elements of Y i 1 whose images in Y i 1 / Y i 2 + im π Y i 1 form a basis. Property (i) of Corollary 6 implies that the set e 1 j k 1 j k I A f 1 1 k 1 1 k I A is a basis for ker π = V 1. Moreover, for all l 2 3 c+ 2, Property (i) of Corollary 6 implies that π e l j k l j k I A π f i l k i l k I A

18 642 simon r. blackburn is a basis for Y l 1 im π +Y l 2 /Y l 2. Hence, for all l 1 2 the set π e i j k i j k I A 2 i l π f i j k i j k I A 2 j l g i k 2 i l 1 k d i is a basis for Y l 1. We now choose our generating set X for P as follows. Let i j k I A. Since π e i j k Y i 1, there exists a representative x for π e i j k in Z i 1. We define x i j k P to be a representative for e i j k such that x p i j k = x; such an element exists since as we run over all the representatives y P of e i j k, we have that y p runs through all the representatives in Z P of π e i j k. For all i j k I A, define y i j k P to be a representative of f i j k such that y p i j k Z j 1; a similar argument to that given above establishes that such a representative exists. Finally, for all i and k such that 2 i c + 2 and 1 k d i, let z i k P be a representative for g i k such that z i k Z i 1 ; such an element exists since g i k Y i 1. For all i 1 2 c + 2 and all y Z i 1 we have that y ps i = 1. Hence, since the elements x i j k, y j i k, and z i k have been chosen so that x p i j k yp j i k z i k Z i 1, we have that the relations (11), (12), and (13) hold. Since P p Z P, we have that the relations (14) and (15) hold. The relations (16) hold as z i k Z i 1 Z P for all integers i and k such that 2 i c + 2 and 1 k d i. Since the elements x a and y a are representatives of the elements e a and f a, we have (by the definition of φ) that the relations (17), (18), and (19) hold, and that x a y a =h for all a I A Suppose that d e+1 = 1 and set b = z e+1 1. Then bz P p generates Z e Z P p modulo Z e 1 Z P p. In particular, b Z e 1,sob has order exactly p e. Hence, b pe 1 is non-trivial. But, by definition of Z e, b pe 1 P. Hence b = h t for some t 1 2 p 1. Letw 1 2 p 1 be such that tw = 1 mod p. Replacing z e+1 1 with z w e+1 1 if necessary, we find that the relations (11) to (20) still hold and in addition that b pe 1 = z pe 1 e+1 1 = h = x a y a for all a I A Thus the relation (20) also holds in this case. Now suppose that d e+1 = 0. Suppose that there exists a (unique) integer j such that e + 1 <j c + 2 and such that α e+1 j = 1. Set b = x p e+1 j 1.As before, we may argue that bz P p generates Z e modulo Z e 1 and hence

19 derived subgroup of prime order 643 that b pe 1 = h t for some t 1 2 p 1. Letw 1 2 p 1 be such that tw = 1 mod p. Replacing x e+1 j 1 by x w e+1 j 1 and replacing y e+1 j 1 by y t e+1 j 1, we find that the relations (11) to (19) are still satisfied, but now the relation (20) holds also. A similar argument shows that we may find a set X of elements satisfying relations (11) to (20) in the case when there exists a (unique) integer j such that 1 j<e+ 1 and such that α j e+1 = 1. We now show that the set X and the relations (11) to (20) form a presentation of P. The set x p a a I A y p a a I A z i k 2 i c k d i generates Z P since the set π e a a I A π f a a I A g i k 2 i c k d i is a basis for Z P /Z P p. Also, the set x a a I A y a a I A generates P modulo Z P, since e a a I A f a a I A is a basis for V. Thus the set X generates P. Suppose P 2 is a group generated by a set X satisfying the relations (11) to (20). We show that P 2 has order at most p n. Consider the subgroup H of P 2 generated by the set x p i j k i j k I A i 2 y p i j k i j k I A i 2 z i k 2 i c k d i (21) The relations (14), (15), and (16) imply that H is central in P 2, and so in particular that H is normal and abelian. By (11), (12), and (13), the elements x p i j k have order at most ps p i, the elements y i j k have order at most p s j, and the elements z i k have order at most p s i, so the order of H is at most p raised to the power s i + c+2 d i c+2 ( c+2 ) i s j + s i = s i α i j + α j i + d i i j k I A i j k I A i=2 k=1 i=2 j=i j=1 c+2 = s i r i i=2 = c Now consider the quotient P 2 /H; this quotient is generated by the set x a a I A y a a I A. The relations (17), (18), (19), and (20) together imply that P 2 /H is abelian. Since x p a y p a H for all a I A, we have that P 2 /H is elementary abelian. Hence P 2 /H p 2 I A = p n c. Thus P 2 = P 2 /H H p n c+c = p n

20 644 simon r. blackburn Hence the generating set X together with the relations (11) to (20) form a presentation for P, as required. Now suppose that P is a group of order p n possessing a presentation given in the statement of the proposition. We show that P and that P has type ρ e A. Consider the group H Z P generated by the set X defined by (21). Since P has order p n and since P/H has order at most p n c, H has order p c. This implies that H is the direct product of the cyclic subgroups generated by the elements in X, that the elements x p i j k have order ps i, that the elements y p i j k have order ps j and that the elements z i j have order p s i. One consequence is that H = Cp k 1 C p k 2 C p kr, where ρ = k 1 k 2 k r. Another consequence is that the element b pe 1 has order p; the commutator relations imply that P = b pe 1, and hence P =p. Thus P. Let ρ e A n be the type of P. We need to show that ρ e A = ρ e A. Now, the relations (17) to (20) imply that the alternating form φ on P/H induced by taking commutators is non-degenerate. Thus H = Z P. Hence ρ = ρ. We have that P = b pe 1 Z P pe 1 However, since b generates a direct factor of Z P of order p e, we have that b pe 1 Z P pe and hence P Z P pe. Thus e = e. Define the subgroups Z i, the subspaces Y i, and the subspaces V i as in the definition of the type of a group in. It is not difficult to check that for all l 1 2 c+ 2, Y l 1 has the set x p i j k Z P p i j k I a 2 i l y p i j k Z P p i j k I a 2 j l z i k Z P p 2 i l 1 k d i as a basis, and that for all l 0 1 c+ 2 the set x i j k Z P i j k I a 1 i l y i j k Z P i j k I a 1 j l is a basis for V l. But now the relations (17) to (20) imply that the flag V 0 V 1 V c+2 is of type A, by Corollary 6. Hence A = A, and so P has type ρ e A as required. We now prove a proposition which implies that the mapping of Theorem 8 is surjective.

21 derived subgroup of prime order 645 Proposition 10. Let p be a prime and let n be a positive integer. Let ρ e A be a fixed element of n. Define the integers c, α i j, d i, and s i as above. Then there exists a group P of order p n that is generated by the set X of Proposition 9 and has the property that the relations (11) to (20) are satisfied. Proof. We construct the group P as follows. Let U X be defined by U = x a a I A y a a I A We define a fixed linear ordering on the set U in some arbitrary way. Let F U be the relatively free group on the set U in the variety defined by the laws x p y, x y z, and x y p.letl be the central subgroup of F U generated by the set u p u U. Define T = F U ; since the laws x y p and x y z hold in, T is elementary abelian and is generated by the set u v u v U u<v. Define Z = TL. Now, F U /Z is the relatively free group on the set U in the variety of abelian groups of exponent p; thus F U /Z is an elementary abelian p-group of rank U. We have that F U /T is the free abelian group on the set U. The image of the set u p u U generates a free abelian group of rank U in this quotient; thus L is torsion-free and L is a free abelian group of rank U. Since all the elements of T have finite order, T L = 1 and so Z = T L. Now, F U /Z p is the relatively free group on the set U in the Higman variety H of p-groups; here H is the variety defined by the laws x p y, x y z, x y p, and x p2. Higman [3] shows that the subgroup D of this relatively free group, which is generated by the set u p u U u v u v U u<v, is elementary abelian of rank 1 U U +1. But D is equal 2 to Z/Z p, hence Z has rank 1 U U +1. This implies that T has rank U U 1 and that the set u v u v U u<v is a generating set of minimal size for T. Let i I A be fixed. Let M 1 be the subgroup of T generated by the set x i y i x a y a 1 a I A \ i x a y a a a I A a a x a x a a a I A x a <x a y a y a a a I A y a <y a (22) Since the set x a y a a a I A x a x a x a <x a y a y a y a <y a forms a minimal generating set for T, we have that the generating set (22) for M 1 has minimal size, and hence that M 1 has index p in T.

22 646 simon r. blackburn Let M 2 be the subgroup of L generated by the set x ps i +1 i j k i j k I A y psj +1 i j k i j k I A (23) Since L is a free abelian group freely generated by the set u p u U, we have that M 2 has index p raised to the power c+2 ( c+2 s i + s j = s i α i j + i j k I A i=1 j=i ) i c+2 α j i = s i r i d i (24) in L. Let D be the group given by the presentation consisting of the set j=1 i=2 z i k 2 i c k d i of generators together with the relations z i k z i k =1 for 2 i i c k d i 1 k d i (25) z ps i i k = 1 for 2 i c k d i (26) Then D is abelian and has order equal to p raised to the power Define b F U D by z e+1 1 b = x p e+1 j 1 We have that y p j e+1 1 c+2 d i s i (27) i=2 if d e+1 = 1 if α e+1 j = 1 where j>e+ 1 if α j e+1 = 1 where j<e+ 1 ( b p e 1 x i y i 1) p = b p e M 1 M 2 but that b pe 1 x i y i 1 Z F U D \M 1 M 2. Hence the subgroup N generated by M 1 M 2 and b pe 1 x i y i 1 is normal in F U D, and M 1 M 2 has index p in N. Let P be the group defined by F U D /N. We identify the elements x i j k, y i j k, z i k, and b with their images in the quotient by N. It is clear that P is generated by the set X of Proposition 9. Relations (11) and (12) are satisfied by the definition of M 2 ; relations (13) are satisfied by (26). Relations (14) and (15) hold since F U is in the variety and the relations (16) hold because D is abelian and because P is a quotient of F U D. The relations (17), (18), and (19) hold by the definition of M 1. The relations (20) hold by definition of M 1 and of N.

23 derived subgroup of prime order 647 Finally, we need to show that P has order p n. Now, since F U /Z is elementary abelian of rank U, F U /Z =p U = p 2 I A = p n c. We also have that the index of M 1 M 2 in Z is p to the power c s i r i d i i=2 by (24) and the fact that M 1 has index p in T. Hence, by (27), the order of F U / M 1 M 2 D is p to the power c+2 c+2 n c s i r i d i + s i d i = n + 1 i=2 Therefore, since M 1 M 2 has index p in N, P has order p n as required. Proof of Theorem 8 When p Is Odd. Define a map from the set of all isomorphism classes of groups in of order p n and the set n by defining the image of an isomorphism class to be the type of any one of its representatives. The map is injective since, by Proposition 9, any two groups in of order p n of the same type have an identical presentation, and hence are isomorphic. We show that is also surjective as follows. Let ρ e A n. By Proposition 10, there exists a group P of order p n having a presentation of the form described in Proposition 9. But now Proposition 9 implies that P and that P has type ρ e A. Thus is surjective, and hence bijective as required. i=2 5. CLASSIFYING THE GROUPS IN WHEN p = 2 This section sketches the proof of Theorem 8 in the case when p = 2. The definition of the type of a group P given in the previous section is not valid when p = 2, because the map π V Z P /Z P p is no longer necessarily linear; in particular, ker π is not always a subspace of V. We must therefore modify our definition of the type of a group P in the case when p = 2. We begin the section by dividing the groups P into four families. We then give a definition of the type of a group P which is valid when p = 2; the definition depends on which of the four families P belongs to. Let n be a positive integer and let P have order 2 n. Define the partition ρ, the integer e, the subgroups Z i Z P, the subspaces Y i Z P /Z P p, the quotient V, and the map π as in the case when p is odd. For all i 2 3 c+ 2, define the subsets V i V by V i = v V π v Y i 1

24 648 simon r. blackburn Now P Z P 2 Y 1, and so xy 2 = x 2 y 2 modulo Y 1. Hence V i is a subspace of V for all i 2 c+ 2. Define V 0 = 0. Define the subset K V by K = ker π = v V π v =0 Let φ be the non-degenerate alternating form induced on V by taking commutators. For all i 2 3 c+ 2, define R i to be the subspace of V corresponding to the radical of the form φ restricted to V i. We have that 1 = R 2 R c+2 R 2 R c+1 R 2 R 3 R 2 V 2 Let B = β i j 1 i j c + 1 be the type of the flag V 0 V 2 V 3 V c+2. Then for all j 2 3 c+ 1 we have that dim R 2 R j dim R 2 R j+1 =β 1 j and dim V 2 dim R 2 = 2β 1 1. Recall that a standard basis e a a I A f a a I B for V 0 V 2 V c+2 satisfies the following properties. (i) For all l 2 3 c+ 2 the set e i j k i j k I B i l 1 f i j k i j k I B j l 1 is a basis for V l. (ii) For all a a I B, we have that (iii) { 1 if a = a φ e a f a = 0 otherwise. For all a a I B we have that φ e a e a =φ f a f a =0 We assign P to one of four families (I) to (IV) depending on which of the four corresponding conditions (I) to (IV) below are satisfied. (I) P Z P 2. (II) P Z P 2. Furthermore, there exists a standard basis e a a I B f a a I B for V 0 V 2 V c+2 with the property that e 1 j k 1 j k I B K f 1 1 k 1 1 k I B K (28) (III) P Z P 2. Furthermore, β and there exists a standard basis e a a I B f a a I B for V 0 V 2 V c+2 with the property that e 1 j k 1 j k I B \ K f 1 1 k 1 1 k I B \ K e f K (29)

25 derived subgroup of prime order 649 (IV) P Z P 2. Furthermore, there exists a standard basis e a a I B f a a I B for V 0 V 2 V c+2 and an integer r 2 3 c+ 1 with the property that β 1 r 1 and e 1 j k 1 j k I B \ 1 r 1 K f 1 1 k 1 1 k I B K e 1 r 1 K (30) We remark that Families (II), (III), and (IV) have analogies in the case when p is odd. Family (II) corresponds to the case when e = 1 and V 1 = V 2. Family (III) corresponds to the case when e = 1 and R 1 strictly contains R 2. Family (IV) corresponds to the case when e = 1 and R 2 strictly contains R 1. The following lemma verifies that the families (I) to (IV) do indeed partition the set. Lemma 11. Let P have order 2 n. Then P is contained in precisely one of the families (I) to (IV). Moreover, if P is contained in family (IV), then there is a unique choice for the integer r 2 c+ 1 such that there exists a standard basis for V 0 V 2 V c+2 satisfying (30). Proof. Let P have order 2 n. Define the integer e, the vector space V, the subspaces V 0 V 2 V c+2, the set K, the form φ, and the map π as above. We first show that P is contained in at least one of the families (I) to (IV). If P Z P 2, then P is contained in Family (I). Thus it is sufficient to consider the case when P Z P 2. The fact that xy 2 = x 2 y 2 y x implies that π is a homomorphism when restricted to any subspace of V, where the form φ is trivial. Furthermore since P =2 this fact also implies that for all v 1 v 2 V 2 such that φ v 1 v 2 =0 we have that v 1 + v 2 K if and only if precisely one of the elements v 1 and v 2 is an element of K. Let B = β i j 1 i j c + 1 be the type of the flag V 0 V 2 V c+2. Let e a a I B f a a I B be a standard basis for this flag. Suppose that e 1 j k 1 j k I B j 2 K (31) We show that P is contained in Family (II) or Family (III). For all k 1 2 β 1 1 such that e 1 1 k K and f 1 1 k K, we replace f 1 1 k by f 1 1 k + e 1 1 k. For all k 1 2 β 1 1 such that e 1 1 k K and f 1 1 k K, we replace e 1 1 k by e 1 1 k + f 1 1 k. These alterations result in a standard basis for the flag V 0 V 2 V c+2

26 650 simon r. blackburn such that for all k 1 2 β 1 1 either e 1 1 k f 1 1 k K or e 1 1 k f 1 1 k K. Suppose that there exist l l 1 2 β 1 1 such that l l and such that e 1 1 l f 1 1 l e 1 1 l f 1 1 l K. We replace e 1 1 l by e 1 1 l + f 1 1 l + e 1 1 l, replace f 1 1 l by f 1 1 l + e 1 1 l, replace e 1 1 l by e 1 1 l + f 1 1 l + e 1 1 l, and replace f 1 1 l by f 1 1 l + e 1 1 l. This revised standard basis for V 0 V 2 V c+2 has the property that e 1 1 l f 1 1 l e 1 1 l f 1 1 l K. This revision therefore reduces the number of integers k such that e 1 1 k f 1 1 k K by 2. Repeating as often as is necessary, we may assume without loss of generality that there exists at most one element k 1 2 β 1 1 such that e 1 1 k f 1 1 k K. If no such k exists, we have [by (31)] that our basis satisfies (28) and hence that P lies in Family (II). If one such k exists, clearly β 1 1 1; furthermore, we may assume that k = 1 by swapping the elements of the pairs e e 1 1 k and f f 1 1 k if necessary. But now this basis satisfies (29), and hence P lies in Family (III). Suppose now that e 1 j k 1 j k I B j 2 K We show that P lies in Family (IV). Let r 2 c+ 1 be the largest integer such that e 1 r l K for some l. (Thus, e 1 j k 1 j k I B j r + 1 K and β 1 r 1). By exchanging the elements of the pairs e 1 r l e 1 r 1 and f 1 r l f 1 r 1 if necessary, we may assume without loss of generality that e 1 r 1 K. Suppose that there exist integers j and k such that 1 j k I B \ 1 r 1, such that j r and such that e 1 j k K. Then we replace e 1 j k by e 1 j k + e 1 r 1 and replace f 1 r 1 by f 1 r 1 + f 1 j k to produce a standard basis for V 0 V 2 V c+2 with the property that e 1 j k K. By repeating this process as often as necessary, we obtain a standard basis e a a I B e a a I B for V 0 V 2 V c+2 with the property that e 1 j k 1 j k I B \ 1 r 1 j r K For all k 1 2 c+ 2 such that f 1 1 k K we replace f 1 1 k by f 1 1 k + e 1 1 k ; this process produces a standard basis for V 0 V 2 V c+2 with the additional property that f 1 1 k 1 1 k I B K Hence we have obtained a basis satisfying (30), and so P lies in Family (IV). We have shown that a group P is contained in at least one of the families (I) to (IV). We now show that these families are disjoint.

27 derived subgroup of prime order 651 Clearly Family (I) is disjoint from the remaining families, since a group P in Family (I) has the property that P Z P 2, whereas a group P in Family (II), (III), or (IV) has the property that P Z P 2. Define R 2 to be the radical of φ restricted to V 2. Since the form φ is trivial when restricted to R 2, the map π is a homomorphism when restricted to R 2.IfP is in Family (II) or (III), the equations (28) and (29) imply that e 1 j k 1 j k I B j 2 K Since this set is a basis for R 2 and π is a homomorphism on R 2, we have that R 2 K when P is contained in Family (II) or Family (III). However, if P lies in Family (IV), the conditions (30) imply that the element e 1 r 1 K; but e 1 r 1 R 2 and so π restricted to R 2 is not trivial in this case. Thus Family (IV) is disjoint from Families (II) and (III). A simple counting argument (using, for example, induction on β 1 1 ) shows that for any group in Family (II) we have that V 2 \ K has order β β c+1 j=2 β 1 j whereas for any group in Family (III) we have that V 2 \ K has order β β c+1 j=2 β 1 j Thus, since any group in Family (III) must have the property that β 1 1 1, we have that Family (II) and Family (III) are disjoint. Finally, we show that there is at most one choice for the element r in the definition of Family (IV). Fix r to be one such choice. For i 2 3 c + 2, define R i to be the radical of φ restricted to V i. Now, R 2 R r+1 is generated by the set e 1 j k 1 j k I B j r + 1 and this set is contained in K by (30). Hence, since π is a homomorphism when restricted to R 2, R 2 R r+1 K. Since R 2 R j R 2 R r+1 for all j r + 1, we have that R 2 R j K for all j r + 1. However, R 2 R r K, since e 1 r 1 R 2 R r and e 1 r 1 K. Hence r is defined uniquely as the greatest integer such that R 2 R r K. We now define four subsets of n : I n II n = III n IV n = ρ e A n e 1 { ρ e A n e = 1 and c+2 2 α 2 j + j=2 = ρ e A n e = 1 and α 1 2 = 1 i=1 } α i 2 = 0 = ρ e A n e = 1 and α 2 r+1 = 1 for some r

28 652 simon r. blackburn Since c+2 j=2 α 2 j + 2 i=1 α i 2 1 whenever e = 1, it is not difficult to check that the four sets defined above form a partition of n. Let P have order 2 n. We define the type of P to be an element ρ e A n as follows. We define the partition ρ and the integer e exactly as in the case when p is odd. Our definition of A depends on the family to which P belongs. Suppose that P is in Family (I). In this case, we define A exactly as in the case when p is odd; this definition makes sense since P Z P 2, which implies that the mapping π is linear. Note that P Z P 2 if and only if e 1. Hence the type of P is an element of n I. Suppose that P does not lie in Family (I). We say that we are in case (II), (III), or (IV), depending on whether P is in Family (II), (III), or (IV). When P lies in Family (IV), let r be such that the conditions (30) hold. Since P Z P 2, we have that e = 1. Let B = β i j 1 i j c + 1 be the type of the flag V 0 V 2 V c+2 defined above. We define A = α i j 1 i j c + 2 as follows. For all i and j such that 3 i j c + 2, we define α i j = β i 1 j 1 We define, for all j 2 3 c+ 2, 0 in case (II) or (III), α 2 j = 0 in case (IV), if j r in case (IV), if j = r + 1 In case (II), we define β 1 1 if j = 1 α 1 j = 0 if j = 2 if j 3 β 1 j 1 In case (III), we define β if j = 1 α 1 j = 1 if j = 2 if j 3 β 1 j 1 (Note that in case (III) we must have β 1 1 1, and so the integers α i j are all non-negative.) Finally, in case (IV), we define β 1 1 if j = 1 0 if j = 2 α 1 j = β 1 j 1 if j 3 and j r + 1 β 1 j 1 1 if j = r + 1