Contemporary Mathematics Volume 00, 0000

The Stable Homotopy Type of Rank Two p-Groups

JILL DIETZ AND STEWART PRIDDY

Abstract. Complete p-local stable splittings are given for the classifying spaces of rank 2 p-groups for p an odd prime greater than three. This is accomplished using the existing theory of splittings and computing the relevant outer automorphism groups modulo a normal p-subgroup.

0. Introduction and Statement of Main Results

Let p be an odd prime. In this paper we determine the stable homotopy type of the classifying spaces of rank 2 p-groups, i.e., those whose largest elementary abelian subgroup has dimension 2. If p > 3 these groups have been classified by N. Blackburn into three types: metacyclic, C(r), and G(r, e) of orders p^{r}. (See [B] or [Hu,] where the classification is given under the stronger assumption that every abelian normal subgroup is generated by at most two elements.) The classifying spaces of metacyclic p-groups were studied by the first author [D1, D2]; further results for p = 2 are given in [M]. Our results concern the latter two types for all p > 3. Presentations are given by

C(r) = ⟨a, b, c | a^{p} = b^{p} = c^{p^{r−2}} = 1, [a, b] = c^{p^{r−3}}, c ∈ Z(C)⟩

where r ≥ 3, [a, b] = a^{−1}b^{−1}ab, Z(C) is the center and

G(r, e) = ⟨a, b, c | a^{p} = b^{p} = c^{p^{r−2}} = [b, c] = 1, [a, b^{−1}] = c^{ep^{r−3}}, [a, c] = b⟩

where r ≥ 4 and e ≠ 0 is a quadratic nonresidue mod p.

Stable splittings for p-groups have proved useful in classifying the stable type of BG for all groups G with Sylow p-subgroup P (see [MP3], [M], [D1], [D2]).

\begin{matrix}c \\ ◯ \end{matrix}0000 American Mathematical Society 0000-0000/00 $1.00 + $.25 per page

1991 Mathematics Subject Classification. Primary 55R35; Secondary 20J06, 55P42.

This paper is in final form and no version of it will be submitted for publication elsewhere.

The second author is partially supported by NSF Grant DMS-880067.

In many cases the possible cohomology rings of such G have been determined. We plan to return to this topic in a later paper.

Here we will show BC(r) has p(p − 1) original summands, that is, summands corresponding to the p(p − 1) simple F_{p}Out(C(r)) modules. These summands are denoted by X_{q,k}(r) for 0 ≤ q ≤ p − 1, 0 ≤ k ≤ p − 2. Further, BG(r, e) has 2(p − 1) original summands X_{i}(r, e), 1 ≤ i ≤ 2(p − 1), corresponding to the 2(p − 1) simple F_{p}Out(G(r, e)) modules.

We recall that B(Z/p × Z/p) has original summands L(2, k) corresponding to the simple F_{p}GL_{2}(F_{p}) modules St_{2} ⊗ (det_{2})^{k} where St_{2} is the Steinberg module of dimension p, det_{2} is the determinant representation, and 0 ≤ k ≤ p − 2 [HK, MiP]. In the case of Z/p, we have the classical splitting B(Z/p) = ⋁_{0≤k≤p−2} L(1, k) where L(1, 0) = BΣ(p).

We can now state our main results.

Theorem 0.1. For primes p > 3, complete p-local stable splittings of BC(r) are given by

(1) If r > 3,

BC(r) = ⋁_{q,k}(q + 1)X_{q,k}(r) ∨ ⋁_{k} 2L(1, k)

where 0 ≤ q ≤ p − 1, 0 ≤ k ≤ p − 2.

(2)

BC(3) = ⋁_{q,k}(q + 1)X_{q,k}(3) ∨ ⋁_{k}(p + 1)L(2, k) ∨ ⋁_{k} 2L(1, k) ∨ (p − 1)L(1, 0)

where 0 ≤ q ≤ p − 1, 0 ≤ k ≤ p − 2.

Theorem 0.2. For primes p > 3, complete p-local stable splittings of BG(r, e) are given by

(1) If r > 4,

BG(r, e) = ⋁_{i} X_{i}(r, e) ∨ ⋁_{k} X_{p−1,k}(r − 1) ∨ ⋁_{k} L(1, k)

where 0 ≤ i ≤ 2(p − 1), 0 ≤ k ≤ p − 2.

(2)

BG(4, e) = ⋁_{i} X_{i}(4, e) ∨ ⋁_{k} X_{p−1,k}(3) ∨ ⋁_{k} L(2, k) ∨ ⋁_{k} L(1, k)

where 1 ≤ i ≤ 2(p − 1), 0 ≤ k ≤ p − 2.

We remark that part (2) of Theorem 0.1 corrects an error in Example 5.C.1 of [MP1].

Information on the cohomology of these groups can be found in [Y] although we do not need any of these results.

The paper is organized as follows: In Section One we gather some group theoretic preliminaries and information on the automorphisms of C(r) and G(r, e). Representation theory is used in Section Two to describe the original summands of BC(r) and BG(r, e). Finally in Section Three we determine the remaining (non-original) summands completing the proofs of Theorems 0.1 and 0.2.

1. Outer Automorphism Groups of C(r) and G(r,e)

In this section we present some preliminaires on C(r), G(r, e) and their outer automorphisms groups. Since our splittings involve idempotents constructed from these automorphism groups, it is only necessary to determine them modulo a normal p-subgroup.

We begin by describing some useful group theoretic properties.

Lemma 1.1. Let C = C(r). Then

(1) C is a regular p-group of class 2.

(2) Γ_{2}(C) = ⟨c^{p^{r−3}}⟩.

(3) Commutators are determined by [a^{i}b^{j}, a^{s}b^{t}] = [a, b]^{it−sj}.

Proof. (3) follows from the usual commutator identities [Hu; III, §1]. Clearly (3) implies (2). Since p is odd and c is central, (2) implies (1). □

Lemma 1.2. Let G = G(r, e). Then

(1) G is a regular p-group of class 3.

(2) Γ_{2}(G) = ⟨b, c^{q}⟩, Γ_{3}(G) =< c^{q} >= Z(C), where q = p^{r−3}.

(3) Commutators are determined by [a^{i}b^{j}c^{k}, a^{s}b^{t}c^{u}] = b^{iu−sk}c^{n} where n = qe(u\binom{i}{2} + js − it − k\binom{s}{2}).

Proof. As in Lemma 1.1 (3), straightforward calculations with commutator identities yield (3) which in turn implies (2). Thus, G is of class 3 and, hence, regular if p > 3. Since ⟨b, c⟩ is an abelian normal subgroup of G [Hu; III, 10.9(b)] implies G is regular for p = 3. □

The usual method for studying the automorphisms of a group G is to consider the induced automorphisms on the Frattini quotient, i.e. on G/[G, G] ⊗ Z/p. This works well for G(r, e) and is employed below. However, by Lemma 1.1, the Frattini quotient of C(r) has rank 3 which is larger than the rank of C(r) itself. A more efficient procedure uses the following result. We thank the referee for the proof which is more elegant than our own.

Proposition 1.3. If C is a characteristic, central, cyclic subgroup of a pgroup P such that C ∩ [P, P] ≠ 1 then the natural map

ρ : Out(P) ⟶ Out(P/C)

has a p-group as kernel.

Proof. Let α ∈ Aut(P) represent an element in the kernel of ρ. Adjusting α by an inner automorphism we may assume α acts trivially on P/C. Since C is central and characteristic, the commutator map

P/C × P/C → [P, P]

shows α acts trivially on [P, P]. Because C is cyclic and C ∩ [P, P] ≠ 1 we conclude α acts as a p-power automorphism on C, and hence acts trivially on CΦ(P)/Φ(P). By hypothesis it also acts trivially on P/CΦ(P), so considering the extension

1 → CΦ(P)/Φ(P) → P/Φ(P) → P/CΦ(P) → 1

it acts as an element of order p on P/Φ(P). Therefore α acts as a p-power automorphism on P. □

It is easy to see that P = C(r), C = ⟨c⟩ ≈ Z/p^{r−2} satisfies these hypotheses. Moreover, P/⟨c⟩ ≈ (Z/p)^{2}⟨a, b⟩ is elementary abelian of rank 2. Thus Out(P/⟨c⟩) = GL_{2}(F_{p}) and we have

Proposition 1.4. ρ : Out(C(r)) ⟶ GL_{2}(F_{p}) is surjective with kernel a p-group.

Proof. We define automorphisms τ, χ, and β whose images generate GL_{2}(F_{p}). On generators let τ(a) = b, τ(b) = a, τ(c) = c^{−1}; χ(a) = ab, χ(b) = b, χ(c) = c; β(a) = a^{i}, where i is a multiplicative generator of Z/p, β(b) = b, β(c) = c^{i}. Let z = c^{p^{r−3}} then using the identity (ab)^{j} = a^{j}b^{j}z^{\binom{j}{2}}, it is straightforward to verify that these define automorphisms of C(r). Since

ρ(τ) = ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ) , ρ(χ) = ( \begin{matrix}1 & 1 \\ 0 & 1 \end{matrix} ) , ρ(β) = ( \begin{matrix}i & 0 \\ 0 & 1 \end{matrix} ) ,

ρ is surjective. □

Corollary 1.5. ρ induces a one-to-one correspondence

{simple F_{p}Out(C(r)) modules} ⟷ {simple F_{p}GL_{2}(F_{p}) modules}

Proof. If H is a normal p-subgroup of G then ker{F_{p}G → F_{p}(G/H)} is a nilpotent ideal [HK; Lemma 4.9].

We now turn to the outer automorphisms of G = G(r, e). Let P be a p-group and let Φ(P) be its Frattini subgroup. Since Φ(P) is characteristic we have an exact sequence

1 → K → Out(P) →^{ρ} Out(P/Φ(P))

where K = ker(ρ). By a well known theorem of P. Hall [Hu], K is a p-group. We apply this to G setting Φ = Φ(G). Then

G/Φ = G/[G, G] ⊗ Z/p = ⟨\overline{a}, \overline{c}⟩

an elementary abelian p group of rank two and we have

Proposition 1.6. ρ : Out(G(r, e)) → GL_{2}(F_{p}) maps onto the subgroup Z/2 × Z/(p − 1) of diagonal matrices. The kernel of ρ is a p-group.

Proof.

Let α ∈ Out(G). Then α(a) = a^{i}b^{j}c^{k} for some i, j, k. By Lemma 1.2, G is regular and Γ_{2}(G)^{p} = 1, hence 1 = α(a^{p}) = a^{ip}b^{jp}c^{kp}. Thus c^{kp} = 1 and so p^{r−3}|k. Therefore we have α(a) = a^{i}b^{j}c^{ql}. Write α(c) = a^{s}b^{t}c^{u}. Then α(b) = α([a, c]) = [a^{i}b^{j}c^{k}, a^{s}b^{t}c^{u}] = [a^{i}b^{j}c^{ql}, a^{s}b^{t}c^{u}] = [a^{i}b^{j}, a^{s}b^{t}c^{u}] since c^{q} is central. By Lemma 1.2, α(b) = b^{iu}c^{mq} for some m. Thus 1 = α([b, c]) = [b^{ui}c^{mq}, a^{s}b^{t}c^{u}] = c^{q(uis)} and so p|isu. If p|u or p|i, it is easy to see that α is not an automorphism. Therefore p|s and we conclude α(c) = b^{t}c^{u} where p ∤ u.

A restriction on i is obtained using the relation [a, b] = c^{−qe}:

α([a, b]) = [a^{i}b^{j}c^{ql}, b^{ui}c^{mq}] = c^{−qeui^{2}}
α(c^{−qe}) = (b^{t}c^{u})^{−qe}

Therefore c^{uqe(1−i^{2})} = 1 and so p|(i^{2} − 1), i.e. i ≡ ±1. Modulo Φ we have

α(\overline{a}) = \overline{a}^{i} i = ±1
α(\overline{c}) = \overline{c}^{u} u ≠ 0

Thus Imρ ≤ Z/2 × Z/(p − 1).

It remains to show ρ is surjective. Let α_{i}, i = 1, 2 be the automorphisms defined by α_{1}(a) = a, α_{1}(c) = c^{u} and α_{2}(a) = a^{−1}, α_{2}(c) = c^{u}. Then

ρ(α_{1}) = ( \begin{matrix}1 & 0 \\ 0 & u \end{matrix} )

ρ(α_{2}) = ( \begin{matrix}−1 & 0 \\ 0 & u \end{matrix} )

Thus ρ maps onto Z/2 × Z/(p − 1) □

Corollary 1.7. ρ induces a one-to-one correspondence

{simple F_{p}Out(G(r, e)) modules} ⟷ {simple F_{p}(Z/2×Z/(p−1)) modules}.

2. Original Summands

The theory of original summands was introduced by Nishida who used the name dominant summands. The reader is referred to [N, MP2]. We recall the basic idea. Let P be a p-group. Every stable summand X of BP is a summand of BQ for some subgroup Q ≤ P, which is unique up isomorphism subject to the condition that X is not a summand of the classifying space of any smaller subgroup of P. Such an X is said to originate in or be an original summand of BQ. It follows from the solution of the Segal conjecture [LMM] that such X’s are in one-to-one correspondence with the simple F_{p}Out(Q) modules. Further, their multiplicity is the dimension of the corresponding module if F_{p} is a splitting field for Out(Q).

Simple F_{p}G = F_{p}GL_{2}(F_{p}) modules have been classified by Glover [G]. Let V be the natural two dimensional representation of G and let det_{2} : G → F_{p} be the determinant representation. Then there are p(p − 1) simple F_{p}G modules given by

M_{q,k} = V^{⊗q} ⊗ (det_{2})^{k}

for 0 ≤ q ≤ p − 1, 0 ≤ k ≤ p − 2. By convention V^{0} is the trivial one dimensional module. The Steinberg module St_{2} = V^{⊗(p−1)}. Moreover F_{p} is a splitting field for G and dim_{F_{p}}M_{q,k} = q + 1.

Using the notation of the Introduction we have

Theorem 2.1. Stably, BC(r) has p(p−1) distinct homotopy types of original summands, X_{q,k}(r), corresponding to the p(p − 1) simple F_{p}Out(C(r)) modules M_{q,k}. The multiplicity of X_{q,k}(r) is q + 1.

Proof. Corollary 1.5. □

Since Z/2 × Z/(p − 1) has 2(p − 1) simple 1-dimensional modules over F_{p} we have

Theorem 2.2. Stably, BG(r, e) has 2(p − 1) distinct homotopy types of original summands, X_{i}(r, e), corresponding to the 2(p − 1) simple F_{p}Out(G(r, e)) modules. The multiplicity of X_{i}(r, e) is 1.

Proof. Corollary 1.7. □

3. Non-original summands

In this section we complete the proofs of Theorems 0.1 and 0.2 by determining summands which originate in the classifying spaces of proper subgroups of C(r) and G(r, e). We will use the concept of linkage which we now recall (see [MP1],

[MP2] for more details).

Let Q be a p-group and Y a summand of BQ which originates in BR, R ≤ Q. We say Y is linked in BQ if any equivalence of the form

Y → BQ →^{tr} BR → Y

must have R = Q (in which case tr = id). In this case it is necessary that R be a retract of Q [MP2; Prop. 2.9]. Additionally we say Y is linked to X in BQ if X is an original summand of BQ such that Y is a summand of eBQ where e ∈ Z_{p}^{∧}Out(Q) < {BQ, BQ} is a primitive idempotent corresponding to X.

We will often use the following key fact about linkage.

Proposition 3.1 [MP2, Prop.2.8]. Let Q be a p-subgroup of a finite group G and suppose Q is not a subretract of a Sylow p-subgroup of G. If Y is a summand which originates in BR, R ≨ Q, and is linked to a summand X which originates in BQ, then the multiplicity of X in BG is a lower bound for the multiplicity of Y in BG.

\underline{C(r):}

Proposition 3.2. No summand of BC(r), r ≥ 3, originates in the classifying space of a non-abelian subgroup of p-rank two.

Proof. Let Q be a non-abelian, proper subgroup of C = C(r), necessarily of p-rank two. By the classification theorem [B] of such groups for p > 3, Q is either metacyclic, C(s), or G(s + 1, e). If Q is metacyclic, then [D1; Prop. 2.11] shows Out(Q) ≤ B_{2}, the Borel subgroup of GL_{2}(F_{p}). As in the proof of Proposition 1.6, we see that the same is true when Q is of the form G(s, e), s < r. The p-rank of Q equals that of C so Q is not a subretract of C. Thus, by [D1; Prop. 3.18], no summand of BC originates in BQ when Q is either of these two types.

Let Q be of the form C(s), 3 ≤ s < r. If X is a summand of BC originating in BQ, then there is a sequence of normal subgroups

C = K_{0} ▷ K_{1} ▷ ⋯ ▷ K_{t} = Q

where each K_{i} is of index p in K_{i−1}, and X is a summand of each K_{i} [M]. Each K_{i} must be non-abelian of p-rank 2, and by [D1; Note 3.14], we may assume that C_{K_{i−1}}(K_{i}), the centralizer of K_{i} in K_{i−1}, equals the center of K_{i}.

K_{t−1} cannot be metacyclic since [D1] shows that BK_{t−1} does not have summands originating in non-abelian subgroups.

If K_{t−1} is of the form C(s + 1) then C_{K_{t−1}}(Q) = Z(Q) implies

Z/p^{s−1} ≅ Z(K_{t−1}) ≤ C_{K_{t−1}}(Q) = Z(Q) ≅ Z/p^{s−2}

which is impossible.

If K_{t−1} is of the form G(s + 1, e) then K_{t−2} cannot be metacyclic since subgroups of metacyclic groups are always metacyclic. If K_{t−2} = C(s + 2) then

Z/p × Z/p^{s−2} ≅ Φ(K_{t−1}) ≤ Φ(K_{t−2}) ≅ Z/p^{s−1}.

Thus, K_{t−2} must be of the form G(s + 2, e′). The argument repeats, giving K_{0} = G(s + t, e′′). This contradiction shows that no summand of BC originates in the classifying space of a rank 2, non-abelian subgroup. □

Proposition 3.3. Summands of BC(r), r ≥ 3, originate in the classifying space of an abelian subgroup Q of p-rank two if and only if r = 3 and Q = Z/p × Z/p. In this case, the summands L(2, k), 0 ≤ k ≤ p − 2, are the only summands originating in B(Z/p × Z/p) which are also summands of BC(3). The multiplicity of L(2, k) in BC(3) is p + 1.

Proof. Let Q ≅ Z/p^{i} × Z/p^{j} be a subgroup of C = C(r). If i ≠ j, then no dominant summand of BQ is a summand of BC by [D1; Prop. 3.16]. Let i = j and, as above, assume C_{C}(Q) = Z(Q) = Q. Since Z(C) ≅ Z/p^{r−2}, we must have r − 2 ≤ i. We also know that 2i < r since |Q| < |C|. Thus, no summands of BC originate in BQ unless r = 3 and Q ≅ Z/p × Z/p. It follows that Q is of index p and is thus normal in C = C(3). Therefore N_{C}(Q)/Q ≅ Z/p.

Let X be an original summand of BQ corresponding to M_{q,k}. We claim the multiplicity of X in BC is p+1 if q = p−1 and 0 otherwise. Let {Q_{1}, Q_{2}, ..., Q_{n}} be a complete set of representatives for the conjugacy classes of subgroups of C isomorphic to Q. Since each Q_{i} has rank 2, Q_{i} ≥ Z(C) = Φ(C) and so the number of such conjugacy classes equals the number of lines in C/Φ(C) ≅< \overline{a}, \overline{b} >, i.e. n = p + 1.

According to [MP1; Cor. 4.3], the multiplicity of X in BC is equal to

∑_{i=1}^{n} dim_{F_{p}} \overline{W_{i}}M_{q,k}

where

\overline{W}_{i} = ∑_{x∈N_{C}(Q_{i})/Q_{i}} c_{x}.

Since the Sylow p-subgroup of GL_{2}(F_{p}) is cyclic of order p we may assume the

action of a generator of N_{C}(Q_{i})/Q_{i} is given by ( \begin{matrix}1 & 1 \\ 0 & 1 \end{matrix} ). Thus the calculation

of [D1; Prop. 3.16] carries over to show dim_{F_{p}} \overline{W_{i}}M_{q,k} = 1 if q = p − 1 and 0 otherwise. □

It remains to consider summands coming from cyclic subgroups.

Proposition 3.4. BC(r) contains BZ/p ∨ BZ/p. If r > 3, BC(r) contains no other summands originating in Z/p. In all cases BC(r) contains no summands originating in cyclic groups of higher order.

Proof. First we observe that C = C(r) has two non-conjugate retractions onto Z/p and none onto cyclic groups of higher order.

By Lemma 1.1, C/Γ_{2}(C) = (Z/p)^{2} < \overline{a}, \overline{b} > ×Z/p^{r−3} < \overline{c} >. Consider the retractions

ρ_{a} : C → C/Γ_{2}(C) → Z/p < \overline{a} >
ρ_{b} : C → C/Γ_{2}(C) → Z/p < \overline{b} > .

If x = a^{i}b^{j}c^{k} ∈ C then x^{−1}ax = a[a, x] = a[a, b]^{j} = ac^{jp^{r−3}}. Thus a is not conjugate to b and so ρ_{a} is not conjugate to ρ_{b}. Thus B(C) contains BZ/p∨BZ/p via these retractions. No other summands come by retraction since C/Φ(C) = Z/p × Z/p.

Given a retraction ρ : C → Z/p^{s}, let N = ker(ρ). If x = a^{i}b^{j}c^{k} then [x, b] = c^{ip^{r−3}}. Hence N ≥ Γ_{2}(C) and so ρ factors through C/Γ_{2}(C) = (Z/p)^{2} < \overline{a}, \overline{b} > ×Z/p^{r−3} < \overline{c} >. Since c has order p^{r−2} this implies s = 1.

For r > 3 this exhausts the possibilities since any other summands coming from cyclic subgroups must be linked to a summand originating in a rank 2 subgroup but this is impossible by Propositions 3.2-3. □

Proposition 3.5. BC(3) contains BZ/p∨BZ/p∨(p−1)L(1, 0) and no other summands originating in Z/p.

Proof. First we determine if any summands of BZ/p are linked to summands originating in proper subgroups of C(3). By Proposition 3.3 there are possibly other summands linked to L(2, k). However by [HK] this happens only for L(2, 0) in which case L(1, 0) is linked. By Proposition 3.4 it remains to show that the multiplicity of L(1, 0) in BC(3) is p + 1. There are at least p + 1 copies of L(1, 0) by Proposition 3.1. There are at most p + 1 copies of L(1, 0) in BC(3) since by Lemma 1.1 there are p + 1 conjugacy classes of (non-central) cyclic subgroups of order p which gives an upper bound by [MP1; Remark 2(1)]. (Since C(3) is extra-special the center is in the kernel of any projection of C(3) or one of its rank two subgroups onto a Z/p and so the center does not contribute any summands.) Thus there are exactly p + 1 copies of L(1, 0) in BC(3). □

This completes the proof of Theorem 0.1.

\underline{G(r,e):}

Proposition 3.6. Summands of BG(r, e), r ≥ 4, originating in the classifying space of a non-abelian subgroup of p-rank two must originate in BC(r − 1). In this case, the summands X_{p−1,k}(r), 0 ≤ k ≤ p − 2, are the only original summands of BC(r − 1) which are also summands of BG(r, e). The multiplicity of X_{p−1,k}(r) is one.

Proof. Using the ideas and notation in the proof of Proposition 3.2, we see that no summand of BG(r, e) originates in BQ when Q is metacyclic or of the form G(s, e′). We also know that if Q = C(s), s < r, then each K_{i} must be of the form G(s′, e′). But, as was noted above, no summand of BG(r, e) can originate in BG(s′, e′). Thus, we need only consider the case when C(s) is normal and of index p in G = G(r, e).

To fix notation let

Q = ⟨x, y, z|x^{p} = y^{p} = z^{r−3} = 1, z ∈ Z(Q), [x, y] = z^{p^{r−4}}⟩

and let a, b, c be the generators of G = G(r, e) as in the Introduction. We can find exponents so that x = a^{i}b^{j}c^{k} and y = a^{s}b^{t}c^{u}. By Lemma 1.2,

[c, y] = [c, a^{s}b^{t}c^{u}] = b^{−s}c^{−qe\binom{s}{2}}.

We must have c^{p} ∈ Q, thus Q normal in G implies b^{−s} ∈ Q. Similarly, b^{−i} ∈ Q.

If both i and s are divisible by p, then Q is abelian. Assume i ≢ 0 mod p, then b ∈ Q. Since G is regular, Γ_{2}(G)^{p} = 1 implies 1 = x^{p} = a^{ip}b^{jp}c^{kp} so that k ≡ 0 mod p. We see that c^{k} ∈ Q, thus, a^{i} ∈ Q. In fact, Q = ⟨a, b, c^{p}⟩.

In this case, W_{G}(Q) = N_{G}(Q)/Q = ⟨cQ⟩ ≅ Z/p. We include W_{G}(Q) in Out(Q) by the homomorphism defined by cQ ↦ c_{c}, where c_{c}(·) = c^{−1}(·)c. Computing conjugation by c gives c_{c}(a) = ab and c_{c}(b) = b, and

ρ(c_{c}) = ( \begin{matrix}1 & 1 \\ 0 & 1 \end{matrix} ) .

Thus, ρ(W_{G}(Q)) = U_{2}, the unipotent subgroup of GL_{2}(F)_{p}. The proof of [D1, Prop. 3.17] shows that the summands of BQ corresponding to the Steinberg modules, M_{p−1,k}, 0 ≤ k ≤ p − 2, are the only original summands of BQ which are also summands of BG. The above analysis shows Q is unique; thus the multiplicity of X_{p−1,k} in BG is one [MP1; Cor. 4.6]. □

Proposition 3.7. BG(r, e), r > 4, does not have summands originating in the classifying spaces of abelian subgroups of p-rank two.

Proof. Let Q ≅ Z/p^{i} × Z/p^{j}. As in Proposition 3.3, if i ≠ j, then no dominant summand of BQ is a summand of BG(r, e). Let i = j and assume that C_{G}(Q) = Q. Then Z(G) ≤ C_{G}(Q) implies r − 3 ≤ i. Since Q is a proper subgroup of G, we also have 2i ≤ r − 1. Thus, r = 5. It is easy to see that Q cannot be a subgroup of G(5, e) when i = 2, and when i = 1, no such Q is self-centralizing. □

Proposition 3.8. L(2, k), 0 ≤ k ≤ p−2, are the only summands of BG(4, e) originating in the classifying space of an abelian subgroup of p-rank two. The multiplicity of L(2, k) is one.

Proof. Let Q ≅ Z/p^{i} × Z/p^{j} be a subgroup of G = G(4, e). As in the proof of Proposition 3.3, no original summand of BQ is a summand of BG unless i = j. Since G is non-abelian of order p^{4} the only possibility is i = j = 1. Since the rank of G is two, Q is not a retract and Q ≥ Z(G) =< c^{p} >. Therefore it is easy to see that Q =< c^{p}, a^{i}b^{j} > for some i, j. If i ≠ 0, Q is self-centralizing. Further c^{−1}ac = ab implies all such Q′s are conjugate. Therefore we may assume Q =< c^{p}, a > in which case W = N_{G}(Q)/Q =< \overline{b} > acting via b^{−1}ab = ab. Thus as in the proof of Proposition 3.3, we have dim_{F_{p}} \overline{W} M_{q,k} = 1 if q = p − 1 and 0 otherwise. By [MP1, Cor.4.6] the multiplicity of the corresponding summand L(2, k) is one. □

Proposition 3.9. BG(4, e) contains exactly one copy of BZ/p and no summands originating in cyclic groups of higher order.

Proof. Let X be a summand originating in a cyclic group D ≤ G. Then X is linked to a subgroup Q ≤ G which properly contains D. By Proposition 3.6 if Q ≤ G is non-abelian then X must be linked in BC(3) to X_{p−1,k} for some k, contradicting Theorem 0.1. If Q ≤ G is abelian then then Q ≅ Z/p × Z/p and X is linked to L(2, k) for some k by Proposition 3.8. X = L(1, 0) is the only possibility and its multiplicity is one since there is one conjugacy class of such Q. Finally we consider the possibility that X is linked in BG itself which, by definition, occurs only via retraction. By Lemma 1.1, G/Γ_{2}(G) ≅ Z/p < \overline{a} > ×Z/p^{q} < \overline{c} >. Thus arguing as in the proof of Proposition 3.4 we see that G has a retraction onto Z/p but none onto a cyclic subgroup of higher order. Therefore, there is exactly one copy of X = L(1, k) for 0 ≤ k ≤ p − 2. □

This completes the proof of Theorem 0.2.

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Department of Mathematics, University of Washington, Seattle, WA 98195 E-mail address: dietz@math.washington.edu

Department of Mathematics, Northwestern University, Evanston, Il 60208 E-mail address: s priddy@math.nwu.edu