RANK THREE p-GROUPS AND FREE ACTIONS ON THE HOMOTOPY
PRODUCTS OF THREE SPHERES
MICHAEL A. JACKSON
Abstract. A classic result of Swan states that a finite group G acts freely on a homotopy
sphere if and only if every abelian subgroup of G. This has led to the question of which finite
groups act on the homotopy product of n spheres. In this paper we will show that every rank
three p-group for an odd prime p acts on a homotopy product of three spheres. In order to
demonstrate these free actions we develop a new obstruction theory for cosimplicial monoids.
This new theory allows for obstructions in torsion obstruction groups to be multiplied away
by taking multiples of the element wanted.
1. Introduction
Recall that the p-rank of a finite group G, rkp (G), is the largest dimension of an elementary
abelian p-subgroup of G as a vector space over Fp . In addition recall that the rank of a finite
group G, rk(G), is the maximum of rkp (G) taken over all primes p. We can define the homotopy rank of a finite group G, h(G), to be the minimal integer k such that G acts freely on a
finite CW-complex Y ' Sn1 × Sn2 × · · · × Snk . There is a conjecture of Benson and Carlson
[4] which states that for any finite group G, rk(G) = h(G). The case of this conjecture when
rk(G) = 1 is a direct result of Swan’s theorem [15]. This conjecture has also been verified for
rank two p-groups and rank two finite simple groups, excluding P SL3 (Fp ) for odd primes p,
by Adem and Smith [1, 2]. Subsequently the author has shown that if G is a rank two group
that is Qd(p)-free for each odd prime p, then h(G) = 2 [12].
In this paper we will verify this same conjecture when G is a rank three p-group for p an odd
prime. Our main theorem is the following:
Theorem 1. Let G be a finite p-group for some odd prime p. If rk(G) = 3, then h(G) = 3.
Heller [9] has shown that if h(G) = 2, then rk(G) = 2; therefore we only need to demonstrate
a free action of G on a finite complex homotopic to the product of three spheres. A portion
of Theorem 1 is then an immediate result of the following theorem of Adem and Smith.
Theorem 2 (Adem and Smith [2, Theorem 4.11]). Let G denote a p-group such that r =
rk(G) ≤ rk(Z(G)) + 1. Then G acts freely on a finite complex Y ' SN × S2[G:Z(G)]−1 )r−1 . In
particular, h(G) ≤ r.
In light of Theorem 2, the main theorem follows from the following proposition:
1
2
MICHAEL A. JACKSON
Proposition 3. Let G be a finite p-group for some odd prime p. If rk(G) = 3 and rk(Z(G)) =
1, then h(G) = 3.
2. isotropy groups, actions and obstruction
The central theorem which has allowed for the recent progress in groups acting on homotopy
products of spheres is due to Adem and Smith.
Theorem 4 (Adem and Smith [2]). Let G be a finite group and let X be a finitely dominated,
simply connected G-CW complex such that every isotropy subgroup has rank one. Then for
some integer N > 0 there exists finite CW-complex Y ' SN × X and a free action of G on
Y such that the projection Y → X is G-equivariant.
Theorem 4 can be applied in the case of a rank three p-group G, by showing that G acts on
N
a finite CW complex X ' SN
1 × S2 where Ni > 1, such that the isotropy groups of the action
are rank one.
Throughout this paper, G is a finite p-group for an odd prime P , such that rk(G) = 3
and rk(Z(G)) = 1. Thus there is a unique central subgroup of order p which we will call
Z. Let χ be the complex character of Z such that χ(1) = p − 1 and χ(z) = −1 for each
other element of Z. Now let φ = IndG
Z χ. The complex character φ of G corresponds to
a unitary representation V of G which gives an action of G on the representation sphere
|G|
S(V ) ' S2( p (p−1))−1 . The isotropy groups of the given action on S(V ) will be exactly the
subgroups H ⊂ G such that H ∩ Z(G) = {1}. (See [2].)
We may view the given action of G on S(V ) as a fibration:
S(V ) → S(V ) ×G EG → BG.
In order to find an action of G on a finite CW-complex X ' S2(
complete the following diagram which begins a tower:
SN ' X̄ −−−→
S(V )
Ē


y
−−−→ S(V ) ×G EG


y
BG.
So we need to create a fibration:
SN ' X̄ → Ē → S(V ) ×G EG.
|G|
(p−1))−1
p
× SN we aim to
RANK THREE p-GROUPS AND FREE ACTIONS
3
In creating this fibration, we will use the notation and the results of Jackowski and Slominska
[11]. Let S(G) be the G-poset of all G subgroups. Let W be a poset constructed by the cellular
decomposition of the G-CW complex S(V ), making S(V ) ' |W |. We make an isotropy
presheaf on W , which is a morphism of G-posets d : W → S(V ) taking each w ∈ W to the
isotropy group Gw . We now construct a small category Wd from the isotropy presheaf d, such
that the objects of Wd are the same as those of W and a morphism w → w0 are equivalence
classes modulo d(w0 ) of elements of g ∈ G such that w ≤ gw0 . Let α : S(G) → Spaces be
the functor taking P to G/P and αd = α ◦ d. Then Jackowski and Slominska have shown
that the map
hocolimWd EG ×G αd → EG ×G |W | ' EG ×G S(V )
is a homotopy equivalence. Notice that EG ×G αd (w) ' BGw and so the homotopy colimit
hocolimWd EG ×G αd will be of the homotopy type hocolimP ∈I(G) BP , where I(G) is the
G-poset of all isotropy subgroups of the action of G on S(V ).
Notice that I(G) as a collection is closed under conjugation in G. Let mathcalOI be the
I(G)-orbit category, wich is the category with objects G/P for P inI(G) and with G-maps
as morphisms. Let i be the inclusion functor from mathcalOI to the category of G − spaces.
Composing i with the Borel construction (−)hG give a functor βI : mathcalOI → Spaces,
where βI (G/P ) has the homotopy type of BP for any P ∈ mathcalI(G). Compose the functor βI with with the functor Hom(−, BU (n)∧p ) to get a functor Dd,n : OI → Spaces, which
takes G/P ∈ OI to a space with the homotopy type of Hom(BP, BU (n)∧p ). We now define
the map ψ : [EG ×G S(V ), BU (n)∧p ] → lim←P ∈I(G) [BP, BU (n)∧p ] given by the obvious restrictions once you pass to the space [hocolimP ∈I(G) BP, BU (n)∧p ] which is homotopy equivalent to
[EG ×G S(V ), BU (n)∧p ]. The obstructions to to an element of lim←P ∈I(G) [BP, BU (n)∧p ] lying
in the image of ψ lie in limi+1 πi Dd,n .
To see where the obstructions lie, we will need to apply the cosimplicial replacement functor
Q∗
Q
Q∗
πt Dd,n for each t ≥ 0 and lims πt Dd,n ∼
. There are natural isomorphisms πt ∗ Dd,n ∼
=
=
Q
∗
s
π
πt Dd,n for each s, t ≥ 0 [5, chapter XI]. These together give the natural isomorphism
Q∗
s
lim πt Dd,n ∼
Dd,n for each s, t ≥ 0.
= π s πt
Q
An element γ ∈ lim←P ∈I(G) [BP, BU (n)∧p ] can be thought of as an element of π0 Tot1 ∗ Dd,n
and in turn the obstructions to γ being in the image of the map ψ lie in the cohomotopy
groups π i+1 πi Π∗ Dd,n ∼
= π i+1 Π∗ πi Dd,n for i ≥ 1.
Recall that for any representation ρ : P → U (m) with P a p-group,
πn (Dd,n (w), Bρ) ∼
= πn (BCU (m) (Im(ρ))) ⊗ Z∧p (see [7]).
4
MICHAEL A. JACKSON
Since the centralizer of a finite p-subgroup of U (m) is the product of various U (i), it is clear
that BCU (m) (Im(ρ)) is simply connected; therefore,
π 2 π1 Π∗ Dd,n = 0.
Now that we have seen that the first obstruction vanishes, the rest of the obstructions will
lie in limi+1 πi Dd,n for i ≥ 2.
To deal with these obstructions we will use a monoidal structure and the torsion of these
obstructions. We will lay out the general obstruction theory for cosimplicial monoids in the
next section and show that the given obstructions must be torsion in the subsequent section.
3. Cosimplicial monoids
In this section we will be looking at the obstruction theory for cosimplicial spaces and developing an obstruction theory for a monoid of cosimplicial spaces in which the obstructions
add under the operation of the monoid. The primary references for this section are Bousfield
[6], Bousfield and Kan [5], and Goerss and Jardine [8]. We start with a definition.
Definition 5. A cosimplicial space X is called a cosimplicial monoid if there is a set of
`
cosimplicial spaces {X(n)}n≥0 with X = r≥0 X(r) such that the following hold:
(1)
(2)
(3)
(4)
X(0) is a point,
there is a natural associative operation M : X(q) × X(r) → X(q + r),
M makes the space X into a monoid with identity X(0), and
M is a cosimplicial map by defining (X(q) × X(r))n = X(q)n × X(r)n .
Fixing a cosimplicial monoid X, we need to define several maps. Given a vertex b ∈ Tot1 X,
let a denote the projected vertex b0 ∈ Tot0 X. We will define the following two sets of maps:
aM
n
: X n → X n,
where a M n (f ) = M ((d1 )n a, f ), and
Man : X n → X n ,
where Man (f ) = M (f, (d1 )n a).
Lemma 6. The following diagrams commute for each n ≥ j ≥ 0.
M n+1
X n+1 −a−−−→ X n+1




sj y
sj y
Xn
Mn
−a−−→
Xn
M n+1
X n+1 −−a−→ X n+1




sj y
sj y
Xn
Mn
−−−a→ X n
RANK THREE p-GROUPS AND FREE ACTIONS
5
Lemma 7. The following diagrams commute for each n ≥ i > 0. However they only commute
up to homotopy if n ≥ i = 0.
M n−1
M n−1
X n−1 −a−−−→ X n−1




i
i
dy
dy
X n−1 −−a−→ X n−1




i
i
dy
dy
Mn
Mn
Xn
−a−−→
Xn
X n −−−a→ X n
These two lemmas are an easy exercise in cosimplicial spaces.
Since some of the diagrams in the two previous lemmas do not commute, we do not get cosimplicial maps a M : X → X and Ma : X → X. However, we do get a nice map of cosimplicial
groups.
For t ≥ 1, πt (X • , c) = {πt (X m , c)}m≥0 is a cosimplicial group for c ∈ Tot1 (X) if the spaces
X m have simply connected components. Since the diagrams in the previous lemmas commute
up to homotopy, the induced maps
(a M n )∗ : πn (X • , c) → πn (X • ,a M 1 ◦ c)
and
(Man )∗ : πn (X • , c) → πn (X • , Ma1 ◦ c)
give maps of cosimplicial groups (a M )∗ and (Ma )∗ . These maps induce well behaved maps
on obstruction groups.
Let us recall how the obstructions are defined. Let F 0 ∈ π0 Tot0 X and let n be fixed. Suppose
F 0 lifts to F n ∈ π0 Totn X. This gives F n : skn ∆ → X, which extends F n−1 : skn−1 ∆ → X.
Now F n gives a map F n : ∂∆n+1 → X n+1 , which is the obstruction cocycle associated
to F n . This determines [F n ] ∈ π n+1 πn (X, F n−1 (0)) [8, pp. 417-429]. If [F n ] = 0 in
π n+1 πn (X, F n−1 (0)) then F n−1 can be extended to a map F n+1 : skn+1 ∆ → X, which extends
an F̃ n : skn ∆ → X with
[F̃ n ] = [F n ] ∈ π n+1 πn (X, F n−1 (0)).
The maps of cosimplicial groups (a M )∗ and (Ma )∗ given above, in turn induce the following
maps:
(a M )] : π n+1 πn (X, F n−1 (0)) → π n+1 πn (X,a M n−1 ◦ F n−1 (0))
and
(Ma )] : π n+1 πn (X, F n−1 (0)) → π n+1 πn (X, Man−1 ◦ F n−1 (0)).
Given F i ∈ π0 Toti X, let F 0 ∈ π0 Tot0 X be the component of the projected vertex of a
representative of F i .
Theorem 8. Let X be a cosimplicial monoid with each component of X n simply connected
for each n > 1. Let F 0 , G0 ∈ π0 Tot0 X and let n be fixed. Let n ≥ 1. Suppose F 0 and
G0 have been lifted to F n , Gn ∈ π0 Totn X respectively. Let f ∈ Tot0 X be a representative
6
MICHAEL A. JACKSON
of F 0 and g ∈ Tot0 X a representative of G0 . Let K 0 ∈ π0 Tot0 X be given by M (f, g).
Let f˜n , g̃n ∈ Totn X be representatives of F n and Gn respectively, such that f, g are projected
vertices of f˜n and g̃n respectively. If K 0 has be lifted to K n ∈ π0 Totn X where K n = M (f˜n , g̃n ),
then (Mg )] [F n ] + (f M )] [Gn ] = [K n ].
Proof. Since these maps are induced maps
(Mg )] [F n ] + (f M )] [Gn ] = [Mgn+1 ◦ F n ] + [f M n+1 ◦ Gn ].
Notice that ∂∆n+1 has the same homotopy type as Sn , and Sn is an H-cospace; therefore,
∂∆n+1 is an H-cospace also. So there exists a coproduct
θ : ∂∆n+1 → ∂∆n+1 × ∂∆n+1
making the following diagram commutes up to homotopy, where k is the natural inclusion.
θ
∂∆n+1 −−−→ ∂∆n+1 ∨ ∂∆n+1


ky
∆
∂∆n+1 −−−→ ∂∆n+1 × ∂∆n+1
We would like to show that the following diagram commutes up to homotopy.
θ
(Mgn+1 ∨f M n+1 )◦(F n ∨Gn )
∆
M ◦(F n ×Gn )
∂∆n+1 −−−→ ∂∆n+1 ∨ ∂∆n+1 −−−−−−−−−−−−−−−→ X n+1 ∨ X n+1




∇
k
y
y
−−−−−−−→
∂∆n+1 −−−→ ∂∆n+1 × ∂∆n+1
n
Since the bottom is K and the top and right hand side represent
X n+1
[Mgn+1 ◦ F n ] + [f M N +1 ◦ Gn ] ∈ πn (X n+1 ),
if this diagram commutes up to homotopy, then the theorem is proven.
Define maps
ι1 , ι2 : ∂∆n+1 → ∂∆n+1 × ∂∆n+1
by ι1 (x̄) = (x̄, (0, 0, . . . , 0)) and ι2 (x̄) = ((0, 0, . . . , 0), x̄). Now by our definitions, it follows
that
M ◦ (F n × Gn ) ◦ ι1 ' Mgn+1 ◦ F n
and
M ◦ (F n × Gn ) ◦ ι2 ' f M n+1 ◦ Gn .
This implies that
∇ ◦ (Mgn+1 ∨ f M n+1 ) ◦ (F n ∨ Gn ) ' M ◦ (F n × Gn ) ◦ k.
Therefore, the diagram above commutes up to homotopy and the theorem is proven.
RANK THREE p-GROUPS AND FREE ACTIONS
7
Define φn1 : X n → X n to be the identity map and inductively define φni : X n → X n for i ≥ 2
by φni (F n ) = M (φni−1 (F n ), F n ). Since M is a cosimplicial map, these maps give a cosimplicial
map φi : X → X for each i ≥ 1. From these maps we get induce maps for each i ≥ 1
(φi )] : π n+1 πn (X, F n−1 (0)) → π n+1 πn (X, φi ◦ F n−1 (0)).
Now also define ιn1 : X n → X n to be the identity map and define using induction on i,
ι ni : X n → X n for each i ≥ 2 by ιni (F n ) = Mfnˆ (ιni−1 (F n )). As before, these maps induce
maps for each i ≥ 1
(ιi )] : π n+1 πn (X, F n−1 (0)) → π n+1 πn (X, ιi ◦ F n−1 (0)).
Using all of these new maps, we get the following corollary.
Corollary 9. Let X be a cosimplicial monoid with each component of X n simply connected
for each n > 1. Let F 0 ∈ π0 Tot0 X and suppose that F 0 has been lifted to F n ∈ π0 Totn X.
So [F n ] ∈ π n+1 πn (X, F n−1 (0)). If the two maps, Mfn and f M n , are homotopic for all n ≥ 0,
then (φk )] ([F n ]) = k((ιk )] [F n ]) for each k ≥ 1.
Corollary 10. Let X be a cosimplicial monoid with each component of X n simply connected
for each n > 1. Assume that Mfn and f M n are homotopic for all n ≥ 0. Let F 0 ∈ π0 Tot0 X
and suppose that F 0 has been lifted to F n ∈ π0 Totn X. If
[F n ] ∈ π n+1 πn (X, F n−1 (0))
is torsion (i.e. k[F n ] = 0 for some k ∈ Z>0 ), then (φk )] ([F n ]) = 0.
Proof. It is clear that k((ιk )] [F n ]) = (ιk )] (k[F n ]) since (ιk )] is a homomorphism. Since the
right hand side is 0 by hypothesis, the left hand side is also and the corollary follows.
Putting these two corollaries together, we get the following theorem.
Theorem 11. Let X be a cosimplicial monoid with each component of X n simply connected
for each n > 1. Let F 0 ∈ π0 Tot0 X and suppose that F 0 has been lifted to F n ∈ π0 Totn X.
Also suppose that [F n ] ∈ π n+1 πn (X, F n−1 (0)) is torsion (i.e. k[F n ] = 0 for some k ∈ Z>0 ).
If the two maps, Mfn and f M n , are homotopic for all n ≥ 0, then (φk )] ([F 0 ]) can be lifted to
an element of π0 Totn+1 X, which extends (φk )] ([F n−1 ]).
4. Torsion obstructions
In the previous section, we saw that for a cosimplicial monoid and an element with only
finitely many torsion obstructions, some multiple of our element would have no obstructions
at all. In this section we will give a sufficient condition for every element of an obstruction
group to be torsion. We start off with a definition of the depth of a partially ordered category.
Definition 12. For a category O with finitely many objects and a partial order, we will say
that the depth of O is
depth(O) = max{n ∈ Z|∃ xi ∈ Obj(O) with x0 < x1 < · · · < xn and xi−1 6= xi }.
8
MICHAEL A. JACKSON
For a functor F : O → P, we will say that depth(F ) = depth(O).
This allows us to show the following proposition and its corollary, which are the central points
of this section.
Proposition 13. Let O be a category with finitely many objects, a partial order and such
that for any x ∈ Obj(O), M orO (x, x) is a finite group. Let F : O → Q − mod be a functor.
If n = depth(F ), then limi F = 0 for each i > n.
Proof. The proof will use induction on the depth n. For n = 0, we note that for x, y ∈ Obj(O),
M orO (x, y) is empty if x 6= y and is a finite group if x = y. So limi F = 0 for each i > 0. Now
let n > 0 and assume the proposition is true for each k < n. Let P be the full subcategory
of O where the set of objects in P is the set
{x ∈ Obj(O)| 6 ∃ y ∈ Obj(O) with y 6= x and y < x}.
Let G : F |P → Q − mod be the functor that is F |P . It is obvious that depth(P) = 0, so
limi G = 0 for each i > 0. Let R(G) be the right Kan extension of G along the inclusion
P ,→ O. The natural map F → R(G) is an isomorphism when restricted to P.
Claim: limi R(G) = 0 for each i > 0.
Note that any finitely-generated QP-module is projective.
limi R(G) = ExtiQO (Q, R(G))
where Q is the constant functor in QO − mod. Let P• be a projective resolution of Q over
QO. We then have an exact sequence
Pn → Pn−1 → · · · → P1 → P0 → Q.
Now for P ∈ QO − mod, we will write P |P for the restriction of P to a QP-module.
This restriction preserves exact sequences. Therefore, since every QP-module is projective,
(P |P )• is a projective resolution of Q|P over QP. We get the equality ExtiQP (Q|P , G) =
ExtiQO (Q, R(G)) since G = R(G)|P . Therefore, limi G = limi R(G) and the claim is proven.
Let K = ker(F → R(G)) and L = coker(F → R(G)). There is an exact sequence of
functors into Q − mod
0 → K → F → R(G) → L → 0.
Since F → R(G) is an isomorphism when restricted to P, depth(K) ≤ n − 1 and depth(L) ≤
n − 1. This implies by the inductive hypothesis that
limi K = limi L = 0 for each i > n − 1.
Let A = ker(R(G) → L) = coker(K → F ). This gives a short exact sequence
0 → A → R(G) → L → 0,
RANK THREE p-GROUPS AND FREE ACTIONS
9
which in turn gives a long exact sequence
· · · → limi R(G) → limi A → limi+1 L → · · · .
Therefore, if i > n then limi+1 L = limi R(G) = 0 and thus, limi A = 0. We also get a short
exact sequence
0 → K → F → A → 0,
which gives rise to a long exact sequence
· · · → limi A → limi F → limi K → · · · .
If i > n > 0 then limi A = limi K = 0 and so limi F = 0.
Corollary 14. Let O be a category with finitely many objects, a partial order and such that
for any x ∈ Obj(O), M orO (x, x) is a finite group. Let D : O → Spaces be a functor such
that for each x ∈ Obj(O), π0 D(x) = 0 and π1 D(x) = 0. If n = depth(D), then every
non-zero element of limi πt D is torsion when i > n and t > 1.
Proof. Suppose this is not true. Let α ∈ limi πt D where α 6= 0 is not torsion for some i > n
and t > 1. Fix i and t. Let F be the functor given by composing πt D with the functor
− ⊗Z Q. So F : O → Q − mod. By the last proposition, limi G = 0 since i > n.
There is an obvious map of functors πt D → F which gives a map
Θ : limi πt D → limi F.
Since Θ maps into a rational vector space, it can be extended to a map
Θ̄ : (limi πt D) ⊗Z Q → limi F.
Notice that Θ̄ is a natural map and recall that
(limi πt D) ⊗ Q = (limi πt D) ⊗ Q ∼
= (π i Π∗ πt D) ⊗ Q.
Since O has only finitely many objects, (Π∗ πt D) ⊗ Q and Π∗ ((πt D) ⊗ Q) are naturally isomorphic; therefore,
π i Π∗ ((πt D) ⊗ Q) ∼
= limi (πt D ⊗ Q) = limi F = 0.
Putting all of this together with the naturality of Θ̄, we see that Θ̄ is an isomorphism. So
α ⊗ 1 = 0 implying α = 0. This is a contradiction and so the corollary is proven.
10
MICHAEL A. JACKSON
5. Obstruction theory for torsion obstructions
We now reexamine the situation from Section 2. First we construct a monoid in the following
way:
Define maps µm,n : U (n) × U (m) → U (n + m) for each pair of nonegative integers n and m
induced by the isomorphism Cm ⊕ Cn ∼
= Cm+n , where U (0) is taken to be the trivial group.
`
Let U (−) = n≥0 U (n). Applying the multiplication operation given by µm,n to the set U (−)
gives a monoid. This monoidal structure induces monoidal structure on the space BU (−)
and in turn on [X, BU (−)] and [X, BU (−)∧p ] for any space X.
Recall the map ψ : [EG ×G S(V ), BU (n)∧p ] → lim←P ∈I(G) [BP, BU (n)∧p ] from Section 2.
We can think of this a a set of maps of spaces giving the monoid homomorphism Ψ :
[EG ×G S(V ), BU (−)∧p ] → lim←P ∈I(G) [BP, BU (−)∧p ]. Again adjusting from the notation
of Section 2 we let the functor Dd : OI → Spaces take G/P ∈ OI to a space with the
homotopy type of Hom(BP, BU (−)∧p ). So in this new notation the obstructions to to an
element of lim←P ∈I(G) [BP, BU (−)∧p ] lying in the image of ψ lie in limi+1 πi Dd .
Q
As in Section 2, after applying the the cosimplicial replacement functor ∗ , an element
Q
γ ∈ lim←P ∈I(G) [BP, BU (−)∧p ] can be thought of as an element of π0 Tot1 ∗ Dd . So as before the obstructions to γ being in the image of the map Ψ lie in the cohomotopy groups
π i+1 πi Π∗ Dd ∼
= π i+1 Π∗ πi Dd for i ≥ 2, since π 2 π1 Π∗ Dd = 0.
Now we can apply the cosimplicial monoid structure of Section 3. Let X be the cosimQ
Q
plicial replacement of Dd , so X = ∗ Dd . Similarly let X(n) = ∗ Dd,n (for the functors
`
defined in Section 2). Looking at the definitions we see that X = n≥0 X(n) as a space and
that it inherits the monoidal structure of BU (−). So we now have our obstructions lying in
a cosimplicial monoid.
Proposition 15. The following diagram commutes for each k ∈ Z>0 where µk takes the k th
power in the monoid [Y, BU (−)∧p ] for any space Y .
µ
k
→ [EG ×G S(V ), BU (−)∧p ]
[EG ×G S(V ), BU (−)∧p ] −−−




Ψy
Ψy
lim [BP, BU (n)∧p ]
←P ∈I(G)
µ
k
−−−
→
lim [BP, BU (n)∧p ]
←P ∈I(G)
This diagram also induces a map on obstruction groups for each m ≥ 1:
(µk )∗ : limm+1 πm Dd → limm+1 πm Dd .
Proof. The diagram commutes by the definition of Ψ. The second statement follows from the
first.
RANK THREE p-GROUPS AND FREE ACTIONS
11
Theorem 16. Using the notation of Proposition 15, let θ ∈ limm+1 πm Dd . If k · θ = 0, then
(µk )∗ θ = 0 where (µk )∗ is given in the last proposition.
Proof. Notice that the maps Mfn and f M n , as defined in Section 3, are homotopic in our case
since conjugate inclusions into U (m) give homotopic maps on BU (m). The theorem then
follows from Proposition 15 and Corollary 10.
Before we state our theorem we need to guarantee that the obstructions that arise are torsion.
In light of Section 4, we need only show that the depth of our category is less than or equal
to two. Recall that our G-poset I(G) contains all of the isotropy subgroups of G, or more
specifically all subgroups of G which have trivial intersection with the center. We want to
replace the category I(G) by a G-homotopy equivalent poset where the depth is known.
Let Ap (I(G)) be the full G-subposet of I(G) consisting of non-trivial elementary abelian
subgroups. So Ap (I(G)) is the G-poset containing all of the non=trivial elementary abelian
isotropy subgroups of G, or more specifically all non-trivial elementary abelian subgroups of
G which have trivial intersection with the center. It is clear that since G has rank three and
the center of G is cyclic, Ap (I(G)) contains only rank two and rank one elementary abelian
subgroups. Clearly the depth of Ap (I(G)) is two.
It is only left to show that Ap (I(G)) is G-homotopy equivalent to I(G). This follows from
similar argument to Quillen who showed that for an arbitrary finite group the poset of all nontrivial p-subgroups is homotopy equivalent to the poset of all nontrivial elementary abelian
p-subgroups. (See [13, Proposition 2.1], [16, Theorem 2], and [3, 6.6.1]).
Replacing the poset I(G) with the G-homotopy equivalent poset Ap (I(G)) in the diagram
from Proposition 15 gives the following diagram:
µ
k
[EG ×G S(V ), BU (−)∧p ] −−−
→ [EG ×G S(V ), BU (−)∧p ]




Ψy
Ψy
lim
←P ∈Ap (I(G))
µ
k
[BP, BU (n)∧p ] −−−
→
lim
←P ∈Ap (I(G))
[BP, BU (n)∧p ]
??? can we use Ap (I(G)) or do we need to use another decomposition??
Theorem 17. Given an element γ ∈ lim←P ∈Ap (I(G)) [BP, BU (n)∧p ], there exists a positive
integer k such that µ̄k (γ̄) is in the image of φ̄.
(PROOF IS OLD)
Proof. Recall that a map is an F-surjection if every element of the cokernal is nilpotent.
This is equivalent to saying that every element of the range has some power that is in the
image. Let χ be a complex character of Q that respects fusion in G. We want to show that
12
MICHAEL A. JACKSON
k · χ ∈ Im(Ψ̄) for some k ∈ Z>0 . Let
m = max{n ∈ Z≥0 |pn |[NG (P ) : P ] for some P ∈ C}.
It is clear that m < ∞, and limi+1 πi DC = 0 for all i ≥ m by Corollary ??.
Assume that χ 6∈ Im(Ψ̄) (otherwise we are done). This implies that there is a first obstruction
θ to χ being in this image. θ ∈ limi+1 πi DC = 0 for some 1 < i < m. We will use induction
on this i. By Corollary 14, θ is torsion and so there exists j ∈ Z>0 such that jθ = 0.
If j · χ ∈ Im(Ψ̄), we are done. Otherwise, there is a first obstruction θj to j · χ being in this
image. By the last theorem, θj ∈ limij +1 πij DC = 0 where i < ij < m. Since θj is torsion, we
apply the same argument as above. After a finite number of steps we will be at least to m
and we are done since there will be no more obstructions.
Corollary 18. Let G be a finite p-group with rk(G) = 3 and rk(Z(G)) = 1. Let Z be
the unique central subgroup of order p and χ the complex character of Z such that χ(1) =
p − 1 and χ(z) = −1 for each other element of Z. Let V be the unitary representation of
G, corresponding to the character IndG
Z (χ). Then we have an action on the representation
sphere S(V ) such that the isotropy subgroups are all subgroups of the form H ⊂ G such
that H ∩ Z(G) = {1}. Suppose there exists a class function β : G → C such that for any
subgroup H ⊂ G with H ∩ Z(G) = {1}, β|H is a character of H and such that if H is also a
rank 2 elementary abelian subgroup [β|H , 1H ] = 1. In this situation, G acts freely on a finite
CW-complex homotopy equivalent to the product of three spheres.
6. p-groups
Throughout this paper we will assume that G is a p-group p an odd prime. In addition
we will assume that rk(G) = 3 and rk(Z(G)) = 1. Note that G has an abelian normal
subgroup of type (p, p). Call one such group Q. Let Z = Q ∩ Z(G). The subgroups Q and
Z will be important in our discussion. We will also will need the following finite p-group
n−1
n−2
M (pn ) = hx, y|xp
= y p = 1, y −1 xy = x1+p i.
Proposition 19. Let G be a finite p-group for some odd prime p such that rk(G) = 3. Let Q
be an abelian normal subgroup of type (p, p). Suppose that H ⊆ G such that H ∩ Z(G) = {1}.
Let n be such that |H| = pn . Then either H is cyclic, H ⊆ CG (Q), H is abelian of type
(p, pn−1 ), or H ∼
= M (pn ).
Proof. Notice that rk(H) ≤ 2. If rk(H) = 1 then H is a cyclic group. So we will assume for
now that rk(H) = 2. Let Z = Q ∩ Z(G). It follows that |Z| = p. Notice that |Q/Z| = p.
Therefore Q/Z ⊆ Z(G/Z) and CG/Z (Q/Z) = G/Z (see [10, Corollary 2.24] ). So we see that
|G|/p = |CG/Z (Q/Z)| ≤ |CG (Q)|.
In the case that H ∩ Q 6= 1, obviously |H ∩ Q| = p. Let q ∈ H ∩ Q such that q 6= 1.
Suppose that q 6∈ Z(H). This implies that there exists h ∈ H with h−1 qh ∈ Q. And so
hh−1 qh, qi = Q implying that Q ⊆ H which is a contradiction. So q ∈ Z(H) which implies
RANK THREE p-GROUPS AND FREE ACTIONS
13
that H ⊆ CG (Q). So we have seen that if H ∩ Q 6= {1}, then H ⊆ CG (Q).
Now we will assume that H ∩ Q = {1}. If H ⊆ CG (Q), then H × Q ⊂ G. This cannot
be the case since this would imply that rk(G) > 3. So we may assume that H 6⊆ CG (Q). Let
K = H ∩ CG (Q). So [H : K] = p, since [G : CG (Q)] = p. Also notice that K H and that
K is cyclic. (If K were not cyclic K × Q ⊆ G and so rk(G) > 3.) Since H has a maximal
cyclic subgroup, either H is abelian of type (p, pn−1 ) or H = M (pn ).
Proposition 20. Let G be a finite p-group for some odd prime p such that rk(G) = 3. There
exists a class function β : G → C such that for any subgroup H ⊂ G with H ∩ Z(G) = {1},
β|H is a complex character of H and if in addition H is a rank 2 elementary abelian subgroup
then [β|H , 1H ] = 0.
Proof. Let Q be an abelian normal subgroup of type (p, p) and let Z = Q ∩ Z(G). First we
will define the class function β of G by giving the image of β for each element of G. Let
β(1) = (p2 − p)|G|. For each z ∈ Z \ {1}, let β(z) = 0. For each q ∈ Q \ Z, let β(q) = −p|G|.
For each x ∈ CG (Q) \ Q, β(x) = 0. For each y ∈ G \ CG (Q), β(y) = −|G| if the order of y is
p and β(y) = 0 if the order of y is not p.
It is easy to see that β defined in this way gives a class function since we have defined
the following chain of subgroups each normal in G:
{1} Z Q CG (Q) G.
Now let H be any subgroup of G with H ∩ Z(G) = {1}. We will use the characterization of
such subgroups from Proposition 19 to show that in each case β|H is a complex character of H.
Case 1: H ∩ Q 6= {1}. From the proof of Proposition 19, we know that H ⊆ CG Q. Let
K = H ∩ Q, so |K| = p. Let χ be the character of the cylic group K which is p − 1 on the
p|G|
identity element and −1 for each other element of K. Then β|H = |H:K|
IndH
K χ.
Case 2: H ∩ Q = {1} and H ⊆ CG (Q). As before H is cyclic. Let χ be the character of H which is |H| on the identity element and 0 for each other element of H. Then
|G|
β|H = |H|
(p2 − p)χ.
Case 3: H ∩ Q = {1} and H is cyclic. If then β|H is zero for all non-identity elements
and we may use the same character χ as in case 2. If on the other hand H ∩ CG (Q) = {1}
then H has order p. Let χ be the character of H which is (p2 − p) on the identity element
and −1 on each other element. Then β|H = |G|χ.
n−1
Case 4: H ∩ Q = {1} and H is abelian of type (p, pn−1 ). Let H = hx, y|xp = y p
=
ipn−2
1, [x, y] = 1i. Notice that hyi = CG (Q) ∩ H.For each i, with 1 ≤ i ≤ p − 1, Hi = hxy
i. It
14
MICHAEL A. JACKSON
is clear that for each i, |Hi | = p and Hi ⊂ G \ CG (Q) ∪ {1}. Also if i 6= j, then Hi ∩ Hj = {1}.
For each i, let χi be the character of Hi which is p − 1 on the identity element and −1 on
each other element of Hi . Now let
χ=
p−1
X
IndH
Hi χi .
i=1
So we see that χ(1) = |H|(p − 1); for each z ∈ H \ hyi, χ(z) = − |H|
; and for each z ∈ hyi,
p
χ(z) = 0. So β|H =
p|G|
χ.
|H|
n−1
Case 5: H ∩ Q = {1} and H ∼
= y p = 1, y −1 xy =
= M (pn ). So we may let H = hx, y|xp
n−2
n−2
x1+p i. Let N = hxp , yi H. So we see that N ∼
= (Zp )2 . Let χ be the character of N
j
such that χ(1) = p − 1, χ(y ) = −1 for each 1 ≤ j ≤ p − 1, and χ(z) = 0 for all z ∈ N \iyh.
p|G|
Then β|H = IndH
N |H:N | χ.
So we see that for each possible type of subgroup H, β|H is a complex character of H.
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RANK THREE p-GROUPS AND FREE ACTIONS
15
Department of Mathematics, Hylan Building, University of Rochester, Rochester, NY 14627,
USA, [email protected]