PERIODIC RESOLUTIONS FOR CERTAIN FINITE GROUPS

DAVE BENSON

Abstract. Let p and q be distinct odd primes, and let

G = ⟨x, y | x^{p}y = y^{−1}x^{p}, xy = y^{2q−1}x^{−1}⟩ ≅ ℤ/pq ⋊ Q_{8}. Using this deficiency zero presentation discovered by Bernard Neumann, we investigate the beginning of a projective resolution of ℤ as a ℤG-module in the case p ≢ q (mod 4). This gives enough information to compute H^{∗}(G, M) for any ℤG-module M.

1. A family of presentations

This paper is concerned with the calculation of the beginning of a projective resolution of ℤ as a ℤG-module, where G belongs to a certain class of periodic groups. A finite group G has periodic cohomology if and only if every Sylow subgroup of G is cyclic or generalised quaternion. See for example Cartan and Eilenberg [2], Section XII.11. The classification of such groups can be found in Theorem 6.1.11 and Section 6.3 of Wolf [11], as well as on page 103 of Thomas and Wall [10].

In this paper, we concentrate on groups expressible as a semidirect product

ℤ/pq ⋊ Q_{8} = (ℤ/p × ℤ/q) ⋊ Q_{8}.

Here, p and q are distinct odd primes, and Q_{8} acts on ℤ/p and ℤ/q via two different quotients of order two, so that the kernel of the action is equal to the centre of Q_{8}.

We begin by investigating the group G defined by two generators x and y, and two relations

x^{p}y = y^{−1}x^{p} x^{r}y^{s} = y^{2q−s}x^{−r}.

Here, p, q, r and s are odd (but not necessarily prime), and (p, r) = (q, s) = 1. In the case r = s = 1, this presentation was investigated by B. H. Neumann [7].

Since x^{2p}y = x^{p}y^{−1}x^{p} = yx^{2p}, we have

x^{2p} ∈ Z(G),

where Z(G) denotes the centre of G. Also,

x^{r}y^{2q}x^{−r} = (x^{r}y^{s})(y^{2q−s}x^{−r}) = (y^{2q−s}x^{−r})(x^{r}y^{s}) = y^{2q}

so y^{2q} commutes with both x^{r} and x^{2p}. Since r is coprime to 2p, it follows that

y^{2q} ∈ Z(G).

We set z = y^{2q} ∈ Z(G). Since zx^{p} = x^{p}z = x^{p}y^{2q} = y^{−2q}x^{p} = z^{−1}x^{p}, we have

z^{2} = 1.

This means we can rewrite the second relation in the form

x^{r}y^{s} = zy^{−s}x^{−r}.

We have, using the first relation,

x^{p−r}(x^{r}y^{s}) = (y^{−s}x^{−r})x^{p+r} = (zx^{r}y^{s})x^{p+r} = (zx^{2p})(x^{r}y^{s})x^{−p+r}

and so

x^{p(p−r)}(x^{r}y^{s}) = (zx^{2p})^{p}(x^{r}y^{s})x^{−p(p−r)}.

But also, since p − r is even and x^{2p} ∈ Z(G) we have

x^{p(p−r)}(x^{r}y^{s}) = (x^{r}y^{s})x^{p(p−r)}.

It follows that x^{2p(p−r)} = (zx^{2p})^{p}, so x^{2pr} = z^{−p} = z. Set v = x^{4p} and w = x^{r}. Then v^{r} = 1, v ∈ Z(G), and w^{2p} = z. Furthermore, since (4p, r) = 1, we have G = ⟨v, w, y⟩, and

w^{p}y = y^{−1}w^{p} wy^{s} = zy^{−s}w^{−1}.

Since wy^{s}wy^{s} = z ∈ Z(G), conjugating by w^{p} gives

wy^{s}wy^{s} = w^{p}(wy^{s}wy^{s})w^{−p} = wy^{−s}wy^{−s}

so that y^{2s}w = wy^{−2s}. Since y^{2q} = z ∈ Z(G) and (q, s) = 1 we deduce that y^{2}w = wy^{−2} so that w^{2} and y^{2} commute. Since q − s is even, this gives

wy^{q} = wy^{s}y^{q−s} = zy^{−s}w^{−1}y^{q−s} = zy^{−s}y^{s−q}w^{−1} = zy^{−q}w^{−1} = y^{q}w^{−1}.

Set i = y^{q}, j = w^{−p}, k = ij and

Q = ⟨i, j⟩ ⊆ G.

Since we have i^{2} = j^{2} = z, ij = ji^{−1}, and z is central with z^{2} = 1, it follows that Q satisfies the relations of a quaternion group of order eight. Writing ī, j̄, k̄ for i^{−1}, j^{−1} and k^{−1}, we have the usual quaternionic relations

ij = j̄i = k, jk = k̄j = i, ki = īk = j,
i^{2} = j^{2} = k^{2} = ijk = z ∈ Z(Q), z^{2} = 1.

Since (4, p) = (4, q) = 1, we have

G = ⟨v, w^{4}, y^{4}, i, j⟩,

and these elements satisfy the quaternion relations together with

iy^{4} = y^{4}i,  jw^{4} = w^{4}j,  iw^{4} = w^{−4}i,  jy^{4} = y^{−4}j,
w^{4}y^{4} = y^{4}w^{4},  (w^{4})^{p} = 1,  (y^{4})^{q} = 1,  v ∈ Z(G),  v^{r} = 1.

It follows that there is a surjective homomorphism

ℤ/r × (ℤ/p × ℤ/q) ⋊ Q_{8} → G

sending the generator of ℤ/r to v, the generator of ℤ/p to w^{4}, the generator of ℤ/q to y^{4}, and the generators of Q_{8} to i and j. Finally, expressing x and y in terms of these generators, it is not hard to check that the corresponding elements of the group on the left satisfy the defining relations, so that this map is an isomorphism and |G| = 8pqr.

One way of keeping track of this isomorphism is as follows. By replacing p with −p if necessary, we can assume that p ≡ 3 (mod 4). Similarly, we can assume that q ≡ 1 (mod 4). Then we can set a = w^{1+p} and b = y^{1−q}, so that ⟨a⟩ = ⟨w^{4}⟩ and ⟨b⟩ = ⟨y^{4}⟩, and write N for the normal subgroup ⟨a, b⟩ of G. Then G = ⟨v⟩×(N ⋊Q), and in terms of these generators we have w = ja and y = ib. These elements satisfy

ab = ba, a^{p} = 1, b^{q} = 1, ia = a^{−1}i, ja = aj, ib = bi, jb = b^{−1}j.

Remarks. (i) It follows from the above discussion that the isomorphism type of G depends only on p, q and r, and not on s.

(ii) It would be interesting to know whether the periodic groups of the form

ℤ/m ⋊ (Q_{2^{n}} × ℤ/r)

with m and r odd and coprime, all have deficiency zero presentations.

2. The resolution

Let G be the group discussed in the last section. We assume from now on that r = s = 1, so that w = x and v = 1. We shall also assume that p and q are distinct odd primes. Thus

G = ⟨x, y | x^{p}y = y^{−1}x^{p}, xy = y^{2q−1}x^{−1}⟩ ≅ (ℤ/p × ℤ/q) ⋊ Q_{8}.

By the analysis of Section 1, every element of G has a unique expression of the form x^{α}y^{β} with 0 ≤ α ≤ 4p − 1, 0 ≤ β ≤ 2q − 1. The element x^{2p} = y^{2q} = z is central, and x^{4p} = y^{4q} = z^{2} = 1. In order to multiply such expressions, one uses these facts together with the rules

y^{β}x^{α} = \begin{cases}x^{α}y^{β} & α, β even \\ x^{α}y^{−β} & α odd, β even \\ x^{−α}y^{β} & α even, β odd \\ x^{−α}y^{−β}z & α, β odd. \end{cases}

Using Fox’s free differential calculus [4] on the relators of the above presentation, namely yx^{p}yx^{−p} and xyxy^{1−2q}, we can form the beginning of a free resolution of ℤ as a ℤG-module. The following sequence is exact:

ℤGγ ⊕ ℤGγ′ ⟶^{d_{2}} ℤGβ ⊕ ℤGβ′ ⟶^{d_{1}} ℤGα ⟶^{ε} ℤ → 0.

Here, the maps are given as follows:

ε(α) = 1, d_{1}(β) = (x − 1)α, d_{1}(β′) = (y − 1)α
d_{2}(γ) = \frac{∂(yx^{p}yx^{−p})}{∂x}β + \frac{∂(yx^{p}yx^{−p})}{∂y}β′
= (y − 1)(x^{p−1} + ⋯ + x + 1)β + (yx^{p} + 1)β′
d_{2}(γ′) = \frac{∂(xyxy^{1−2q})}{∂x}β + \frac{∂(xyxy^{1−2q})}{∂y}β′
= (xy + 1)β + (x − (y^{2q−2} + ⋯ + y + 1))β′.

We begin by examining the degenerate case where p = 1 rather than a prime. In this case, the equation for d_{2}(γ) reduces to

d_{2}(γ) = (y − 1)β + (yx + 1)β′

while the equation for d_{2}(γ′) remains unaltered. The kernel of d_{2} is generated by the element

(x − 1)γ + (y − 1)γ′

whose annihilator is the group sum.

Similarly, in the degenerate case where q = 1 rather than a prime, the equation for d_{2}(γ) remains unaltered, while the equation for d_{2}(γ′) reduces to

d_{2}(γ′) = (xy + 1)β + (x − 1)β′.

The kernel of d_{2} is generated by the element

(x − 1)γ + (yx^{p−1} − 1)γ′

whose annihilator is the group sum.

In each of these two cases, we obtain an exact sequence of the form

0 → ℤ → ℤGδ ⟶^{d_{3}} ℤGγ ⊕ ℤGγ′ ⟶^{d_{2}} ℤGβ ⊕ ℤGβ′ ⟶^{d_{1}} ℤGα ⟶^{ε} ℤ → 0

where d_{3}(δ) is equal to the given generator for the kernel of d_{2}. Our purpose here is to try to understand under what conditions on primes p and q we can get a sequence like this one. This will happen precisely when the kernel of d_{2} is generated by a single element. We do not completely succeed in this goal. What we do succeed in doing is obtaining a pair of generators (Theorem 4.8) for the kernel of d_{2}. The next step will be to find conditions under which these may be reduced to a single generator. We describe in Section 5 why one generator will not always suffice; there are delicate congruence conditions on p and q in order for the Swan obstruction to vanish. It seems plausible that these are exactly the conditions for reducing to a single generator; necessity is clear, but sufficiency is not.

Theorem 4.8 gives enough information to compute H^{∗}(G, M) for any ℤG-module M. Namely, since H^{∗}(G, ℤ) is periodic with period four, it suffices to compute the Tate cohomology groups Ĥ^{i}(G, M) for −1 ≤ i ≤ 2. This can be achieved by applying Hom_{ℤG}(−, M) to the segment of a Tate resolution given by

ℤGδ ⊕ ℤGδ′ ⟶^{d_{3}} ℤGγ ⊕ ℤGγ′ ⟶^{d_{2}} ℤGβ ⊕ ℤGβ′ ⟶^{d_{1}} ℤGα ⟶^{ε^{∗}∘ε}

ℤGα^{∗} ⟶^{d_{1}^{∗}} ℤGβ^{∗} ⊕ ℤGβ^{′∗} ⟶^{d_{2}^{∗}} ℤGγ^{∗} ⊕ ℤGγ^{′∗}

and taking the cohomology of the resulting complex. Here, δ and δ′ are taken to be a pair of generators for the kernel of d_{2}. The maps ε^{∗}, d_{1}^{∗} and d_{2}^{∗} are the ℤ-duals of ε, d_{1} and d_{2}, ℤGα^{∗} denotes Hom_{ℤ}(ℤGα, ℤ), a free ℤG-module of rank one with basis element α^{∗}, and so on.

3. The kernel of d_{2}

In order to find the kernel of d_{2}, we write out the equations in the last section in greater detail. We shall assume that p ≡ 3 (mod 4) and q ≡ 1 (mod 4). As in Section 1, we write i = y^{q}, j = x^{−p}, k = ij, a = x^{1+p}, b = y^{1−q}. Thus x = ja and y = ib, and the generators i, j, a and b satisfy

i^{2} = j^{2} = k^{2} = ijk = z ∈ Z(G), z^{2} = 1,
ab = ba, a^{p} = b^{q} = 1, ia = a^{−1}i, ja = aj, ib = bi, jb = b^{−1}j.

Each element of G may be written uniquely as a product of an element of Q and an element of N. Substituting ja for x and ib for y, we get

x^{p−1} + ⋯ + x + 1 = za^{p−1} + ja^{p−2} + 1a^{p−3} + j̄a^{p−4} + za^{p−5} + ⋯ + ja + 1
y^{2q−2} + ⋯ + y + 1 = 1b^{2q−2} + īb^{2q−3} + zb^{2q−4} + ib^{2q−5} + ⋯ + ib + 1,

so that

yx^{p} + 1 = k̄b^{−1} + 1, xy + 1 = k̄a^{−1}b + 1,

(y − 1)(x^{p−1} + ⋯ + x + 1) = −1(a^{p−3} + a^{p−7} + ⋯ + 1)

− z(a^{p−1} + a^{p−5} + ⋯ + a^{2}) + i(a^{p−3} + a^{p−7} + ⋯ + 1)b
+ ī(a^{p−1} + a^{p−5} + ⋯ + a^{2})b − j(a^{p−2} + a^{p−6} + ⋯ + a)
− j̄(a^{p−4} + a^{p−8} + ⋯ + a^{3}) + k(a^{p−2} + a^{p−6} + ⋯ + a)b^{−1}
+ k̄(a^{p−4} + a^{p−8} + ⋯ + a^{3})b^{−1},

and

x − (y^{2q−2} + ⋯ + y + 1) = −1(b^{2q−2} + b^{2q−6} + ⋯ + 1) − z(b^{2q−4} + b^{2q−8} + ⋯ + b^{2})

− i(b^{2q−5} + b^{2q−9} + ⋯ + b) − ī(b^{2q−3} + b^{2q−7} + ⋯ + b^{3}) + ja.

Write the general element ϕγ − ϕ′γ′ of the kernel of d_{2} in the form

(1ϕ_{1} + zϕ_{z} + iϕ_{i} + īϕī + jϕ_{j} + j̄ϕj̄ + kϕ_{k} + k̄ϕk̄)γ

− (1ϕ′_{1} + zϕ′_{z} + iϕ′_{i} + īϕ′ ; ī + jϕ′_{j} + j̄ϕ′ ; j̄ + kϕ′_{k} + k̄ϕ′ ; k̄)γ′,

where each of the coefficients ϕ_{1}, …, ϕk̄, ϕ′_{1}, …, ϕ′ ; k̄ is an element of the group algebra ℤ[ℤ/pq]. Then the equations describing the kernel of d_{2} are

ϕ(y − 1)(x^{p−1} + ⋯ + x + 1) = ϕ′(xy + 1)  (1)
ϕ(yx^{p} + 1) = ϕ′(x − (y^{2q−2} + ⋯ + y + 1))  (2)

These equations give 16 conjugate linear equations in these 16 variables. In fact, it turns out that by conjugating each of these equations by one of 1, i, j or k, these 16 equations become linear in the 16 new variables

ϕ_{1} = ϕ_{1}, ϕ_{2} = ϕ_{z}, ϕ_{3} = iϕ_{i}ī, ϕ_{4} = īϕīi,

ϕ_{5} = jϕ_{j}j̄, ϕ_{6} = j̄ϕj̄j, ϕ_{7} = kϕ_{k}k̄, ϕ_{8} = k̄ϕk̄k,

ϕ′_{1} = ϕ′_{1}, ϕ′_{2} = ϕ′_{z}, ϕ′_{3} = iϕ′_{i}ī, ϕ′_{4} = īϕ′ ; īi,

ϕ′_{5} = jϕ′_{j}j̄, ϕ′_{6} = j̄ϕ′ ; j̄j, ϕ′_{7} = kϕ′_{k}k̄, ϕ′_{8} = k̄ϕ′ ; k̄k.

In order to solve these equations, we work first over the quotient of the group algebra ℤ[ℤ/pq] given by setting a^{p−1} + ⋯ + a + 1 = 0 and b^{q−1} + ⋯ + b + 1 = 0. This is the ring O_{K} = ℤ[ζ_{pq}] of integers in the cyclotomic number field K = ℚ(ζ_{pq}), where ζ_{pq} = e^{2πi/pq} is a primitive pqth root of unity. Thus a and b are primitive pth and qth roots of unity respectively.

In this quotient, we may multiply the equations by any non-zero element of K, without altering the space of solutions. We multiply the equation (1) by a^{3}+a^{2}+a+1, and the equation (2) by b^{2} + 1, to obtain

ϕ(−1 − za^{2} + ib + īa^{2}b − ja + j̄(a^{2} + a + 1) + kab^{−1} − k̄(a^{2} + a + 1)b^{−1})

= ϕ′(k̄a^{−1}b + 1)(a^{3} + a^{2} + a + 1) (3)

ϕ(k̄b^{−1} + 1)(b^{2} + 1) = ϕ′(−1 + z + ib^{−1} + īb + ja(b^{2} + 1)).  (4)

We then conjugate as above to make the equations linear. The resulting 16 equations are as follows:

\begin{matrix}−ϕ_{1} − ϕ_{2}a^{2} + ϕ_{3}a^{2}b + ϕ_{4}b + ϕ_{5}(a^{2} & + a + 1) − ϕ_{6}a \\ −ϕ_{7}(a^{2} + a + 1)b^{−1} + ϕ_{8}ab^{−1} & = (ϕ′_{1} + ϕ′_{7}a^{−1}b)(a^{3} + a^{2} + a + 1) \end{matrix}  (5)

\begin{matrix}ϕ_{1}b + ϕ_{2}a^{−2}b − ϕ_{3} − ϕ_{4}a^{−2} + ϕ_{5} & a^{−1}b^{−1} − ϕ_{6}(1 + a^{−1} + a^{−2})b^{−1} \\ +ϕ_{7}(1 + a^{−1} + a^{−2}) − ϕ_{8}a^{−1} & = (ϕ′_{3} + ϕ′_{6}ab)(1 + a^{−1} + a^{−2} + a^{−3}) \end{matrix} (6)

\begin{matrix}−ϕ_{1}a + ϕ_{2}(a^{2} + a + 1) − ϕ_{3}(a^{2}+ & a + 1)b + ϕ_{4}ab − ϕ_{5} − ϕ_{6}a^{2} \\ +ϕ_{7}b^{−1} + ϕ_{8}a^{2}b^{−1} & = (ϕ′_{3}a^{−1}b^{−1} + ϕ′_{5})(a^{3} + a^{2} + a + 1) \end{matrix} (7)

\begin{matrix}ϕ_{1}a^{−1}b − ϕ_{2}(1 + a^{−1} + a^{−2})b− & ϕ_{3}a^{−1} + ϕ_{4}(1 + a^{−1} + a^{−2}) + ϕ_{5}a^{−2}b^{−1} \\ +ϕ_{6}b^{−1} − ϕ_{7} − ϕ_{8}a^{−2} & = (ϕ′_{2}ab^{−1} + ϕ′_{7})(1 + a^{−1} + a^{−2} + a^{−3}) \end{matrix} (8)

\begin{matrix}−ϕ_{1}a^{2} − ϕ_{2} + ϕ_{3}b + ϕ_{4}a^{2}b − ϕ_{5}a+ & ϕ_{6}(a^{2} + a + 1) + ϕ_{7}ab^{−1} \\ −ϕ_{8}(a^{2} + a + 1)b^{−1} & = (ϕ′_{2} + ϕ′_{8}a^{−1}b)(a^{3} + a^{2} + a + 1) \end{matrix}  (9)

\begin{matrix}ϕ_{1}a^{−2}b + ϕ_{2}b − ϕ_{3}a^{−2} − ϕ_{4} − ϕ_{5}( & 1 + a^{−1} + a^{−2})b^{−1} + ϕ_{6}a^{−1}b^{−1} \\ −ϕ_{7}a^{−1} + ϕ_{8}(1 + a^{−1} + a^{−2}) & = (ϕ′_{4} + ϕ′_{5}ab)(1 + a^{−1} + a^{−2} + a^{−3}) \end{matrix}  (10)

\begin{matrix}ϕ_{1}(a^{2} + a + 1) − ϕ_{2}a + ϕ_{3}ab − ϕ_{4} & (a^{2} + a + 1)b − ϕ_{5}a^{2} − ϕ_{6} \\ +ϕ_{7}a^{2}b^{−1} + ϕ_{8}b^{−1} & = (ϕ′_{6} + ϕ′_{4}a^{−1}b^{−1})(a^{3} + a^{2} + a + 1) \end{matrix}  (11)

\begin{matrix}−ϕ_{1}(1 + a^{−1} + a^{−2})b + ϕ_{2}a^{−1}b & + ϕ_{3}(1 + a^{−1} + a^{−2}) − ϕ_{4}a^{−1} + ϕ_{5}b^{−1} \\ +ϕ_{6}a^{−2}b^{−1} − ϕ_{7}a^{−2} − ϕ_{8} & = (ϕ′_{1}ab^{−1} + ϕ′_{8})(1 + a^{−1} + a^{−2} + a^{−3}) \end{matrix}  (12)
ϕ_{1}(b^{2} + 1) + ϕ_{7}(b + b^{−1}) = −ϕ′_{1} + ϕ′_{2} + ϕ′_{3}b + ϕ′_{4}b^{−1} + ϕ′_{6}a(b^{2} + 1)  (13)
ϕ_{3}(b^{2} + 1) + ϕ_{6}(b + b^{−1}) = ϕ′_{1}b^{−1} + ϕ′_{2}b − ϕ′_{3} + ϕ′_{4} + ϕ′_{8}a^{−1}(b^{2} + 1)  (14)
ϕ_{3}(b + b^{−1}) + ϕ_{5}(1 + b^{−2}) = ϕ′_{1}a(1 + b^{−2}) − ϕ′_{5} + ϕ′_{6} + ϕ′_{7}b + ϕ′_{8}b^{−1}  (15)
ϕ_{2}(b + b^{−1}) + ϕ_{7}(1 + b^{−2}) = ϕ′_{3}a^{−1}(1 + b^{−2}) + ϕ′_{5}b^{−1} + ϕ′_{6}b − ϕ′_{7} + ϕ′_{8}  (16)
ϕ_{2}(b^{2} + 1) + ϕ_{8}(b + b^{−1}) = ϕ′_{1} − ϕ′_{2} + ϕ′_{3}b^{−1} + ϕ′_{4}b + ϕ′_{5}a(b^{2} + 1)  (17)
ϕ_{4}(b^{2} + 1) + ϕ_{5}(b + b^{−1}) = ϕ′_{1}b + ϕ′_{2}b^{−1} + ϕ′_{3} − ϕ′_{4} + ϕ′_{7}a^{−1}(b^{2} + 1)  (18)
ϕ_{4}(b + b^{−1}) + ϕ_{6}(1 + b^{−2}) = ϕ′_{2}a(1 + b^{−2}) + ϕ′_{5} − ϕ′_{6} + ϕ′_{7}b^{−1} + ϕ′_{8}b (19)
ϕ_{1}(b + b^{−1}) + ϕ_{8}(1 + b^{−2}) = ϕ′_{4}a^{−1}(1 + b^{−2}) + ϕ′_{5}b + ϕ′_{6}b^{−1} + ϕ′_{7} − ϕ′_{8}.  (20)

We simplify this set of equations as follows. Adding equations (13) and (17), and equations (14) and (18), and dividing by b^{2} + 1, we get

ϕ_{1} + ϕ_{2} + ϕ_{7}b^{−1} + ϕ_{8}b^{−1} = ϕ′_{3}b^{−1} + ϕ′_{4}b^{−1} + ϕ′_{5}a + ϕ′_{6}a.  (21)
ϕ_{3} + ϕ_{4} + ϕ_{5}b^{−1} + ϕ_{6}b^{−1} = ϕ′_{1}b^{−1} + ϕ′_{2}b^{−1} + ϕ′_{7}a^{−1} + ϕ′_{8}a^{−1}  (22)

Similarly, adding equations (15) and (19), and equations (16) and (20), and dividing by b + b^{−1}, we get

ϕ_{3} + ϕ_{4} + ϕ_{5}b^{−1} + ϕ_{6}b^{−1} = ϕ′_{1}ab^{−1} + ϕ′_{2}ab^{−1} + ϕ′_{7} + ϕ′_{8}  (23)

ϕ_{1} + ϕ_{2} + ϕ_{7}b^{−1} + ϕ_{8}b^{−1} = ϕ′_{3}a^{−1}b^{−1} + ϕ′_{4}a^{−1}b^{−1} + ϕ′_{5} + ϕ′_{6}.  (24)

Comparing equations (22) and (23), and using the fact that a − 1 is non-zero, we see that each side of equation (22) is separately zero. Similarly, comparing equations (21) and (24), we see that each side of (21) is separately zero. So the last four equations are equivalent to the following four equations:

ϕ_{1}b + ϕ_{2}b + ϕ_{7} + ϕ_{8} = 0  (25)
ϕ_{3}b + ϕ_{4}b + ϕ_{5} + ϕ_{6} = 0  (26)
ϕ′_{1}a + ϕ′_{2}a + ϕ′_{7}b + ϕ′_{8}b = 0  (27)
ϕ′_{3} + ϕ′_{4} + ϕ′_{5}ab + ϕ′_{6}ab = 0.  (28)

Dividing the equations (13)+(14)b and (13)b−(14) by b^{2} +1, and dividing equations (15) + (16)b and (15)b − (16) by b + b^{−1}, we get:

ϕ_{1} + ϕ_{3}b + ϕ_{6} + ϕ_{7}b^{−1} = ϕ′_{2} + ϕ′_{4}b^{−1} + ϕ′_{6}a + ϕ′_{8}a^{−1}b  (29)
ϕ_{1}b − ϕ_{3} − ϕ_{6}b^{−1} + ϕ_{7} = −ϕ′_{1}b^{−1} + ϕ′_{3} + ϕ′_{6}ab − ϕ′_{8}a^{−1}  (30)
ϕ_{2}b + ϕ_{3} + ϕ_{5}b^{−1} + ϕ_{7} = ϕ′_{1}ab^{−1} + ϕ′_{3}a^{−1} + ϕ′_{6}b + ϕ′_{8}  (31)
−ϕ_{2} + ϕ_{3}b + ϕ_{5} − ϕ_{7}b^{−1} = ϕ′_{1}a − ϕ′_{3}a^{−1}b^{−1} − ϕ′_{5} + ϕ′_{7}b  (32)

Our sixteen equations are now (5)–(12) and (25)–(32).

Next, we observe that we have the following linear relations between the equations:

(5) − (8)a^{2}b + (9) − (12)a^{2}b = (25)(a^{2} + 1)(b − b^{−1})
(6)a^{2}b − (7) + (10)a^{2}b − (11) = (25)(a^{2} + 1)(b − b^{−1})
(5) + (9) = (−(25)b^{−1} + (26) + (27)(1 + a^{−1}))(a^{2} + 1)

(6) + (10) = (25)(1 + a^{−2}) − (26)(1 + a^{−2})b^{−1} + (28)(1 + a^{−1} + a^{−2} + a^{−3})

−(6)a^{3} + (12)a^{2} = (29)(a^{3} + a^{2} + a + 1)
(9) + (11)a = −(25)(a^{2} + 1)b^{−1} − (26)(a^{3} + a) + (30)(a^{3} + a^{2} + a + 1)
(5)a − (7) = (31)(a^{3} + a^{2} + a + 1)
(6)a^{2} + (12) = −(25)(a^{3} + a) − (26)(a^{2} + 1)b^{−1} + (32)(a^{3} + a^{2} + a + 1)

These relations imply that equations (5)–(12) are redundant, so that we are left with equations (25)–(32). It is convenient to express these in matrix form as follows:

( \begin{matrix}b & b & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & b & b & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & a & a & 0 & 0 & 0 & 0 & b & b \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & ab & ab & 0 & 0 \\ 1 & 0 & b & 0 & 0 & 1 & b^{−1} & 0 & 0 & 1 & 0 & b^{−1} & 0 & a & 0 & a^{−1}b \\ b & 0 & −1 & 0 & 0 & −b^{−1} & 1 & 0 & −b^{−1} & 0 & 1 & 0 & 0 & ab & 0 & −a^{−1} \\ 0 & b & 1 & 0 & b^{−1} & 0 & 1 & 0 & ab^{−1} & 0 & a^{−1} & 0 & 0 & b & 0 & 1 \\ 0 & −1 & b & 0 & 1 & 0 & −b^{−1} & 0 & a & 0 & −a^{−1}b^{−1} & 0 & −1 & 0 & b & 0 \end{matrix} )

A basis for the null space of this matrix is given by the columns of the following matrix:

( \begin{matrix}−1 & 0 & 0 & 0 & ab^{−2} & 0 & 0 & ab^{−1} \\ −1 & 0 & 0 & 0 & −b^{−2} & (a − 1)b & a − 1 & −b^{−1} \\ 0 & −1 & 0 & 0 & b^{−1} & 0 & (1 − a)b & −ab^{−2} \\ 0 & −1 & 0 & 0 & −ab^{−1} & 1 − a & 0 & b^{−2} \\ 0 & b & 0 & 0 & 0 & −b & −b^{2} & 0 \\ 0 & b & 0 & 0 & a − 1 & ab & ab^{2} & (a − 1)b^{−1} \\ b & 0 & 0 & 0 & 0 & −ab^{2} & b & 1 − a \\ b & 0 & 0 & 0 & (1 − a)b^{−1} & b^{2} & −ab & 0 \\ 0 & 0 & −1 & 0 & 0 & 0 & 1 & b^{−1} \\ 0 & 0 & −1 & 0 & 0 & 0 & −b^{2} & −b \\ 0 & 0 & 0 & ab & ab^{−1} & a & 0 & 0 \\ 0 & 0 & 0 & ab & −ab & −ab^{2} & 0 & 0 \\ 0 & 0 & 0 & −1 & 1 & −b^{−1} & 0 & 0 \\ 0 & 0 & 0 & −1 & −b^{−2} & b & 0 & 0 \\ 0 & 0 & ab^{−1} & 0 & 0 & 0 & ab & −ab^{−2} \\ 0 & 0 & ab^{−1} & 0 & 0 & 0 & −ab^{−1} & a \end{matrix} )

Now a − 1 and a + 1 are coprime in O_{K}, and b^{2} + 1 = (b^{4} − 1)/(b^{2} − 1) is a unit in O_{K}. Using these facts, we easily see that the minor of the above matrix given by rows 1–4 and rows 9–12 has determinant a unit in O_{K}. It follows that the kernel has the columns of the above matrix as an O_{K}-basis.

The first column of this matrix, for example, tells us that the element

(−1 − z + kb^{−1} + k̄b^{−1})γ = (1 + z)(yx^{p} − 1)γ

is in the kernel of d_{2}, modulo the cyclotomic relations. We write

N_{a} = 1 + a + ⋯ + a^{p−1},
N_{b} = 1 + b + ⋯ + b^{q−1},
N_{x} = x^{4p−1} + ⋯ + x + 1 = (1 + z + j + j̄)N_{a},
N_{y} = y^{4q−1} + ⋯ + y + 1 = (1 + z + i + ī)N_{b}.

Then we have

d_{2}(1 + z)(yx^{p} − 1)γ = (1 − y)N_{x}β.  (33)

The remaining columns of this matrix lift to the following equations:

d_{2}y^{q}(1 + z)(yx^{p} − 1)γ = y^{q}(1 − y)N_{x}β  (34)
d_{2}(1 + z)(xy − 1)γ′ = (1 − x)N_{y}β′  (35)
d_{2}x^{p}(1 + z)(xy − 1)γ′ = x^{p}(1 − x)N_{y}β′  (36)
d_{2}x^{−p}((1 − x^{1+p} + x^{−p}y^{−2} − xy^{−2}z)(y − z)γ + (1 + xy^{−1})(1 − y^{2})γ′) = 0  (37)
d_{2}x^{1+p}y^{q}((1 − x^{1+p} + x^{−p}y^{−2} − xy^{−2}z)(y − z)γ + (1 + xy^{−1})(1 − y^{2})γ′) = 0  (38)
d_{2}((1 − x^{1+p} + x^{−p}y^{−2} − xy^{−2}z)(y − z)γ + (1 + xy^{−1})(1 − y^{2})γ′) = 0  (39)
d_{2}xy^{q}((1 − x^{1+p} + x^{−p}y^{−2} − xy^{−2}z)(y − z)γ + (1 + xy^{−1})(1 − y^{2})γ′) = 0  (40)

Now equation (34) is a y^{q} times equation (33), equation (36) is x^{p} times equation (35), and equations (37), (38) and (40) are obtained by multiplying equation (39) by x^{−p}, x^{1+p}y^{q} and xy^{q} respectively. So we may restrict our attention to equations (33), (35) and (39). We have proved the following.

Theorem 3.1. The kernel of

d̄_{2} : (ℤG/I)γ ⊕ (ℤG/I)γ′ → (ℤG/I)β ⊕ (ℤG/I)β′

is generated by the elements (1 + z)(yx^{p} − 1)γ, (1 + z)(xy − 1)γ′ and

(1 − x^{1+p} + x^{−p}y^{−2} − xy^{−2}z)(y − z)γ + (1 + xy^{−1})(1 − y^{2})γ′.

Here, I is the ideal of ℤG generated by the elements a^{p−1} + ⋯ + a + 1 and b^{q−1} + ⋯ + b + 1. □

4. The switchback map

To make use of the above theorem, we examine the switchback map for the snake lemma applied to the following diagram.

\xymatrix{0 \ar[r] & I ⊕ I \ar[r] \ar[d] & ℤGγ ⊕ ℤGγ′ \ar[r] \ar[d]^{d_{2}} & ℤG/I ⊕ ℤG/I \ar[r] \ar[d]^{d̄_{2}} & 0 \\ 0 \ar[r] & I ⊕ I \ar[r] & ℤGβ ⊕ ℤGβ′ \ar[r] & ℤG/I ⊕ ℤG/I \ar[r] & 0}

Lemma 4.1. As a ℤG-module, the image of the switchback map for the above diagram is a direct sum M_{p} ⊕ M_{q}. Here, M_{p} is generated by (1 − y)N_{x}β, killed by multiplication by p, and M_{q} is generated by (1 − x)N_{y}β′, killed by multiplication by q.

Proof. The generation statement follows by examining the images under d_{2} of the elements of ℤGγ ⊕ℤGγ′ named in Theorem 3.1, using equations (33), (35) and (39).

The image of (1−y)N_{x}β is annihilated by multiplication by p, since equation (33) gives

d_{2}N_{a}(1 + z)(yx^{p} − 1)γ = N_{a}(1 − y)N_{x}β = p(1 − y)N_{x}β.

Similarly, the image of (1 − x)N_{y}β′ is annihilated by multiplication by q, since equation (35) gives

d_{2}N_{b}(1 + z)(xy − 1)γ′ = N_{b}(1 − x)N_{y}β′ = q(1 − x)N_{y}β′.  □

Lemma 4.2. (i) The annihilator in ℤG of (1 − y)N_{x}β is the left ideal generated by the elements 1 − x^{1+p}, yx^{p} + 1 and y^{2q−1} + ⋯ + y + 1.

(ii) The annihilator in ℤG of (1−x)N_{y}β′ is the left ideal generated by the elements 1 − y^{2}, xy + 1 and x^{2p−1} + ⋯ + x + 1.

Proof. (i) We have

(1 − x^{1+p})(1 − y)N_{x}β = (1 − x^{1+p} − y + yx^{−1−p})N_{x}β
= (1 + yx^{−1−p})(1 − x^{1+p})N_{x}β = 0,
(yx^{p} + 1)(1 − y)N_{x}β = (yx^{p} + 1 − x^{p} − y)N_{x}β
= (1 − y)(1 − x^{p})N_{x}β = 0
(y^{2q−1} + ⋯ + y + 1)(1 − y)N_{x}β = (1 − y^{2q})N_{x}β = (1 − x^{2p})N_{x}β = 0,

so these elements are in the annihilator. Since ℤG has a basis consisting of the elements y^{n}x^{m} with 0 ≤ n ≤ 2q − 1 and 0 ≤ m ≤ 4p − 1, the elements y^{n} and y^{n}x^{p} span ℤG/ℤG(1 − x^{1+p}). Then the elements y^{n} with 0 ≤ n ≤ 2q − 1 span ℤG/ℤG(1 − x^{1+p}, yx^{p} + 1) and finally the elements y^{n} with 0 ≤ n ≤ 2q − 2 span

ℤG/ℤG(1 − x^{1+p}, yx^{p} + 1, y^{2q−1} + ⋯ + y + 1).

Since the submodule of ℤGβ generated by (1−y)N_{x}β has ℤ-rank 2q−1 by Lemma 4.1, these elements must generate the annihilator.

(ii) We have

(1 − y^{2})(1 − x)N_{y}β′ = (1 − y^{2} − x + xy^{−2})N_{y}β′
= (1 + xy^{−2})(1 − y^{2})N_{y}β′ = 0,
(xy + 1)(1 − x)N_{y}β′ = (xy + 1 − y^{2q−1} − x)N_{y}β′
= (x − (y^{2q−2} + ⋯ + y + 1))(y − 1)N_{y}β′ = 0,
(x^{2p−1} + ⋯ + 1)(1 − x)N_{y}β′ = (1 − x^{2p})N_{y}β′ = (1 − y^{2q})N_{y}β′ = 0,

so these elements are in the annihilator. The rest of the proof is parallel to the proof of (i). □

Proposition 4.3. The kernel of d_{2} on ℤGγ ⊕ ℤGγ′ is generated as a ℤG-module by the elements

(1 − x^{1+p})(1 + z)(yx^{p} − 1)γ  (41)
(1 − x^{1+p} + x^{−p}y^{−2} − xy^{−2}z)(y − z)γ + (1 + xy^{−1})(1 − y^{2})γ′  (42)

together with possibly some further elements of Iγ ⊕ Iγ′.

Proof. The image in ℤG/I ⊕ℤG/I of the kernel of d_{2} is the same as the kernel of the switchback map on the kernel of d̄_{2}. The kernel of d̄_{2} is calculated in Theorem 3.1, and the switchback map is described in Lemma 4.1. Thus using Lemma 4.2, we see that the kernel is generated by the elements (41) and (42) together with

(yx^{p} + 1)(1 + z)(yx^{p} − 1)γ  (43)
(y^{2q−1} + ⋯ + y + 1)(1 + z)(yx^{p} − 1)γ  (44)
(N_{a} − p)(1 + z)(yx^{p} − 1)γ  (45)
(1 − y^{2})(1 + z)(xy − 1)γ′  (46)
(xy + 1)(1 + z)(xy − 1)γ′  (47)
(N_{b} − q)(1 + z)(xy − 1)γ′  (48)
(x^{2p−1} + ⋯ + x + 1)(1 + z)(xy − 1)γ′.  (49)

The elements (43) and (47) are equal to zero, the elements (44) and (49) are in Iγ ⊕ Iγ′, and the element (46) is equal to (1 + yx^{p−1}) times (41) minus (1 + z) times (42). The element (45) is equal to (1 + 2a + 3a^{2} + ⋯ + pa^{p−1}) times (41) (note that x^{1+p} = a), and the element (48) is equal to (1 + 2b^{2} + 3b^{4} + ⋯ + qb^{2(q−1)}) times (46). □

In order to make use of Lemma 4.1 for rank calculations, it is necessary next to calculate Tor_{1}^{ℤG}(ℤ, ℤG/I).

Lemma 4.4. We have ℤ ⊗_{ℤG} ℤG/I = 0 and Tor_{1}^{ℤG}(ℤ, ℤG/I) = 0.

Proof. The short exact sequence of ℤG-modules

0 → I → ℤG → ℤG/I → 0

gives rise to an exact sequence

0 → Tor_{1}^{ℤG}(ℤ, ℤG/I) → ℤ ⊗_{ℤG} I → ℤ ⊗_{ℤG} ℤG → ℤ ⊗_{ℤG} ℤG/I → 0.

Now I is generated as a submodule of ℤG by the elements

u_{1} = a^{p−1} + ⋯ + a + 1
u_{2} = b^{q−1} + ⋯ + b + 1

which satisfy the relations (a − 1)u_{1} = 0, (b − 1)u_{2} = 0 and

(b^{q−1} + ⋯ + b + 1)u_{1} = (a^{p−1} + ⋯ + a + 1)u_{2}.

A count of ℤ-ranks shows that this is a presentation of the ℤG-module I.

Tensoring over ℤG with ℤ via the augmentation map, we obtain generators ū_{1} = 1 ⊗ u_{1} and ū_{2} = 1 ⊗ u_{2} subject only to the relation qū_{1} = pū_{2}. Choose integers r and s so that rp + sq = 1, and set ū_{3} = rū_{1} + sū_{2}. Then pū_{3} = ū_{1} and qū_{3} = ū_{2}. So ℤ ⊗_{ℤG} I is a copy of ℤ generated by ū_{3}. The image in ℤ ⊗_{ℤG} ℤG ≅ ℤ of ū_{1} is p and of ū_{2} is q, so the image of ū_{3} is rp + sq = 1. It follows that the middle arrow in the above exact sequence is an isomorphism, and the two end terms are zero. □

Proposition 4.5. The sequence

ℤG/I ⊕ ℤG/I ⟶^{d̄_{2}} ℤG/I ⊕ ℤG/I ⟶^{d̄_{1}} ℤG/I → 0

is exact.

Proof. This follows directly from the lemma, since the homology of this sequence calculates ℤ ⊗_{ℤG} ℤG/I and Tor_{1}^{ℤG}(ℤ, ℤG/I). □

Corollary 4.6. The kernel and cokernel of d̄_{2} both have ℤ-rank 8(p − 1)(q − 1).

Proof. Since O_{K} has ℤ-rank (p − 1)(q − 1) and ℤG/I has O_{K}-rank 8, it follows that ℤG/I has ℤ-rank 8(p − 1)(q − 1). □

Corollary 4.7. The kernel and cokernel of d_{2} as a map from I ⊕ I to I ⊕ I both have torsion-free ℤ-rank equal to 8p + 8q − 9. The kernel is torsion-free, while the cokernel has a p-torsion summand M_{p} and a q-torsion summand M_{q}.

Proof. According to Lemma 4.1 the image of the switchback map is a sum of a p-torsion module M_{p} and a q-torsion module M_{q}. So the torsion-free rank of the cokernel of d_{2} on I ⊕ I is given by subtracting the torsion-free rank of the cokernel of d̄_{2} on ℤG/I ⊕ ℤG/I from the torsion-free rank of the cokernel of d_{2} on ℤG ⊕ ℤG. By Corollary 4.6, this gives a torsion-free rank of

(8pq − 1) − 8(p − 1)(q − 1) = 8p + 8q − 9.

The torsion-free rank of the kernel is the same as that of the cokernel. □

Theorem 4.8. The kernel of d_{2} on ℤGγ ⊕ℤGγ′ is generated as a ℤG-module by the elements (41) and (42), together with N_{a}N_{b}[(j − 1)γ + (ī− 1)γ′]. The latter element may be added into element (41) or into element (42) to give just two generators.

Proof. Multiplying N_{a} by the element (42) gives

N_{a}(k − k̄ − j + j̄)b^{−2}γ + N_{a}(kb^{−1} + 1 − k̄b − zb^{2})γ′

which generates a ℤG-submodule M_{1} of I⊕I of ℤ-rank 8q−4 annihilated by N_{b}(1+z). Multiplying N_{b} by the element (42) gives

N_{b}(i − z + k − j − ai + az − ak̄ + aj̄)γ + N_{b}(1 + ak − z − ak̄)γ′,

which generates a ℤG-submodule M_{2} of I ⊕ I of ℤ-rank 6p − 3 annihilated by (1 + z)(1 − j) and by N_{a}(1 + z). The intersection M_{1} ∩ M_{2} consists of the ℤGsubmodule generated by N_{a}N_{b} times (42), annihilated by (1 + z), of ℤ-rank 4, so M_{1} + M_{2} has ℤ-rank 6p + 8q − 11.

Multiplying N_{b} by the element (41) gives

N_{b}(1 − a)(−1 − z + k + k̄)γ.

This is annihilated by (1 − z), and (1 + j) times this element equals (1 + z)N_{b} times (42). Letting M_{3} be the ℤG-module generated by this element, we see that (M_{1} + M_{2} + M_{3})/(M_{1} + M_{2}) is annihilated by (1 − z) and (1 + j) and has ℤ-rank 2p−2. So M_{1} +M_{2} +M_{3} has ℤ-rank 8p+8q −13. It is not hard to verify that it is a ℤ-summand of I ⊕ I, so it follows from Corollary 4.7 that M_{1} + M_{2} + M_{3} is equal to the kernel of d_{2} on I ⊕ I modulo multiples of N_{a}N_{b}. The ℤG-submodule generated by N_{a}N_{b}[(j − 1)γ + (ī− 1)γ′] is easily checked to be in the kernel, and has ℤ-rank 7. Its intersection with M_{1} + M_{2} + M_{3} has ℤ-rank 3, so we are up to 8p + 8q − 9 as required. The theorem now follows from Proposition 4.3. □

5. The Swan obstruction

Swan [8] has proved that a finite group G acts freely and cellularly on some finitely dominated CW complex with the homotopy type of an (n − 1)-sphere S^{n−1} if and only if the integral cohomology of G is periodic with period dividing n. This means that H^{n}(G, ℤ) ≅ ℤ/|G|, and that cup product with an additive generator of H^{n}(G, ℤ) induces an isomorphism on Tate cohomology Ĥ^{i}(G, ℤ) → Ĥ^{i+n}(G, ℤ) for all i ∈ ℤ. The complex may be taken to be finite if and only if an obstruction called the Swan invariant vanishes. This invariant lies in a quotient of K̃_{0}(ℤG) by a subgroup T_{G} called the Swan subgroup. This section contains an explanation of the Swan invariant, based on Jon Carlson’s modules L_{ζ}.

Suppose that G is a periodic group. For a finitely generated ℤG-module M, we write Ω^{n}M for the nth kernel in a resolution of M by finitely generated free ℤGmodules. This is only well defined up to adding and removing finitely generated free summands (by the extended version of Schanuel’s lemma). If ζ ∈ Ĥ^{n}(G, ℤ), we can represent ζ by a cocycle ζ̂: Ω^{n}ℤ → ℤ. By adding a free summand to Ω^{n}ℤ if necessary, we may assume that this map is surjective (even if ζ = 0!), and we write L_{ζ} = Ker(ζ̂). So there is a short exact sequence

0 → L_{ζ} → Ω^{n}ℤ ⟶^{ζ̂} ℤ → 0.  (50)

An element ζ ∈ Ĥ^{n}(G, ℤ) lies in the multiplicative group Ĥ^{∗}(G, ℤ)^{×} of invertible elements if and only if L_{ζ} is projective. In this case, we get a well defined element [L_{ζ}] ∈ K̃_{0}(ℤG). Here, K̃_{0}(ℤG) denotes the quotient of the Grothendieck ring K_{0}(ℤG) of finitely generated projective ℤG-modules by the subgroup of finitely generated free ℤG-modules, so that K_{0}(ℤG) ≅ ℤ × K̃_{0}(ℤG) and K̃_{0}(ℤG) is finite (Swan [9]).

The modules L_{ζ} with ζ ∈ Ĥ^{0}(G, ℤ)^{×} are called Swan modules. The Swan module corresponding to m ∈ (ℤ/|G|)^{×} ≅ Ĥ^{0}(G, ℤ)^{×} can be described as the (projective) submodule of ℤG generated by the sum of the group elements and by the elements divisible by n where n is an integer satisfying mn ≡ 1 modulo |G|. In other words, the Swan modules are exactly the projective modules which are generated by a single element after quotienting out the image of the sum of the group elements.

Lemma 5.1. If ζ and η are elements of Ĥ^{∗}(G, ℤ)^{×} of degrees n and m, then [L_{ζ}][L_{η}] = 0 and [L_{ζ}] + [L_{η}] = [L_{ζη}] in K̃_{0}(ℤG).

Proof. As elements of K̃_{0}(ℤG), we have [ΩL_{η}] = −[L_{η}]. Tensoring the exact sequence (50) with L_{η} we get

0 → L_{ζ} ⊗ L_{η} → Ω^{n}L_{η} → L_{η} → 0.

This is an exact sequence of projective modules, so it splits. Using the fact that an invertible element of Tate cohomology has to lie in even degree, we get

[L_{η}] = [Ω^{n}L_{η}] = [L_{η}] + [L_{ζ} ⊗ L_{η}]

so that

[L_{ζ}][L_{η}] = [L_{ζ} ⊗ L_{η}] = 0.

The diagram

[figure: 0; Ω^{n}L_{η}; 0; L_{ζη}; Ω^{n+m}ℤ; ζ̂η; Ω^{n}(η); ℤ; 0; 0; L_{ζ}; Ω^{n}ℤ; ζ̂; ℤ; 0; 0; 0]

gives us a short exact sequence

0 → Ω^{n}L_{η} → L_{ζη} → L_{ζ} → 0

so that

[L_{ζη}] = [L_{ζ}] + [Ω^{n}L_{η}] = [L_{ζ}] + [L_{η}].  □

This lemma implies that there is a well defined homomorphism

σ: Ĥ^{∗}(G, ℤ)^{×} → K̃_{0}(ℤG)
ζ ↦ −[L_{ζ}]

called the Swan homomorphism. In positive degrees, it has the following interpretation. We form the beginning of a free resolution and form the pushout

[figure: 0; 0; L_{ζ}; L_{ζ}; 0; Ω^{n}ℤ; ζ̂; F_{n−1}; F_{n−2}; …; F_{0}; ℤ; 0; 0; ℤ; F_{n−1}/L_{ζ}; F_{n−2}; …; F_{0}; ℤ; 0.; 0; 0]

The obstruction to rechoosing the resolution so that F_{n−1}/L_{ζ} is free, is the element

[F_{n−1}/L_{ζ}] = −[L_{ζ}] = σ(ζ) ∈ K̃_{0}(ℤG).

This accounts for the negative sign in the description of the map.

The image under σ of Ĥ^{0}(G, ℤ)^{×} ≅ ℤ/|G|^{×} is called the Swan subgroup T_{G} ⊆ K̃_{0}(ℤG). It should be noted that {±1} is in the kernel of σ on Ĥ^{0}(G, ℤ)^{×}. We obtain a well defined map

σ̄: Ĥ^{∗}(G, ℤ)^{×}/Ĥ^{0}(G, ℤ)^{×} (≅ ℤ) → K̃_{0}(ℤG)/T_{G}.

If Ĥ^{∗}(G, ℤ)^{×}/Ĥ^{0}(G, ℤ)^{×} contains a non-zero element of degree n, then that element is unique, and we write σ_{n} for its image in K̃_{0}(ℤG)/T_{G}. The interpretation of T_{G} here is that it is the indeterminacy caused by the freedom to replace ζ by a multiple by an integer coprime to |G|.

If G acts freely on a finite CW complex X with the homotopy type of an (n − 1)- sphere, then the cellular chains on X form a complex of finitely generated free ℤG-modules with homology only in degrees zero and n − 1:

0 → C_{m} → C_{m−1} → ⋯ → C_{0} → 0.

If m ≥ n, then since projective ℤG-modules are weakly injective (Cartan and Eilenberg [2], Section XII.1), we may strip off the terms in this complex in degrees bigger than n and insert the homology of the complex to make an exact sequence

0 → ℤ → F_{n−1} → ⋯ → F_{0} → ℤ → 0

with F_{i} free (i < n − 1) and F_{n−1} stably free. It follows that the element σ_{n} ∈ K̃_{0}(ℤG)/T_{G} is zero. The element σ_{n} is called the Swan obstruction, because it is the obstruction to the existence of such a complex.

If we invert |G| then we have Ĥ^{∗}(G, ℤ(1/|G|)) = 0. It follows that the sequence of ℤ(1/|G|)G-modules

0 → L_{ζ}(1/|G|) → Ω^{n}ℤ(1/|G|) → ℤ(1/|G|) → 0

necessarily splits, giving

0 = [L_{ζ}(1/|G|)] ∈ K̃_{0}(ℤ(1/|G|)G).

If p is a prime dividing |G| and ℤp^{∧} is the ring of p-adic integers, then K̃_{0}(ℤp^{∧}G) is a finitely generated free abelian group, by the Krull–Schmidt theorem. Since K̃_{0}(ℤG) is finite, it follows that the inclusion ℤ → ℤp^{∧} induces the zero map K̃_{0}(ℤG) → K̃_{0}(ℤp^{∧}G). Now examine the long exact sequence in K-theory of the arithmetic square

K_{1}(ℤ(1/|G|)G) ⊕ ⨁_{p| |G|} K_{1}(ℤp^{∧}G) → ⨁_{p| |G|} K_{1}(ℚp^{∧}G) →

K_{0}(ℤG) → K_{0}(ℤ(1/|G|)G) ⊕ ⨁_{p| |G|} K_{0}(ℤp^{∧}G) → ⨁_{p| |G|} K_{0}(ℚp^{∧}G)

(see Davis and Milgram [3], page 246). Using the fact that K̃_{0}(ℤG) is finite and K_{0}(ℤp^{∧}G) is ℤ-free, we see that a representative of σ_{n} in K̃_{0}(ℤG) goes to zero in the next term in this sequence. So σ_{n} lifts to an element of ⨁_{p| |G|} K_{1}(ℚp^{∧}G) that is well defined modulo the inverse image of T_{G}.

The calculations in K_{1} have been carried out for the groups considered in this paper by Milgram [6] and later by Madsen [5] and Bentzen and Madsen [1]. The result of these calculations is as follows. Let A denote the ring ℤ[η_{p}, η_{q}] where η_{p} = ζ_{p} + ζ_{p}^{−1} and η_{q} = ζ_{q} + ζ_{q}^{−1}, and ζ_{p} and ζ_{q} are primitive pth and qth roots of unity respectively, with p and q distinct odd primes. Denote by A^{×} the units in A and by A^{∗} the totally positive units in A. Consider the reductions modulo p and q

Φ_{A}: A^{×} → (A/pA)_{(2)}^{×} ⊕ (A/qA)_{(2)}^{×}.

Here, (2) denotes 2-primary component. Then σ_{4} = 0 if and only if (i) p and q are not both congruent to 3 modulo 8, and (ii) (4, 4) is in the image of Φ_{A} restricted to A^{∗}

(Proposition 5.2 of [1]).

Condition (i) is guaranteed since we have assumed that q ≡ 1 (mod 4). Condition (ii) is therefore necessary, in order that the two generators given in Theorem 4.8 can be replaced by a single generator. It seems plausible that this arithmetic condition is also sufficient, but the author has not succeeded in proving this.

References

1. S. Bentzen and I. Madsen, On the Swan subgroup of certain periodic groups, Math. Ann. 264 (1983), 447–474.

2. H. Cartan and S. Eilenberg, Homological algebra, Princeton Mathematical Series, no. 19, Princeton Univ. Press, 1956.

3. J. F. Davis and R. J. Milgram, A survey of the spherical space form problem, Math. Reports, Harwood academic publishers, 1985.

4. R. H. Fox, Free differential calculus I, Ann. of Math. 57 (1953), 547–560.

5. I. Madsen, Reidemeister torsion, surgery invariants and spherical space forms, Proc. London Math. Soc. 46 (1983), 193–240.

6. R. J. Milgram, Odd index subgroups of units in cyclotomic fields and applications, Algebraic K-theory, Evanston, 1980, Lecture Notes in Mathematics, vol. 854, 1981, pp. 269–298.

7. B. H. Neumann, Yet more on finite groups with few defining relations, Group theory (Singapore, 1987), de Gruyter, Berlin/New York, 1989, pp. 183–193.

8. R. G. Swan, Periodic resolutions for finite groups, Ann. of Math. 72 (1960), 267–291.

9. , K-theory of finite groups and orders, notes by E. G. Evans, Lecture Notes in Mathematics, vol. 149, Springer-Verlag, Berlin/New York, 1970.

10. C. B. Thomas and C. T. C. Wall, The topological spherical space form problem, I, Compositio Math. 23 (1971), 101–114.

11. J. A. Wolf, Spaces of constant curvature, McGraw-Hill, 1967.

Department of Mathematics, University of Aberdeen, Aberdeen AB24 3UE, Scotland.