Subgroups in which a Cohomology Class Splits

(Second Rough Draft – must reorder, finish last section.)

William A. Stein^{∗}

January 25, 1999

Abstract

Let G be a group, M a Z[G]-module, and c ∈ H^{q}(G, M). We study the indexes of subgroups H ⊂ G for which c restricts to 0. In particular, we investigate the situation when the greatest common divisor of the indexes is itself an index.

1 Introduction

Let G be a (profinite) group and let M be a (discrete) G-module. For any integer q ≥ 0 let H^{q}(G, M) denote the q-th cohomology group in the sense of [A]. Thus H^{q}(G, M) is an additive abelian group associated covariantly to M and contravariantly to G. If H is a finite index subgroup of G there is a restriction map res_{H} : H^{q}(G, M) → H^{q}(H, M). In this paper we study the set of indexes of subgroups H ⊂ G for which a given cohomology class restricts to 0.

Suppose q ≥ 1. Define the index of c ∈ H^{q}(G, M) to be

ind(c) = gcd{[G : H] : res_{H}(c) = 0}.

Say that c is a sharp cohomology class if there exists a subgroup H such that [G : H] = ind(c) and res_{H}(c) = 0. In words, “the index is attained,” or “the greatest common divisor of the indexes is itself an index,” or even “the gcd equals the minimum.” The triple (q, G, M) is called a sharp triple if every c ∈ H^{q}(G, M) is sharp. A group G is a sharp group if (q, G, M) is sharp for every G-module M and q ≥ 1. A Z-module M is a sharp module if (q, G, M) is sharp for every q ≥ 1 and group G with an action on M.

∗UC Berkeley, Department of Mathematics, Berkeley, CA 94720, USA.

In the first section we develop basic properties of the index of a cohomology class. Then strongly Sylow groups are introduced and shown to be sharp. We next derive a criterion for sharpness and use it to prove that quotients of sharp groups are sharp. Then we show that F_{p} × F_{p} is often a sharp module. Next we give examples showing how sharpness can fail, and some questions we were unable to answer. In the last section, sharpness is applied to arithmetic questions.

2 Sharp Groups

Let G be a profinite group, M a G-module, and H^{q}(G, M) the q-th cohomology group. In this section we assume that q ≥ 1. A subgroup H of G splits c if res_{H}(c) = 0. Let ord(c) denote the order of c as a group element.

Lemma 2.1. Let G be a finite group, M a G-module, and q ≥ 1. Then H^{q}(G, M) is annihilated by #G.

Proof. This is proved as Corollary 1 of Section 6 of [A] by using that cores ∘ res_{{1}} = #G.

Proposition 2.2. ord(c) | ind(c) and they have the same prime factors.

Proof. Suppose H ⊂ G splits c. There is a map cores_{H} : H^{q}(H, M) → H^{q}(G, M) such that cores_{H} ∘ res_{H}(c) = [G : H] · c. Thus [G : H] · c = 0 and hence the order of c divides ind(c). Let p be a prime and let G_{p} be a Sylow p-subgroup of G, so the (generalized) index [G : G_{p}] is coprime to p and G_{p} is a (pro) p-group. If p does not divide the order of c then 0 = res_{G_{p}}(c) ∈ H^{q}(G_{p}, M) so that ind(c)|[G : G_{p}] is not divisible by p.

Let t ∈ G. Consider the inner automorphism f : G → G given by f(s) = tst^{−1}. This turns M into a new G-module, denoted by M^{t}, and gives a homorphism f^{∗} : H^{q}(G, M) → H^{q}(G, M^{t}). The map g : M^{t} → M sending a to t^{−1}a defines an isomorphism M^{t} → M and induces a map g_{∗} : H^{q}(G, M^{t}) → H^{q}(G, M).

Proposition 2.3. g_{∗} ∘ f^{∗} is the identity map H^{q}(G, M) → H^{q}(G, M).

Proof. This is Proposition 3 of [A]. The proof uses dimension shifting. (There is a typo in (4.2) of [A]: the first A should be replaced by A^{t}.)

Lemma 2.4. If H splits c then every G-conjugate of H splits c.

Proof. Let t ∈ G, let f : G → G be the inner automorphism f(s) = tst^{−1} and let g be the map M → M sending a to t^{−1}a. As in section 4 of [A] we use functoriality of H^{q}( , ) to obtain a commuting diagram

H^{q}(G, M) ⟶^{res_{H}}  H^{q}(H, M)
↓ f^{∗}  ↓ f^{∗}
H^{q}(G, M^{t}) ⟶^{res_{f^{−1}(H)}} H^{q}(f^{−1}(H), M^{t})
↓ g_{∗}  ↓ g_{∗}
H^{q}(G, M) ⟶^{res_{f^{−1}(H)}} H^{q}(f^{−1}(H), M)

By (2.3), the vertical map g_{∗} ∘ f^{∗} on the left is the identity map. Thus if res_{H}(c) = 0 then so must res_{f^{−1}(H)}(c) = 0, which proves the Lemma.

The following lemma shows that if, for each prime p, we let G_{p} be a Sylow p-subgroup of G, then ind(c) = ∏_{p} ind(res_{G_{p}}(c)). For n an integer and p a prime, let part_{p}(n) = p^{ord_{p}(n)}.

Lemma 2.5. Let G_{p} be a Sylow p-subgroup of G. Then part_{p}(ind(c)) = ind(res_{G_{p}}(c))

Proof. Suppose H ⊂ G splits c and let H_{p} ⊂ H be a Sylow p-subgroup. Then ord_{p}([G : H]) = ord_{p}([G : H][H : H_{p}]) = ord_{p}([G : H_{p}]). By Sylow’s theorems, there exists t ∈ G so that tH_{p}t^{−1} ⊂ G_{p}. By (2.4), tH_{p}t^{−1} also splits c. Thus ind(res_{G_{p}}(c))|[G_{p} : t^{−1}H_{p}t] so, since ord_{p}([G : t^{−1}H_{p}t]) = ord_{p}([G : H_{p}]) = ord_{p}([G : H]), it follows that ind(res_{G_{p}}(c))| ind(c). On the other hand, ord_{p}(ind(c)) ≤ ord_{p}(res_{G_{p}}(c)) because we can compute an upper bound on ord_{p}(ind(c)) using subgroups of G_{p}.

Proposition 2.6. For any subgroup H of G, ind(res_{H}(c))| ind(c).

Remark: If H′ ⊂ G splits c so does H ∩ H′. There is a natural injection of sets H/(H ∩ H′) ↪ G/H′ so [H : H ∩ H′] ≤ [G : H′]. But, it need not be the case that [H : H ∩ H′] | [G : H′], for example, let H and H′ be distinct order 3 subgroups of S_{4}. Then 3 = [H : H ∩ H′] whereas [G : H′] = \frac{24}{3} = 8.

Proof. It suffices to show the divisibility for each prime p. Let H_{p} be a Sylow p-subgroup of H, and use the Sylow theorems to extend H_{p} to a Sylow psubgroup G_{p} of G. If H′ is a subgroup of G_{p} which splits c, then H′ ∩ H_{p} splits res_{H}(c). Since [H_{p} : H′ ∩ H_{p}] ≤ [G_{p} : H′] and each is a power of p, [H_{p} : H′ ∩ H_{p}] | [G_{p} : H′]. Thus

part_{p}(ind(res_{H}(c))) = ind(res_{H_{p}}(c)) | ind(res_{G_{p}}(c)) = part_{p}(ind(c)).

Using the technique of dimension shifting we now show that is is only necessary to require sharpness of (1, G, M) in the definition.

Proposition 2.7. If (1, G, M) is sharp for every M then G is sharp.

Proof. We proceed by induction. Let M be a G-module and q > 1. Assume that (q − 1, G, M′) is sharp for every G-module M′. Consider the G-module M^{∗} = Hom(Z[G], M). There is a natural injection M → M^{∗} which maps a ∈ A to φ_{a}, where φ_{a} is defined by φ_{a}(g) = ga. Hence we have an exact sequence of G-modules

0 → M → M^{∗} → M′ → 0.

Since M^{∗} is co-induced, it follows that δ : H^{q−1}(G, M′) → H^{q}(G, M) is an isomorphism. Since a co-induced G-module is also a coinduced H-module for any subgroup H of G we obtain commuting diagrams

H^{q−1}(G, M′) → H^{q}(G, M)
↓ res_{H}  ↓ res_{H}
H^{q−1}(H, M′) → H^{q}(H, M)

In particular, the set of subgroups H of G which split c ∈ H^{q}(G, M) is the same as the set which split δ(c) ∈ H^{q−1}(G, M′). By our inductive assumption, (q − 1, G, M′) is sharp so c is sharp, and hence (q, G, M) is sharp.

Lemma 2.8. If G is a p-group then G is sharp.

Proof. The greatest common divisor of a set of p-powers equals the smallest so (1, G, M) is sharp.

A group G is strongly Sylow if it has the following property: Given subgroups H_{p} ⊂ G_{p} for each prime p, there exists H ⊂ G such that, for all p, a Sylow p-subgroup of H is G-conjugate to H_{p}. Nilpotent groups are strongly Sylow because they are the direct product of their Sylow p-subgroups. The group S_{3} is strongly Sylow but not nilpotent.

Theorem 2.9. If G is strongly Sylow then G is sharp.

Proof. Let M be a G-module. We must show that (1, G, M) is sharp, so let c ∈ H^{1}(G, M). For each prime p, (2.8) gives a subgroup H_{p} of G_{p} so that ind(res_{G_{p}}(c)) = [G_{p} : H_{p}]. Choose H as guaranteed by the property of G being strongly Sylow. For each prime p, (2.4) insures that every p-Sylow subgroup of H splits c, i.e., ind(res_{H_{p}}(c)) = 1. By (2.5), ind(res_{H}(c)) = 1, so by (2.2), res_{H}(c) = 0. Since for each prime p, ord_{p}([G : H]) = ord_{p}([G : H_{p}]) = ord_{p}([G_{p} : H_{p}]) = ord_{p}(ind(c)) it follows that [G : H] = ind(c) so that c is sharp. Note, this proof works equally well for c ∈ H^{q}(G, M).

Let f : G → M be a 1-cocycle and define ker(f) := {σ ∈ G : f(σ) = 0}.

Proposition 2.10. ker(f) is a subgroup of G and [G : ker(f)] = #f(G).

Proof. If σ_{1}, σ_{2} ∈ K then f(σ_{1}σ_{2}) = f(σ_{1}) + σ_{1}f(σ_{2}) = 0 and 0 = f(1) = f(σ_{1}^{−1}σ) = f(σ_{1}^{−1}) + σ_{1}^{−1}f(σ_{1}) = f(σ_{1}^{−1} so K is a subgroup. If τ ∈ G and σ ∈ K, then f(τσ) = f(τ) + τf(σ) = f(τ) so f gives rise to a welldefined map of sets \overline{f} : G/K → M. If τ_{1}, τ_{2} ∈ G and \overline{f}(τ_{1}K) = \overline{f}(τ_{2}K), then f(τ_{1}) = f(τ_{2}) (by definition of \overline{f} so f(τ_{1}^{−1}τ_{2}) = f(τ_{1}^{−1}) + τ_{1}^{−1}f(τ_{2}) = f(τ_{1}^{−1}) + τ_{1}^{−1}f(τ_{1}) = f(τ_{1}^{−1}τ_{1}) = 0. Thus \overline{f} is injective and #f(G) = #f(G/K) = #(G/K) = [G : K].

Example 2.11. Caution: ker(f) need not be a normal subgroup of G. For example, let G = S_{3} and M = Ze_{1}⊕Ze_{2}⊕Ze_{3} with the natural permutation action. Let f be the 1-cocycle determined by e_{1}, so f(σ) = e_{σ(1)} − e_{1}. Then ker(f) = {ι, (23)} is not a normal subgroup of G. Note that 3 = [S_{3} : ker(f)] = #f(S_{3}) = #{0, e_{2} − e_{1}, e_{3} − e_{1}}, as predicted by the Proposition.

Lemma 2.12. Fix c ∈ H^{1}(G, M). A subgroup H of G splits c iff H ⊂ ker(f) for some cocycle f ∈ c.

Proof. If H ⊂ ker(f) then res_{H}(c) is represented by f|_{H} which is the 0cocycle, hence H splits c. Suppose conversely that we know only that H splits c. If f ∈ c, then f|_{H} ∈ res_{H}(c) = 0. Hence there is x ∈ M so that f(σ) = σ(x) − x for all σ ∈ H. The cocycle g : σ ↦ f(σ) − (σ(x) − x) represents c and g|H = 0 so H ⊂ ker(g).

Thus the set of kernels of cocycles is “cofinal” in the set of subgroups which split c, and so they can be used to compute the index.

Proposition 2.13. If c ∈ H^{1}(G, M) then

ind(c) = gcd{[G : ker(f)] : f ∈ c} = gcd{#f(G) : f ∈ c}.

Proof. The first equality follows from (2.12) and the fact that ker(f) splits c, and the second from (2.10).

Let N be a subgroup of ker(G → Aut M), fix c ∈ H^{1}(G, M) and choose f ∈ c.

Lemma 2.14. The image f(N) is a G-submodule of M and ker(f) ∩ N is normal in G, neither depends on the choice of f ∈ c. Further, f(G) is a union of cosets of f(N).

Proof. Since N acts trivially, the cocycle f : N → M is a homomorphism of groups and thus f(N) is a subgroup of M. If σ ∈ G and x ∈ N then, using that x acts trivially on M, we have σf(x) = f(σσ^{−1}) + σf(x) = f(σ) + σf(σ^{−1}) + σf(x) = f(σ) + σf(x) + σxf(σ^{−1}) = f(σx) + σxf(σ^{−1}) = f(σxσ^{−1}) ∈ f(N). This proves that f(N) is a G-submodule of M. If f(x) = 0 then f(σxσ^{−1}) = σf(x) = 0 for all σ ∈ G so ker(f) ∩ N is normal in G. Since N acts trivially on M, coboundaries restrict to 0 on N so that f(N) and ker(f) ∩ N do not depend on the choice of f ∈ c. Finally, f(G) = ∪_{σ∈G}f(Nσ) = ∪_{σ∈G}(f(N) + f(σ)) is a union of cosets of f(N).

Lemma 2.15. Let f be a 1-cocycle and H, H′ subgroups of G which split f. Then the join H.H′ also splits f.

Proof. By (2.10) ker(f) is a subgroup of G. It contains both H and H′ so it contains their join.

Warning: It is not true in general that if a cohomology class c is split by subgroups H and H′ then it is split by H.H′. See (4.5).

If H is a normal subgroup of G then the sequence

0 → H^{1}(G/H, M^{H}) ⟶^{inf} H^{1}(G, M) ⟶^{res} H^{1}(H, M)

is exact. Upon setting H = ker(f) ∩ N we can define several cohomology classes naturally associated to c ∈ H^{1}(G, M).

Theorem 2.16. Let c ∈ H^{1}(G, M), N ⊂ ker(G → Aut M), and f ∈ c. Then the following are equivalent:

1. c ∈ H^{1}(G, M) is sharp.

2. c′ ∈ H^{1}(G/(ker(f) ∩ N), M) is sharp.

3. \overline{c} ∈ H^{1}(G, M/f(N)) is sharp.

4. \overline{c}′ ∈ H^{1}(G/N, M/f(N)) is sharp.

Proof. (1) ⟺ (2): Suppose H ⊂ G splits c. By (2.12) there exists g ∈ c so that H ⊂ ker(g), and by (2.14) ker(g) also contains ker(f)∩N = ker(g)∩N. Thus g, and hence c, is split by the join H.(ker(f) ∩ N). There is a oneto-one index preserving correspondence between maximal subgroups of G splitting c and maximal subgroups of G/(ker(f) ∩ N) splitting the class c′ ∈ H^{1}(G/(ker(f) ∩ N), M) defined by c.

(1) ⟺ (3): Since any two cocycles in \overline{c} differ by the coboundary of an element of M/c(N), and such a coboundary is the image of the coboundary of an element of M, the set \{f : f ∈ c\} surjects onto the set \{g : g ∈ \overline{c}\}. If f ∈ c, (2.14) implies that f(G) is a union of cosets of N′ = f(N) and N′ does not depend on f, hence \{#g(G) : g ∈ \overline{c}\} = \{#f(G)/#f(N) : f ∈ c\}. Sharpness means that the set of integers above satisfies min = gcd. This is true of the left hand side iff it is true of the right hand side iff it is true of {#f(G) : f ∈ c}.

(3) ⟺ (4): If g ∈ \overline{c} then by (2.14) g(N) = f(N) = \{0\} ⊂ M/f(N), so N ⊂ ker(g). Useing (2.12) we see that there is a bijection between the maximal subgroups of G splitting \overline{c} and the maximal subgroups of G/N which split \overline{c}′.

We can now deduce that quotients of sharp groups are sharp.

Corollary 2.17. If G is sharp and G → G′ is a surjective then G′ is sharp.

Proof. Let M be a G′ module. Make M a G-module via G → G′. Let c ∈ H^{1}(G′, M), f ∈ c, and let N := ker(G → G′) ⊂ ker(G → Aut M). The map G → G′ gives rise to a map π : H^{1}(G′, M) → H^{1}(G, M): on cocycles it sends f : G′ → M to the map G → M obtained by composing with G → G′. Since G is sharp, π(c) ∈ H^{1}(G, M) is sharp, so by (2.16) the class it defines in H^{1}(G/N, M/f(N)) = H^{1}(G′, M) is sharp. But the latter class is c, so c is sharp and by (2.7) we are done.

Theorem 2.18. Fix a G-module M and suppose the action of G on M factors through a sharp quotient G′. Then (q, G, M) is sharp for every q.

Proof. We first prove the theorem in the case q = 1 and then apply dimension shifting. By (2.16), (1, G, M) is sharp iff (1, G′, M/f(N)) is sharp for all 1-cocycles f : G → M. This is the case because G′ is sharp.

Next suppose the theorem is true for q − 1. Let M′ and M^{∗} be as in the proof of (2.7) so there is an exact sequence of G-modules

0 → M → M^{∗} → M′ → 0.

The action of G on M^{∗} factors through G′: If h ∈ ker(G → Aut M) and a ∈ M, then for any g ∈ G, h.φ_{a}(g) = φ_{a}(gh) = gha = ga = φ_{a}(g). Thus the action of G on the quotient M′ = M^{∗}/M factors through G′. As in the proof of (2.7), we see that the set of subgroups H of G which split a fixed c ∈ H^{q}(G, M) is the same as the set which split δ(c) ∈ H^{q−1}(G, M′). By our inductive assumption, using that G acts on M’ through G′, we see that (q − 1, G, M′) is sharp. Thus c is sharp, and hence (q, G, M) is sharp.

3 Sharp Modules

Proposition 3.1. If every (compact profinite) subgroup of Aut M is sharp then M is a sharp module.

Proof. We must show that (q, G, M) is sharp for all groups G and G-module structurs on M. This follows from (2.18) since G acts through a sharp subgroup of Aut M.

We call a Z-module M q-sharp if (q, G, M) is sharp for all groups G. Unlike in the case of groups, we do not suspect that q-sharpness for some q is equivalent to q-sharpness for all q. We have only the following week result.

Proposition 3.2. If M is 1-sharp and M′ is a direct summand of M then M′ is also 1-sharp.

Proof. Write M = M′ ⊕ M′′. Suppose G is a group with an action on M′. Extend G to act trivially on M′′. There is an injection H^{1}(G, M′) ⟶^{ι} H^{1}(G, M′ ⊕ M′′). If c ∈ H^{1}(G, M′) and f ∈ c then the set of cardinalities of images of cocycles in ι(c) is {#(ι(f) + δ(a, b))(G) : (a, b) ∈ M} where δ(a, b) is the cocycle determined by (a, b). Since (ι(f) + δ(a, b))(σ) = (f(σ), 0) + (σ(a), b) − (a, b) = (f(σ) + σ(a) − a, 0) we see that this set of cardinalities is the same as {#g(G) : g ∈ c}. Since M is sharp, ι(c) is sharp so by (2.13), c is sharp.

Corollary 3.3. A cyclic abelian group is a sharp module.

Proof. The automorphism group of a cyclic group is finite cyclic, hence every subgroup is sharp. Now apply (3.1).

Let F_{p} = Z/pZ.

Theorem 3.4. If G ⊂ GL_{2}(F_{p}) and q ≥ 1 then (q, G, F_{p} ⊕ F_{p}) is sharp.

Proof. Let M = F_{p} ⊕ F_{p}. If p = 2, GL_{2}(F_{p}) = S_{3} and we are done because S_{3} is sharp. Thus assume p > 2.

Suppose G contains a scalar t ∈ F_{p}^{×}\{1}. By (2.3) the map H^{q}(G, M) → H^{q}(G, M) induced by conjugation by t is the identity map. It is also multiplication by t on the F_{p} vector space H^{q}(G, M) so since t ≠ 1, we conclude that H^{q}(G, M) = 0. If p ∤ #G then (2.1) implies that H^{q}(G, M) = 0. We thus assume G ⊂ SL_{2}(F_{p}) and p | #G.

Case 1: G contains exactly 1 Sylow p-subgroup: Let S be the Sylow p-subgroup of G. Since all Sylow p-subgroups of G are conjugate, S is normal. The cohomology group H^{2}(G/S, S) classifies short exact sequences 0 → S → E → G/S → 1. Since p ∤ [G : S] (2.1) implies that H^{2}(G/S, S) = 0 so the short exact sequence 0 → S → G → G/S → 1 is trivial, hence splits. The image of G/S under this splitting is a subgroup N ⊂ G whose order is coprime to p and whose index equals #S = p. Let c ∈ H^{1}(G, M) be nonzero. Since M = F_{p} × F_{p}, c has order p. Since the order of N is coprime to p (2.1) implies that res_{N}(c) = 0. But [G : N] = p so by (2.2) the index of c is p and this index is attained by restricting to N so that c is sharp.

Case 2: G contains more than 1 Sylow p-subgroup: The subgroup S = \{( \begin{matrix}1 & ∗ \\ 0 & 1 \end{matrix} )\} is a Sylow p-subgroup of SL_{2}(F_{p}). All Sylow p-subgroups of SL_{2}(F_{p}) are conjugate so they are of this form with respect to some basis. Thus a Sylow p-subgroup is determined by the line it fixes and we have a bijection

{ Sylow p-subgroups of SL_{2}(F_{p}) } = { lines in F_{p} × F_{p} }.

There are p+1 lines in F_{p}×F_{p} hence the same number of Sylow p-subgroups. Since G contains more than 1 Sylow p-subgroup, it must contain the subgroup H generated by all of them. The set of Sylow p-subgroups is closed under conjugation so H is a normal subgroup of SL_{2}(F_{p}).

The order n of H divides the order of SL_{2}(F_{p}) which is p(p+1)(p−1). The union of the p+1 Sylow p-subgroups of SL_{2}(F_{p}) has cardinality p(p+1)−p = p^{2} (the identity is counted p + 1 times, but there is no other overcounting). Thus n ≥ p^{2}. Since n | p(p + 1)(p − 1) it is exactly divisible by p so n can’t equal p^{2} and hence n ≥ p^{2} + 1 = p(p + 1) so that p(p + 1)(p − 1)/n ≤ p − 1. Therefore the quotient SL_{2}(F_{p})/H is a group of order ≤ p − 1.

By the remark in VII.6 of [S] there is an exact sequence

0 → H^{1}(SL_{2}(F_{p})/G, M^{G}) ⟶^{inf} H^{1}(SL_{2}(F_{p}), M) ⟶^{res} H^{1}(G, M)
→ H^{2}(SL_{2}(F_{p})/G, M^{G})

The order of SL_{2}(F_{p})/H is ≤ p − 1 so by (2.1) H^{2}(SL_{2}(F_{p})/H, M) = 0. Furthermore, −1 ∈ SL_{2}(F_{p}) so H^{1}(SL_{2}(F_{p}), M) = 0 and hence H^{1}(G, M) = 0. We are now allowed to write down the exact sequence

0 → H^{2}(SL_{2}(F_{p})/G, M^{G}) ⟶^{inf} H^{2}(SL_{2}(F_{p}), M) ⟶^{res} H^{2}(G, M)
→ H^{3}(SL_{2}(F_{p})/G, M^{G})

Argueing in the same way shows that H^{2}(G, M) = 0. Continuing inductively we conclude that H^{q}(G, M) = 0, so (q, G, M) is sharp.

Corollary 3.5. M = F_{p} ⊕ F_{p} is a 1-sharp module.

Proof. Let G be an arbitrary group equipped with an action on M, and fix c ∈ H^{1}(G, M). We must show that c is sharp. Let N = ker(G → Aut M) and f be a cocycle representing c. By (2.14), f(N) is a G-submodule of Aut M and by (2.16) c is sharp iff the class d = \overline{c}′ ∈ H^{1}(G/N, M/f(N)) is sharp. If #f(N) > 1 the quotient M/f(N) = (F_{p} ⊕ F_{p})/f(N) is cyclic so (3.3) implies d is sharp. The remaining case is #f(N) = 1. Since G/N is a subgroup of Aut M = GL_{2}(F_{p}) we may apply (3.4) to conclude that d is sharp.

4 Examples

Let G be a finite group. There is a natural map Z[G] → Z sending ∑ a_{g}g to ∑ a_{g} ∈ Z. The augmentation ideal I_{G} ⊂ Z[G] is the kernel of this map so there is an exact sequence 0 → I_{G} → Z[G] → Z → 0. Since Z[G] is cohomologically trivial, H^{1}(G, I_{G}) = coker(H^{0}(G, Z[G]) → H^{0}(G, Z)) = Z/(#G)Z. If H is a subgroup of G then Z[G] is cohomologically trivial as an H module so we have an isomorphism H^{1}(H, I_{G}) ≅ coker(H^{0}(H, Z[G]) → H^{0}(H, Z)) ≅ Z/(#H)Z = H^{1}(H, I_{H}) Thus the restriction map res_{H} agrees with the natural map Z/(#G)Z → Z/(#H)Z.

Proposition 4.1. (1, G, I_{G}) is sharp iff the set {#H : H ⊂ G} is closed under taking least common multiples of subsets (modulo #G).

Proof. In the following, we work module #G.

First assume that {#H : H ⊂ G} is closed under lcm. Let m ∈ H^{1}(G, I_{G}) = Z/#GZ. Then res_{H}(m) = 0 iff #H | m. The index of m is #G/ lcm{#H : #H | m}, so m is sharp iff lcm{#H : #H | m} lies in {#H : #H | m}. Since the lcm divides m and {#H : H ⊂ G} is closed under lcm, this is the case.

Next assume that (1, G, I_{G}) is sharp. Let a_{1}, …, a_{n} be elements of {#H : H ⊂ G} and let m = lcm(a_{1}, …, a_{n}). By sharpness, lcm{#H : #H | m} lies in {#H : #H | m}. Since a_{1}, …, a_{n} ∈ {#H : #H | m}, this lcm is m so it follows that there is H ⊂ G so that m = lcm{#H : #H | m} = #H, as desired.

Example 4.2. The subgroups of A_{4} have orders 1, 2, 3, 4, 12 so (1, A_{4}, I_{A_{4}}) is not sharp.

Example 4.3. If M and M′ are sharp then M ⊕M′ need not be sharp. Let n be the largest integer so that Z^{n} is sharp. By (3.3), n ≥ 1, and since Z^{11} is the Z-module underlieing the augmentation ideal I_{A_{4}} is not sharp, n ≤ 10. Then Z ⊕ Z^{n} is not sharp but both Z and Z^{n} are.

Example 4.4. Even if (1, G, M) and (1, G, M′) are both sharp, (1, G, M⊕M′) need not be sharp. The subgroups of S_{4} have orders 1, 2, 3, 4, 6, 8, 12, and 24, which is cosed under lcm modulo 24. Thus I = I_{S_{4}}, we have that (1, S_{4}, I) is sharp. Let c ∈ H^{1}(S_{4}, I) correspond to 6 under the isomorphism H^{1}(S_{4}, I) ≅ Z/24Z. Let φ : S_{4} → Z/2Z be the nontrivial (sign) homomorphism and let d = (c, φ) ∈ H^{1}(S_{4}, I ⊕ (Z/2Z)). Then a subgroup which splits d must split φ hence be contained in A_{4}. Thus the subgroups splitting d are the subgroups of A_{4} which split c. But res_{A_{4}}(c) ∈ H^{1}(A_{4}, I_{S_{4}}) = Z/12Z is not sharp, so d is not sharp.

Example 4.5. Let G = S_{3} and c = 2 ∈ Z/6Z = H^{1}(G, I_{G}). If H is any of the 3 subgroups of of order 2 then res_{H}(c) = 0. If H and H′ are distinct subgroups of order 2 they both split c but their join is S_{3} which does not split c. Thus the analogue of (2.15) for cohomology classes is false.

Example 4.6. If G fits into an exact sequence 0 → G′ → G → G′′ → 0 with G′ and G′′ sharp, must G be sharp? No, as the example 0 → V_{4} → A_{4} → C_{3} → 0 shows.

5 Questions

If G is a sharp group is every subgroup of G sharp? Are direct sums of sharp groups sharp? Give a purely group theoretic characterization of the class of finite sharp groups. Are there sharp groups which are not strongly Sylow? Are strongly Sylow groups closed under either direct sums or quotients?

Are either submodules or quotients of sharp modules necessarily sharp? What is the smallest integer n so that Z^{n} is a sharp module? For a given prime p, what is the smallest integer n so that F_{p}^{n} is a sharp module? Are any divisible abelian groups sharp modules, for example Q? If a Z-module M is q-sharp for some q ≥ 1 must it be q-sharp for all q ≥ 1.

6 Arithmetic Applications

This section is under construction.

In the context of abelian varieties, Lichtenbaum [L] has studied the quotient ind(c)/ ord(c). Cassels [C] gave an elliptic curves E/Q and a class c ∈ H^{1}(Q, E(\overline{Q})) for which ord(c) ≠ ind(c), but showed that if c ∈ � (E) then ord(c) = ind(c).

Proposition 6.1. Let M be an abelian group, K a number field, and let Gal(\overline{K}/K) act on M via the cyclotomic character. Then (1, Gal(\overline{K}/K), M) is sharp.

Proposition 6.2. Let K be a number field, a ∈ K, and m a positive integer. Factor the polynomial X^{m} − a as a product ∏ g_{i}(X) of irreducible polynomials g_{i}(X) ∈ K[X]. Then the minimum of the degrees of the g_{i} divide the other degrees.

We were motived to study sharpness by the following consequence of the Riemann-Roch theorem.

Theorem 6.3. Let E be an elliptic curve of a field K and G = Gal(\overline{K}/K). Then (1, G, E(\overline{K})) is sharp.

We were unable to answer the analogous question for higher dimensional abelian varieties.

Question 6.4. Does there exist a nonsingular plane quartic X over a field K possessing no points of degree dividing 3 over any quadratic extension of K, but possessing a degree 6 point over K.

The cohomology group H^{1}(K, Jac(X)) associated to such an X would not be sharp.

References

[A] Atiyah, Wall, Cohomology of Groups, in Algebraic Number Theory, Ed. J.W.S. Cassels, A. Fröhlich, Academic Press (1967).

[C] J.W.S. Cassels, Arithmetic on curves of genus 1. V. Two counterexamples. J. London Math. Soc. 38 (1963) 244–248.

[LT] S. Lang, J. Tate, Principal homogeneous spaces over abelian varieties, 80 (1958) 659–684.

[L] S. Lichtenbaum, Duality theorems for curves over p-adic fields. Invent. Math. 7 (1969) 120–136.

[S] J.P. Serre, Local Fields, Springer GTM 67 (1995).