% sharp.tex
\documentclass[11pt]{article}
\include{macros}
\title{Subgroups in which a Cohomology Class Splits\\
{\large (Second Rough Draft -- must reorder, finish last section.)}}
\author{William A. Stein\footnote{UC Berkeley, Department of
Mathematics, Berkeley, CA  94720, USA.}}

\begin{document}
\maketitle
\begin{abstract}
Let $G$ be a group, $M$ a $\Z[G]$-module, and $c\in H^q(G,M)$.
We study the indexes of subgroups $H\subset G$
for which $c$ restricts to $0$.  In particular, we investigate
the situation when the greatest common divisor of the indexes
is itself an index.
\end{abstract}
\section{Introduction}
Let $G$ be a (profinite) group and let $M$ be a (discrete) $G$-module.
For any integer $q\geq 0$ let
$H^q(G,M)$ denote the $q$-th cohomology group in the sense of \cite{atiyah}.
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 $\res_H:H^q(G,M)\ra H^q(H,M)$.
In this paper we study the set of indexes of subgroups $H\subset G$
for which a given cohomology class restricts to $0$.

Suppose $q\geq 1$. Define the {\bf index} of $c\in H^q(G,M)$ to be
$$\ind(c) = \gcd\{[G:H] : \res_H(c)=0\}.$$
Say that $c$ is a {\bf 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 {\bf sharp triple} if every $c\in H^q(G,M)$
is sharp.
A group $G$ is a {\bf sharp group} if $(q,G,M)$ is sharp for every  $G$-module
$M$ and $q\geq 1$.
A $\Z$-module $M$ is a {\bf sharp module} if $(q,G,M)$ is sharp
for every $q\geq 1$ and group $G$ with an action on $M$.

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 $\Fp\cross \Fp$ 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.

\section{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 $q\geq 1$.
A subgroup $H$ of $G$ {\bf splits} $c$ if $\res_H(c)=0$.
Let $\ord(c)$ denote the order of $c$ as a group element.

\begin{lemma}\label{orderannihilate}
Let $G$ be a finite group, $M$ a $G$-module, and
$q\geq 1$.  Then $H^q(G,M)$ is annihilated by $\# G$.
\end{lemma}
\begin{proof}
This is proved as Corollary 1 of Section 6 of \cite{atiyah} by using that
$\cores\circ \res_{\{1\}} = \#G$.
\end{proof}

\begin{proposition}\label{ordind}
$\ord(c)\mid \ind(c)$ and they have the same prime factors.
\end{proposition}
\begin{proof}
Suppose $H\subset G$ splits $c$.  There is a map $\cores_H:H^q(H,M)\ra H^q(G,M)$ such
that  $\cores_H \circ\res_H (c) = [G:H]\cdot c$. Thus $[G:H]\cdot c=0$ 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
$0=\res_{G_p}(c)\in H^q(G_p,M)$ so that $\ind(c)|[G:G_p]$ is
not divisible by $p$.
\end{proof}

Let $t \in G$. Consider the inner automorphism $f:G\ra G$
given by $f(s)=tst^{-1}$.  This turns $M$ into a new
$G$-module, denoted by $M^t$, and gives a homorphism
$f^{*}:H^q(G,M)\ra H^q(G,M^t)$.  The map $g:M^t\ra M$ sending $a$ to $t^{-1}a$ defines
an isomorphism $M^t\ra M$ and induces a map
$g_{*}:H^q(G,M^t)\ra H^q(G,M)$.
\begin{proposition}\label{innerpair}
$g_{*}\circ f^{*}$ is the identity map $H^q(G,M)\ra H^q(G,M)$.
\end{proposition}
\begin{proof}
This is Proposition 3 of \cite{atiyah}.  The proof uses dimension shifting.
(There is a typo in
(4.2) of \cite{atiyah}: the first $A$ should be replaced by $A^{t}$.)
\end{proof}

\begin{lemma}\label{conjsplit}
If $H$ splits $c$ then every $G$-conjugate of $H$ splits $c$.
\end{lemma}
\begin{proof}
Let $t \in G$, let $f:G\ra G$ be the inner automorphism $f(s)=tst^{-1}$ and
let $g$ be the map $M\ra M$ sending $a$ to $t^{-1}a$.
As in section 4 of \cite{atiyah} we use functoriality of $H^q(\,,\,)$ to
obtain a commuting diagram
$$\begin{matrix} H^q(G,M) & \xrightarrow{\res_H} & H^q(H,M)\\ \downarrow f^{*} & & \downarrow f^{*} \\ H^q(G,M^{t}) & \xrightarrow{\res_{f^{-1}(H)}} & H^q(f^{-1}(H),M^t)\\ \downarrow g_* & & \downarrow g_*\\ H^q(G,M) & \xrightarrow{\res_{f^{-1}(H)}} & H^q(f^{-1}(H),M) \end{matrix}$$

By (\ref{innerpair}),
the vertical map $g_{*}\circ f^{*}$ on the left
is the {\em identity} map.
Thus if $\res_H(c)=0$ then so must
$\res_{f^{-1}(H)}(c)=0$, which proves the Lemma.
\end{proof}

The following lemma shows that if, for each prime $p$,
we let $G_p$ be a Sylow $p$-subgroup of $G$, then
$\ind(c) = \prod_p\ind(\res_{G_p}(c))$.
For $n$ an integer and $p$ a prime, let $\prt_p(n)=p^{\ord_p(n)}$.
\begin{lemma}\label{indpsylow}
Let $G_p$ be a Sylow $p$-subgroup of $G$.
Then $\prt_p(\ind(c))=\ind(\res_{G_p}(c))$
\end{lemma}
\begin{proof}
Suppose $H\subset G$ splits $c$ and let $H_p\subset H$ 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 $t\in G$ so that $t H_pt^{-1} \subset G_p$. By (\ref{conjsplit}),
$tH_p t^{-1}$ 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, $\ord_p(\ind(c))\leq \ord_p(\res_{G_p}(c))$
because we can compute an upper bound on $\ord_p(\ind(c))$
using subgroups of $G_p$.
\end{proof}

\begin{proposition}
For any subgroup $H$ of $G$,
$\ind(\res_H(c))|\ind(c)$.
\end{proposition}
\par\noindent
{\bf Remark: } If $H'\subset G$ splits $c$ so does $H\intersect H'$. There is
a natural injection of sets $H/(H\intersect H')\hookrightarrow G/H'$ so
$[H:H\intersect H'] \leq [G:H']$. But, it need not be the case that
$[H:H\intersect H'] \mid [G:H']$, for example, let $H$ and $H'$
be distinct order $3$ subgroups of $S_4$.  Then
$3=[H:H\intersect H']$ whereas $[G:H']=\frac{24}{3}=8$.
\begin{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 $p$-subgroup $G_p$ of $G$. If $H'$ is a subgroup of $G_p$ which
splits $c$, then $H'\intersect H_p$ splits $\res_{H}(c)$.  Since
$[H_p:H'\intersect H_p] \leq [G_p : H']$ and each is a power of $p$,
$[H_p:H'\intersect H_p] \mid [G_p : H']$.  Thus
$$\prt_p(\ind(\res_{H}(c)))=\ind(\res_{H_p}(c))\mid \ind(\res_{G_p}(c))=\prt_p(\ind(c)).$$
\end{proof}

%\begin{definition} A group $G$ is {\bf sharp} if $(q,G,M)$ is sharp
%for every $G$-module $M$ and $q\geq 1$.
%\end{definition}

Using the technique of dimension shifting we now show that is is only necessary to
require sharpness of $(1,G,M)$ in the definition.
\begin{proposition}\label{sharpshift}
If $(1,G,M)$ is sharp for every $M$ then $G$ is sharp.
\end{proposition}
\begin{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
$M\ra M^{*}$ which maps $a\in A$ to $\vphi_a$, where $\vphi_a$ is defined
by $\vphi_a(g) = ga$.  Hence we have an exact sequence of $G$-modules
$$0\ra M \ra M^{*} \ra M' \ra 0.$$
Since $M^{*}$ is co-induced, it follows that
$\delta:H^{q-1}(G,M')\ra H^q(G,M)$ 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
$$\begin{matrix}H^{q-1}(G,M') & \ra & H^q(G,M)\\ \downarrow \res_H & & \downarrow \res_H\\ H^{q-1}(H,M') & \ra & H^q(H,M) \end{matrix}$$
In particular, the set of subgroups $H$ of $G$ which
split $c\in H^q(G,M)$ is the same as the set which split
$\delta(c)\in H^{q-1}(G,M')$.  By our inductive assumption,
$(q-1,G,M')$ is sharp so $c$ is sharp, and hence $(q,G,M)$ is
sharp.

\end{proof}

\begin{lemma}\label{psharp}
If $G$ is a $p$-group then $G$ is sharp.
\end{lemma}
\begin{proof}
The greatest common divisor of a set of $p$-powers equals the smallest
so $(1,G,M)$ is sharp.
\end{proof}

A group $G$ is {\bf strongly Sylow} if it has the following property:
Given subgroups $H_p\subset G_p$ for each prime $p$, there exists
$H\subset G$ 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.

\begin{theorem}\label{strongsylowone}
If $G$ is strongly Sylow then $G$ is sharp.
\end{theorem}
\begin{proof}
Let $M$ be a $G$-module.
We must show that $(1,G,M)$ is sharp, so
let $c\in H^1(G,M)$.
For each prime $p$, (\ref{psharp}) 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 $p$, (\ref{conjsplit}) insures
that every $p$-Sylow subgroup of $H$ splits $c$, i.e.,
$\ind(\res_{H_p}(c))=1$.
By (\ref{indpsylow}), $\ind(\res_H(c))=1$, so
by (\ref{ordind}),  $\res_H(c)=0$. 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 $c\in H^q(G,M)$.
\end{proof}

Let $f:G\ra M$ be a $1$-cocycle and define
$\ker(f) := \{\sigma\in G : f(\sigma)=0\}$.
\begin{proposition}\label{kersubgroup}
$\ker(f)$ is a subgroup of $G$ and $[G:\ker(f)]=\#f(G)$.
\end{proposition}
\begin{proof}
If $\sigma_1,\sigma_2\in K$ then $f(\sigma_1\sigma_2)=f(\sigma_1)+\sigma_1 f(\sigma_2)=0$
and $0=f(1)=f(\sigma_1^{-1}\sigma)=f(\sigma_1^{-1})+\sigma_1^{-1}f(\sigma_1)=f(\sigma_1^{-1}$
so $K$ is a subgroup.  If $\tau\in G$ and $\sigma\in K$, then
$f(\tau\sigma)=f(\tau)+\tau f(\sigma) = f(\tau)$ so $f$ gives rise to a well-defined map
of sets $\fbar : G/K\ra M$.  If $\tau_1, \tau_2\in G$ and
$\fbar(\tau_1 K) = \fbar(\tau_2 K)$, then $f(\tau_1)=f(\tau_2)$ (by definition
of $\fbar$ so
$f(\tau_1^{-1}\tau_2)=f(\tau_1^{-1})+\tau_1^{-1}f(\tau_2) = f(\tau_1^{-1})+\tau_1^{-1}f(\tau_1) = f(\tau_1^{-1}\tau_1)=0$.
Thus $\fbar$ is injective and $\#f(G) = \# f(G/K) = \#(G/K) = [G:K]$.
\end{proof}

\begin{example}
{\bf Caution:} $\ker(f)$ need {\em not} be a normal subgroup of $G$. For example,
let $G=S_3$ and $M=\Z e_1 \oplus \Z e_2 \oplus \Z e_3$ with the natural
permutation action.  Let $f$ be the $1$-cocycle determined by $e_1$, so
$f(\sigma)=e_{\sigma(1)} - e_1$.  Then $\ker(f) = \{\iota, (23)\}$ is
{\em 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.
\end{example}

\begin{lemma}\label{kersplit}
Fix $c\in H^1(G,M)$.  A subgroup $H$ of $G$
splits $c$ iff $H\subset \ker(f)$ for some cocycle $f \in c$.
\end{lemma}
\begin{proof}
If $H\subset\ker(f)$ then $\res_H(c)$ is represented by $f|_H$ which is
the $0$-cocycle,  hence $H$ splits $c$.  Suppose conversely that we know only
that $H$ splits $c$.  If $f\in c$, then $f|_H \in \res_H(c)=0$.
Hence there is $x\in M$ so that $f(\sigma)=\sigma(x)-x$
for all $\sigma\in H$.  The cocycle $g: \sigma\mapsto f(\sigma)-(\sigma(x)-x)$
represents $c$ and $g|H = 0$ so $H\subset\ker(g)$.
\end{proof}
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.

\begin{proposition}\label{indeximage}
If $c\in H^1(G,M)$ then
$$\ind(c)=\gcd\{[G:\ker(f)] : f \in c\}=\gcd\{\# f(G) : f\in c\}.$$
\end{proposition}
\begin{proof}
The first equality follows from (\ref{kersplit}) and the fact that
$\ker(f)$ splits $c$, and the second from (\ref{kersubgroup}).
\end{proof}

Let $N$ be a subgroup of $\ker(G\ra \Aut M)$, fix $c\in H^1(G,M)$ and choose $f\in c$.
\begin{lemma}\label{fstuff}
The image $f(N)$ is a $G$-submodule of $M$ and $\ker(f)\intersect N$ is
normal in $G$, neither depends on the choice of $f\in c$. Further,
$f(G)$ is a union of cosets of $f(N)$.
\end{lemma}
\begin{proof}
Since $N$ acts trivially, the cocycle $f:N\ra M$ is a homomorphism of
groups and thus $f(N)$ is a subgroup of $M$.
If $\sigma\in G$ and $x\in N$ then, using that $x$ acts trivially on $M$, we have
$\sigma f(x) = f(\sigma \sigma^{-1}) + \sigma f(x) = f(\sigma) + \sigma f(\sigma^{-1}) + \sigma f(x) = f(\sigma) + \sigma f(x) + \sigma x f(\sigma^{-1}) = f(\sigma x) + \sigma x f(\sigma^{-1}) = f(\sigma x \sigma^{-1}) \in f(N).$
This proves that $f(N)$ is a $G$-submodule of $M$.
If $f(x)=0$ then $f(\sigma x\sigma^{-1})=\sigma f(x)=0$ for all $\sigma \in G$ so
$\ker(f)\intersect N$ is normal in $G$.  Since $N$ acts trivially on $M$, coboundaries
restrict to $0$ on $N$ so that $f(N)$ and $\ker(f)\intersect N$ do not depend on the
choice of $f\in c$.
Finally, $f(G)=\union_{\sigma\in G} f(N\sigma) = \union_{\sigma\in G} (f(N)+f(\sigma))$
is a union of cosets of $f(N)$.
\end{proof}

\begin{lemma}\label{joinlemma}
Let $f$ be a 1-cocycle and $H, H'$ subgroups of $G$ which split $f$. Then
the join $H.H'$ also splits $f$.
\end{lemma}
\begin{proof}
By (\ref{kersubgroup}) $\ker(f)$ is a subgroup of $G$.  It contains
both $H$ and $H'$ so it contains their join.
\end{proof}
{\bf Warning:} It is {\em not} true in general that if a cohomology
{\em class} $c$ is split by subgroups $H$ and $H'$ then it is split by $H.H'$.
See (\ref{splitexample}).

If $H$ is a normal subgroup of $G$ then the sequence
$$0\ra H^1(G/H,M^{H})\xrightarrow{\inf} H^1(G,M)\xrightarrow{\res} H^1(H,M)$$
is exact. Upon setting $H=\ker(f)\intersect N$ we can define several
cohomology classes naturally associated to $c\in H^1(G,M)$.

\begin{theorem}\label{equivalents}
Let $c\in H^1(G,M)$, $N\subset \ker(G\ra \Aut M)$, and $f\in c$.
Then the following are equivalent:\\
1. $c\in H^1(G,M)$ is sharp.\\
2. $c'\in H^1(G/(\ker(f)\intersect N), M)$ is sharp.\\
3. $\overline{c}\in H^1(G,M/f(N))$ is sharp.\\
4. $\overline{c}'\in H^1(G/N,M/f(N))$ is sharp.
\end{theorem}
\begin{proof}
(1)$\iff$(2):
Suppose $H\subset G$ splits $c$.  By (\ref{kersplit})
there exists $g\in c$ so that $H\subset\ker(g)$, and by (\ref{fstuff})
$\ker(g)$ also contains $\ker(f)\intersect N=\ker(g)\intersect N$.  Thus
$g$, and hence $c$, is split by the join $H.(\ker(f)\intersect N)$.
There is a one-to-one index preserving correspondence between maximal subgroups of
$G$ splitting $c$ and maximal subgroups of $G/(\ker(f)\intersect N)$ splitting
the class $c'\in H^1(G/(\ker(f)\intersect N), M)$ defined by $c$.\\
(1)$\iff$(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 $\{f : f\in c\}$ surjects onto the set $\{ g : g \in \overline{c}\}$.
If $f \in c$, (\ref{fstuff}) implies that $f(G)$ is a union of cosets
of $N'=f(N)$ and $N'$ does not depend on $f$, hence
$\{ \#g(G) : g\in\overline{c}\} = \{\#f(G)/\#f(N) : f\in c\}$.
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 $\{ \#f(G) : f\in c\}$.\\
(3)$\iff$(4): If $g\in\overline{c}$ then by (\ref{fstuff})
$g(N)=f(N)=\{0\}\subset M/f(N)$, so $N\subset \ker(g)$.  Useing (\ref{kersplit})
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 $\overline{c}'$.
\end{proof}

We can now deduce that quotients of sharp groups are sharp.
\begin{corollary}\label{sharpquotient}
If $G$ is sharp and $G\ra G'$ is a surjective then $G'$ is sharp.
\end{corollary}
\begin{proof}
Let $M$ be a $G'$ module.  Make $M$ a $G$-module via $G\ra G'$.
Let $c\in H^1(G',M)$, $f\in c$, and let $N:=\ker(G\ra G')\subset\ker(G\ra \Aut M)$.
The map $G\ra G'$ gives rise to a map $\pi: H^1(G',M)\ra H^1(G,M)$: on cocycles
it sends $f:G'\ra M$ to the map $G\ra M$ obtained by composing with $G\ra G'$.
Since $G$ is sharp, $\pi(c)\in H^1(G,M)$ is sharp, so by (\ref{equivalents})
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 (\ref{sharpshift}) we are done.

\end{proof}

\begin{theorem}\label{sharpaction}
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 $q$.
\end{theorem}
\begin{proof}
We first prove the theorem in the case $q=1$ and then
apply dimension shifting.
By (\ref{equivalents}), $(1,G,M)$ is sharp
iff $(1,G',M/f(N))$ is sharp for all $1$-cocycles $f:G\ra M$.
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 (\ref{sharpshift})
so there is an exact sequence of $G$-modules
$$0\ra M \ra M^{*} \ra M' \ra 0.$$
The action of $G$ on $M^{*}$ factors through $G'$:
If $h\in \ker (G\ra \Aut M)$ and $a\in M$, then for any $g\in G$,
$h.\vphi_a(g) = \vphi_a(gh)=gha=ga=\vphi_a(g)$.
Thus the action of  $G$ on the quotient $M'=M^{*}/M$ factors through $G'$.
As in the proof of (\ref{sharpshift}), we see that
the set of subgroups $H$ of $G$ which
split a fixed $c\in H^q(G,M)$ is the same as the set which split
$\delta(c)\in H^{q-1}(G,M')$.  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.
\end{proof}

\section{Sharp Modules}

\begin{proposition}\label{sharpsubgroupenough}
If every (compact profinite) subgroup of $\Aut M$ is sharp then $M$ is a sharp module.
\end{proposition}
\begin{proof}
We must show that $(q,G,M)$ is sharp for all groups $G$ and
$G$-module structurs on $M$.  This follows from
(\ref{sharpaction}) since $G$ acts through a sharp subgroup
of $\Aut M$.
\end{proof}

We call a $\Z$-module $M$ {\bf $q$-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.

\begin{proposition}
If $M$ is $1$-sharp and $M'$ is a direct summand of $M$ then $M'$ is
also $1$-sharp.
\end{proposition}
\begin{proof}
Write $M=M'\oplus M''$.
Suppose $G$ is a group with an action on $M'$. Extend $G$ to act
trivially on $M''$. There is an injection
$H^1(G,M')\xrightarrow{\iota} H^1(G,M'\oplus M'').$
If $c\in H^1(G,M')$ and $f\in c$ then the set of
cardinalities of images of cocycles in $\iota(c)$ is
$\{\#(\iota(f)+\delta(a,b))(G) : (a,b)\in M\}$
where $\delta(a,b)$ is the cocycle determined by $(a,b)$.
Since $(\iota(f)+\delta(a,b))(\sigma) = (f(\sigma),0) + (\sigma(a),b)-(a,b) = (f(\sigma)+\sigma(a)-a,0)$ we
see that this set of cardinalities is the same as
$\{ \# g(G) : g\in c\}$. Since $M$ is sharp, $\iota(c)$ is
sharp so by (\ref{indeximage}), $c$ is sharp.
\end{proof}

\begin{corollary}\label{cyclicsharp}
A cyclic abelian group is a sharp module.
\end{corollary}
\begin{proof}
The automorphism group of a cyclic group is finite cyclic, hence every subgroup
is sharp.  Now apply (\ref{sharpsubgroupenough}).
\end{proof}

%In the case $G$ profinite we remark that the map
%$G\ra \Aut M$ is continuous with $M$ given the discrete topology.
%This is because, by definition, the map $G\cross M\ra M$ is continuous with

\comment{
\begin{theorem}
$M=\Z\oplus\Z$ is a sharp module.
\end{theorem}
\begin{proof}
The profinite subgroups of the discrete group $\Aut M=\sltwoz$ are
finite, so it suffices to show that the finite subgroups of
$\sltwoz$ are sharp.
[... BUT IS THIS TRUE? ...]
\end{proof}
}

Let $\Fp=\Z/p\Z$.
\begin{theorem}\label{gltwo}
If $G\subset \GL_2(\Fp)$ and $q\geq 1$ then
$(q,G,\Fp\oplus\Fp)$ is sharp.
\end{theorem}
\begin{proof}
Let $M=\Fp\oplus\Fp$.
If $p=2$, $\GL_2(\Fp)=S_3$ and we are done because $S_3$ is sharp.
Thus assume $p>2$.

Suppose $G$ contains a scalar $t\in \Fp^{\star} \backslash \{1\}$.
By (\ref{innerpair}) the map
$H^q(G,M)\ra H^q(G,M)$ induced by conjugation by $t$
is the identity map. It is also multiplication by $t$ on
the $\Fp$ vector space $H^q(G,M)$ so since $t\neq 1$, we conclude
that $H^q(G,M)=0$.  If $p\nmid \#G$ then (\ref{orderannihilate}) implies that
$H^q(G,M)=0$.  We thus assume $G\subset \sltwo(\Fp)$ and
$p\mid \# G$.

{\em Case 1: $G$ contains exactly 1 Sylow $p$-subgroup:}
Let $S$ be {\em 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
$0\ra S \ra E \ra G/S \ra 1.$   Since $p\nmid [G:S]$
(\ref{orderannihilate}) implies that $H^2(G/S,S)=0$ so the
short exact sequence $0\ra S\ra G \ra G/S\ra 1$ is trivial,
hence splits.  The image of $G/S$ under this splitting is a
subgroup $N\subset G$ whose order is coprime to $p$ and whose
index equals $\#S=p$. Let $c\in H^1(G,M)$ be nonzero. Since
$M=\Fp\cross\Fp$, $c$ has order $p$. Since the order of $N$
is coprime to $p$ (\ref{orderannihilate}) implies that
$\res_N(c)=0$. But $[G:N]=p$ so by (\ref{ordind}) the index
of $c$ is $p$ and this index is attained by restricting to $N$
so that $c$ is sharp.

{\em Case 2: $G$ contains more than 1 Sylow $p$-subgroup:}
The subgroup $S=\{\abcd{1}{*}{0}{1}\}$ is a Sylow $p$-subgroup of $\sltwo(\Fp)$.
All Sylow $p$-subgroups of $\sltwo(\Fp)$ 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
$$\{\text{ Sylow p-subgroups of \sltwo(\Fp) }\} = \{\text{ lines in \Fp\cross\Fp }\}.$$
There are $p+1$ lines in $\Fp\cross\Fp$ 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 {\em normal} subgroup of $\sltwo(\Fp)$.

The order $n$ of $H$ divides the order of $\sltwo(\Fp)$
which is $p(p+1)(p-1)$.
The union of the $p+1$ Sylow $p$-subgroups of $\sltwo(\Fp)$ has
cardinality $p(p+1) - p = p^2$ (the identity is counted $p+1$ times,
but there is no other overcounting). Thus $n\geq p^2$.  Since $n \mid p(p+1)(p-1)$ it is exactly divisible by $p$ so $n$ can't equal $p^2$
and hence $n\geq p^2+1=p(p+1)$ so that $p(p+1)(p-1)/n \leq p-1$.
Therefore the quotient $\sltwo(\Fp)/ H$ is a group of order $\leq p-1$.

\comment{If $p=3$, $\#\sltwo(\Fp)=24$ and the cardinality of the union
of the four Sylow $3$-subgroups is $4\times 3 - 3=9$. Since the order
of $H$ divides $24$, it must be at least $12$ so $[G:H]\mid 2$.
If $p>3$, the group $\psltwo(\Fp)=\sltwo(\Fp)/\{\pm 1\}$ is simple (see \cite{atlas}).
The image of $H$ in $\psltwo(\Fp)$ is normal and nontrivial hence equals
$\psltwo(\Fp)$. Thus $[\sltwo(\Fp):H]\mid 2$ and
$H\subset G$ so $[\sltwo(\Fp):G]\mid 2$ and $G$ is normal in
$\sltwo(\Fp)$. }

By the remark in VII.6 of \cite{serre} there is an exact sequence
\begin{eqnarray*}
0&\ra &H^1(\sltwo(\Fp)/G,M^G)\xrightarrow{\inf}H^1(\sltwo(\Fp),M)
\xrightarrow{\res}H^1(G,M)\\&\ra& H^2(\sltwo(\Fp)/G,M^G)
\end{eqnarray*}
The order of $\sltwo(\Fp)/H$ is $\leq p-1$ so by
(\ref{orderannihilate}) $H^2(\sltwo(\Fp)/H,M)=0$. Furthermore,
$-1\in\sltwo(\Fp)$ so $H^1(\sltwo(\Fp),M)=0$ and hence
$H^1(G,M)=0$.  We are now allowed to write down the exact
sequence
\begin{eqnarray*}
0&\ra &H^2(\sltwo(\Fp)/G,M^G)\xrightarrow{\inf}H^2(\sltwo(\Fp),M)
\xrightarrow{\res}H^2(G,M)\\&\ra& H^3(\sltwo(\Fp)/G,M^G)
\end{eqnarray*}
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.

\end{proof}

\comment{
\begin{remark} If $p\con 1\pmod 4$ then $\sltwo(\Fp)$
is generated by its Sylow $p$-subgroups. Let $H$ be
the subgroup of $\sltwo(\Fp)$ generated by the Sylow $p$-subgroups.
Since $p\con 1\pmod{4}$ there is $a\in \Fp$ so that $a^2=-1$.
Since $H$ surjects onto the simple group $\psltwo(\Fp)$, we
see that $\sltwo(\Fp)$ is generated by $-1$ and $H$ so that
$\sltwo(\Fp)/H$ is generated by $-1$.  Thus one of the matrices
$\abcd{a}{0}{0}{-a}$ or $\abcd{-a}{0}{0}{a}$ lies in $H$.
Both of these have square $-1$, so $-1$ lies in $H$ after all.
%Subgroups of $\sltwo(\Fp)$ are not in general sharp.
\end{remark}
}

\begin{corollary}
$M=\Fp\oplus\Fp$ is a $1$-sharp module.
\end{corollary}
\begin{proof}
Let $G$ be an arbitrary group equipped with an action on $M$, and
fix $c\in H^1(G,M)$.  We must show that $c$ is sharp.  Let $N=\ker(G\ra\Aut M)$
and $f$ be a cocycle representing $c$.  By (\ref{fstuff}), $f(N)$ is a $G$-submodule
of $\Aut M$ and by (\ref{equivalents}) $c$ is sharp iff the class
$d=\overline{c}'\in H^1(G/N,M/f(N))$ is sharp.  If $\# f(N) > 1$ the quotient
$M/f(N)=(\Fp\oplus\Fp)/ f(N)$ is cyclic so (\ref{cyclicsharp}) implies $d$ is sharp.
The remaining case is $\# f(N) = 1$.   Since $G/N$ is a subgroup of
$\Aut M = \GL_2(\Fp)$ we may apply (\ref{gltwo}) to conclude that $d$ is sharp.
\end{proof}

\section{Examples}

Let $G$ be a finite group.
There is a natural map $\Z[G]\ra\Z$ sending $\sum a_g g$ to $\sum a_g\in\Z$.
The {\bf augmentation ideal} $I_G\subset\Z[G]$ is the kernel of this map so there is an exact
sequence $0\ra I_G \ra \Z[G]\ra \Z\ra 0$.  Since $\Z[G]$ is cohomologically trivial,
$H^1(G,I_G)=\coker(H^0(G,\Z[G])\ra H^0(G,\Z)) = \Z/(\# G)\Z$.  If $H$
is a subgroup of $G$ then $\Z[G]$ is cohomologically trivial as an $H$
module so we have an isomorphism
$H^1(H,I_G)\isom \coker(H^0(H,\Z[G])\ra H^0(H,\Z))\isom \Z/(\#H)\Z=H^1(H,I_H)$
Thus the restriction map $\res_H$ agrees with the natural map
$\Z/(\#G)\Z\ra\Z/(\#H)\Z$.

\begin{proposition}
$(1,G,I_G)$ is sharp iff the set $\{ \# H : H\subset G \}$ is
closed under taking least common multiples of subsets (modulo $\#G$).
\end{proposition}
\begin{proof}
In the following, we work module $\#G$.

First assume that $\{ \# H : H\subset G \}$ is
closed under $\lcm$.
Let $m\in H^1(G,I_G)=\Z/\#G \Z$. Then $\res_H(m)=0$ iff $\#H \mid m$.
The index of $m$ is $\#G / \lcm \{\#H : \# H \mid m\}$, so $m$ is
sharp iff $\lcm\{\#H : \# H \mid m\}$ lies in $\{\#H : \# H \mid m\}$.
Since the $\lcm$ divides $m$ and  $\{ \# H : H\subset G \}$ is closed
under $\lcm$, this is the case.

Next assume that $(1,G,I_G)$ is sharp.
Let $a_1,\ldots, a_n$ be elements of $\{ \# H : H\subset G \}$
and let $m=\lcm(a_1,\ldots,a_n)$.  By sharpness,
$\lcm\{\#H : \# H \mid m\}$ lies in $\{\#H : \# H \mid m\}$.
Since $a_1,\ldots,a_n \in \{\#H : \# H \mid m\}$, this $\lcm$ is
$m$ so it follows
that there is $H\subset G$ so that $m=\lcm\{\#H : \# H \mid m\}=\#H$,
as desired.
\end{proof}

\begin{example} The subgroups of $A_4$ have orders $1,2,3,4,12$ so
$(1,A_4,I_{A_4})$ is not sharp.
\end{example}
%Similarly, the subgroups of $A_6$ have orders $1,2,3,4,5,6,8,9,10,12,18,24,36,60,360.$

\begin{example}
If $M$ and $M'$ are sharp then $M\oplus M'$ need not be sharp.
Let $n$ be the largest integer so that $\Z^n$ is sharp.  By
(\ref{cyclicsharp}), $n\geq 1$, and since
$\Z^{11}$ is the $\Z$-module underlieing the augmentation
ideal $I_{A_4}$ is not sharp, $n\leq 10$.
Then $\Z\oplus\Z^n$ is not sharp but both $\Z$ and $\Z^n$ are.
\end{example}

\begin{example}
Even if $(1,G,M)$ and $(1,G,M')$ are both sharp, $(1,G,M\oplus M')$
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 $c \in H^1(S_4,I)$ correspond to $6$ under the isomorphism
$H^1(S_4,I)\isom \Z/24\Z$.  Let $\vphi:S_4\ra \Z/2\Z$ be the
nontrivial (sign) homomorphism and let
$d=(c,\vphi)\in H^1(S_4,I\oplus(\Z/2\Z))$. Then a subgroup
which splits $d$ must split $\vphi$ hence be contained in
$A_4$.  Thus the subgroups splitting $d$ are the subgroups
of $A_4$ which split $c$. But
$\res_{A_4}(c)\in H^1(A_4,I_{S_4})=\Z/12\Z$
is not sharp, so $d$ is not sharp.
\end{example}

\begin{example}\label{splitexample}
Let $G=S_3$ and $c=2\in \Z/6\Z = H^1(G,I_G)$. 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 (\ref{joinlemma}) for cohomology classes is false.
\end{example}

\begin{example}
If $G$ fits into an exact sequence
$0\ra G'\ra G \ra G''\ra 0$ with $G'$ and $G''$ sharp,
must $G$ be sharp?  No, as the example
$0\ra V_4\ra A_4\ra C_3\ra 0$ shows.
\end{example}

\section{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 $\Fp^n$ is a sharp module?
Are any divisible abelian groups sharp modules, for example $\Q$?
If a $\Z$-module $M$ is $q$-sharp for some $q\geq 1$ must it be
$q$-sharp for {\em all} $q\geq 1$.

\section{Arithmetic Applications}
This section is under construction.

In the context of abelian varieties, Lichtenbaum \cite{lichtenbaum} has
studied the quotient $\ind(c)/\ord(c)$.  Cassels \cite{cassels} gave
an elliptic curves $E/\Q$ and a class $c\in H^1(\Q,E(\Qbar))$ for
which $\ord(c)\neq \ind(c)$, but showed that if $c\in\Sha(E)$ then
$\ord(c)=\ind(c)$.

\begin{proposition}
Let $M$ be an abelian group, $K$ a number field, and let
$\Gal(\Kbar/K)$ act on $M$ via the cyclotomic character.
Then $(1,\Gal(\Kbar/K),M)$ is sharp.
\end{proposition}

\begin{proposition}
Let $K$ be a number field, $a\in K$, and $m$ a positive integer.
Factor the polynomial $X^m-a$ as a product $\prod g_i(X)$ of
irreducible polynomials $g_i(X)\in K[X]$.  Then the minimum
of the degrees of the $g_i$ divide the other degrees.
\end{proposition}

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

\begin{theorem}
Let $E$ be an elliptic curve of a field $K$ and $G=\Gal(\Kbar/K)$.  Then
$(1,G,E(\Kbar))$ is sharp.
\end{theorem}

We were unable to answer the analogous question for higher dimensional
abelian varieties.
\begin{question} 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 $K$.
\end{question}
The cohomology group $H^1(K,\Jac(X))$ associated to such an $X$ would not be sharp.

\begin{thebibliography}{HHHHHHH}
\bibitem[A]{atiyah} Atiyah, Wall, {\em Cohomology of Groups},
in {\em Algebraic Number Theory}, Ed. J.W.S. Cassels, A. Fr\"{o}hlich,
\bibitem[C]{cassels} J.W.S. Cassels,
{\em Arithmetic on curves of genus 1. V. Two counterexamples.}
J. London Math. Soc. {\bf 38} (1963) 244--248.
\bibitem[LT]{langtate} S. Lang, J. Tate,
{\em Principal homogeneous spaces over abelian varieties},
{\bf 80} (1958) 659--684.
\bibitem[L]{lichtenbaum} S. Lichtenbaum,
{\em Duality theorems for curves over $p$-adic fields}. Invent. Math.
{\bf 7} (1969) 120--136.
\bibitem[S]{serre} J.P. Serre, {\em Local Fields}, Springer GTM {\bf 67} (1995).

\end{thebibliography}
\end{document}