CoCalc Shared Fileswww / sharp.tex
Author: William A. Stein
1% sharp.tex
2\documentclass[11pt]{article}
3\include{macros}
4\title{Subgroups in which a Cohomology Class Splits\\
5{\large (Second Rough Draft -- must reorder, finish last section.)}}
6\author{William A. Stein\footnote{UC Berkeley, Department of
7Mathematics, Berkeley, CA  94720, USA.}}
8
9\begin{document}
10\maketitle
11\begin{abstract}
12Let $G$ be a group, $M$ a $\Z[G]$-module, and $c\in H^q(G,M)$.
13We study the indexes of subgroups $H\subset G$
14for which $c$ restricts to $0$.  In particular, we investigate
15the situation when the greatest common divisor of the indexes
16is itself an index.
17\end{abstract}
18\section{Introduction}
19Let $G$ be a (profinite) group and let $M$ be a (discrete) $G$-module.
20For any integer $q\geq 0$ let
21$H^q(G,M)$ denote the $q$-th cohomology group in the sense of \cite{atiyah}.
22Thus $H^q(G,M)$ is an additive abelian group associated covariantly to
23$M$ and contravariantly to $G$.
24If $H$ is a finite index subgroup of
25$G$ there is a restriction map $\res_H:H^q(G,M)\ra H^q(H,M)$.
26In this paper we study the set of indexes of subgroups $H\subset G$
27for which a given cohomology class restricts to $0$.
28
29Suppose $q\geq 1$. Define the {\bf index} of $c\in H^q(G,M)$ to be
30$$\ind(c) = \gcd\{[G:H] : \res_H(c)=0\}.$$
31Say that $c$ is a {\bf sharp cohomology class} if there exists a subgroup
32$H$ such that $[G:H]=\ind(c)$ and $\res_H(c)=0$.
33In words, the  index is attained,'' or
34the greatest common divisor of the indexes
35is itself an index,'' or even the gcd equals the minimum.''
36The triple $(q,G,M)$ is called a {\bf sharp triple} if every $c\in H^q(G,M)$
37is sharp.
38A group $G$ is a {\bf sharp group} if $(q,G,M)$ is sharp for every  $G$-module
39$M$ and $q\geq 1$.
40A $\Z$-module $M$ is a {\bf sharp module} if $(q,G,M)$ is sharp
41for every $q\geq 1$ and group $G$ with an action on $M$.
42
43In the first section we develop basic properties of the index of a
44cohomology class.  Then strongly Sylow groups are
45introduced and shown to be sharp. We next derive a criterion for
46sharpness and use it to prove that quotients of sharp groups are
47sharp.  Then we show that $\Fp\cross \Fp$ is often a sharp module.
48Next we give examples showing how sharpness can fail, and some
49questions we were unable to answer.
50In the last section, sharpness is applied to
51arithmetic questions.
52
53\section{Sharp Groups}
54Let $G$ be a profinite group, $M$ a $G$-module, and $H^q(G,M)$ the
55$q$-th cohomology group.  In this section we assume that $q\geq 1$.
56A subgroup $H$ of $G$ {\bf splits} $c$ if $\res_H(c)=0$.
57Let $\ord(c)$ denote the order of $c$ as a group element.
58
59\begin{lemma}\label{orderannihilate}
60Let $G$ be a finite group, $M$ a $G$-module, and
61$q\geq 1$.  Then $H^q(G,M)$ is annihilated by $\# G$.
62\end{lemma}
63\begin{proof}
64This is proved as Corollary 1 of Section 6 of \cite{atiyah} by using that
65$\cores\circ \res_{\{1\}} = \#G$.
66\end{proof}
67
68\begin{proposition}\label{ordind}
69$\ord(c)\mid \ind(c)$ and they have the same prime factors.
70\end{proposition}
71\begin{proof}
72Suppose $H\subset G$ splits $c$.  There is a map $\cores_H:H^q(H,M)\ra H^q(G,M)$ such
73that  $\cores_H \circ\res_H (c) = [G:H]\cdot c$. Thus $[G:H]\cdot c=0$ and hence
74the order of $c$ divides $\ind(c)$.
75Let $p$ be a prime and let
76$G_p$ be a Sylow $p$-subgroup of $G$, so the (generalized) index
77$[G:G_p]$ is coprime to $p$ and $G_p$ is a (pro) $p$-group.
78If $p$ does not divide the order of $c$ then
79$0=\res_{G_p}(c)\in H^q(G_p,M)$ so that $\ind(c)|[G:G_p]$ is
80not divisible by $p$.
81\end{proof}
82
83Let $t \in G$. Consider the inner automorphism $f:G\ra G$
84given by $f(s)=tst^{-1}$.  This turns $M$ into a new
85$G$-module, denoted by $M^t$, and gives a homorphism
86$f^{*}:H^q(G,M)\ra H^q(G,M^t)$.  The map $g:M^t\ra M$ sending $a$ to $t^{-1}a$ defines
87an isomorphism $M^t\ra M$ and induces a map
88$g_{*}:H^q(G,M^t)\ra H^q(G,M)$.
89\begin{proposition}\label{innerpair}
90$g_{*}\circ f^{*}$ is the identity map $H^q(G,M)\ra H^q(G,M)$.
91\end{proposition}
92\begin{proof}
93This is Proposition 3 of \cite{atiyah}.  The proof uses dimension shifting.
94(There is a typo in
95(4.2) of \cite{atiyah}: the first $A$ should be replaced by $A^{t}$.)
96\end{proof}
97
98
99\begin{lemma}\label{conjsplit}
100If $H$ splits $c$ then every $G$-conjugate of $H$ splits $c$.
101\end{lemma}
102\begin{proof}
103Let $t \in G$, let $f:G\ra G$ be the inner automorphism $f(s)=tst^{-1}$ and
104let $g$ be the map $M\ra M$ sending $a$ to $t^{-1}a$.
105As in section 4 of \cite{atiyah} we use functoriality of $H^q(\,,\,)$ to
106obtain a commuting diagram
107$$\begin{matrix} 108 H^q(G,M) & \xrightarrow{\res_H} & H^q(H,M)\\ 109 \downarrow f^{*} & & \downarrow f^{*} \\ 110 H^q(G,M^{t}) & \xrightarrow{\res_{f^{-1}(H)}} & H^q(f^{-1}(H),M^t)\\ 111 \downarrow g_* & & \downarrow g_*\\ 112H^q(G,M) & \xrightarrow{\res_{f^{-1}(H)}} & H^q(f^{-1}(H),M) 113\end{matrix}$$
114
115By (\ref{innerpair}),
116 the vertical map $g_{*}\circ f^{*}$ on the left
117is the {\em identity} map.
118Thus if $\res_H(c)=0$ then so must
119$\res_{f^{-1}(H)}(c)=0$, which proves the Lemma.
120\end{proof}
121
122The following lemma shows that if, for each prime $p$,
123we let $G_p$ be a Sylow $p$-subgroup of $G$, then
124$\ind(c) = \prod_p\ind(\res_{G_p}(c))$.
125For $n$ an integer and $p$ a prime, let $\prt_p(n)=p^{\ord_p(n)}$.
126\begin{lemma}\label{indpsylow}
127Let $G_p$ be a Sylow $p$-subgroup of $G$.
128Then $\prt_p(\ind(c))=\ind(\res_{G_p}(c))$
129\end{lemma}
130\begin{proof}
131Suppose $H\subset G$ splits $c$ and let $H_p\subset H$ be a Sylow $p$-subgroup.
132Then $\ord_p([G:H])=\ord_p([G:H][H:H_p])=\ord_p([G:H_p])$.  By Sylow's theorems,
133there exists $t\in G$ so that $t H_pt^{-1} \subset G_p$. By (\ref{conjsplit}),
134$tH_p t^{-1}$ also splits $c$.  Thus
135$\ind(\res_{G_p}(c))| [G_p:t^{-1}H_p t]$
136so, since
137$\ord_p([G:t^{-1}H_p t]) = \ord_p([G:H_p])=\ord_p([G:H])$,
138it follows that
139$\ind(\res_{G_p}(c)) | \ind(c)$.
140On the other hand, $\ord_p(\ind(c))\leq \ord_p(\res_{G_p}(c))$
141because we can compute an upper bound on $\ord_p(\ind(c))$
142using subgroups of $G_p$.
143\end{proof}
144
145\begin{proposition}
146For any subgroup $H$ of $G$,
147   $\ind(\res_H(c))|\ind(c)$.
148\end{proposition}
149\par\noindent
150{\bf Remark: } If $H'\subset G$ splits $c$ so does $H\intersect H'$. There is
151a natural injection of sets $H/(H\intersect H')\hookrightarrow G/H'$ so
152$[H:H\intersect H'] \leq [G:H']$. But, it need not be the case that
153$[H:H\intersect H'] \mid [G:H']$, for example, let $H$ and $H'$
154be distinct order $3$ subgroups of $S_4$.  Then
155$3=[H:H\intersect H']$ whereas $[G:H']=\frac{24}{3}=8$.
156\begin{proof}
157It suffices to show the divisibility for each prime $p$.  Let $H_p$ be a
158Sylow $p$-subgroup of $H$, and use the Sylow theorems to extend $H_p$ to
159a Sylow $p$-subgroup $G_p$ of $G$. If $H'$ is a subgroup of $G_p$ which
160splits $c$, then $H'\intersect H_p$ splits $\res_{H}(c)$.  Since
161$[H_p:H'\intersect H_p] \leq [G_p : H']$ and each is a power of $p$,
162$[H_p:H'\intersect H_p] \mid [G_p : H']$.  Thus
163   $$\prt_p(\ind(\res_{H}(c)))=\ind(\res_{H_p}(c))\mid \ind(\res_{G_p}(c))=\prt_p(\ind(c)).$$
164\end{proof}
165
166
167%\begin{definition} A group $G$ is {\bf sharp} if $(q,G,M)$ is sharp
168%for every $G$-module $M$ and $q\geq 1$.
169%\end{definition}
170
171Using the technique of dimension shifting we now show that is is only necessary to
172require sharpness of $(1,G,M)$ in the definition.
173\begin{proposition}\label{sharpshift}
174If $(1,G,M)$ is sharp for every $M$ then $G$ is sharp.
175\end{proposition}
176\begin{proof}
177We proceed by induction. Let $M$ be a $G$-module and $q>1$.
178Assume that $(q-1,G,M')$ is sharp for every $G$-module $M'$.
179Consider the $G$-module $M^{*}=\Hom(\Z[G],M)$.  There is a natural injection
180$M\ra M^{*}$ which maps $a\in A$ to $\vphi_a$, where $\vphi_a$ is defined
181by $\vphi_a(g) = ga$.  Hence we have an exact sequence of $G$-modules
182$$0\ra M \ra M^{*} \ra M' \ra 0.$$
183Since $M^{*}$ is co-induced, it follows that
184$\delta:H^{q-1}(G,M')\ra H^q(G,M)$ is
185an isomorphism.  Since a co-induced $G$-module is also a coinduced $H$-module
186for any subgroup $H$ of $G$ we obtain commuting diagrams
187$$\begin{matrix}H^{q-1}(G,M') & \ra & H^q(G,M)\\ 188 \downarrow \res_H & & \downarrow \res_H\\ 189 H^{q-1}(H,M') & \ra & H^q(H,M) 190\end{matrix}$$
191In particular, the set of subgroups $H$ of $G$ which
192split $c\in H^q(G,M)$ is the same as the set which split
193$\delta(c)\in H^{q-1}(G,M')$.  By our inductive assumption,
194$(q-1,G,M')$ is sharp so $c$ is sharp, and hence $(q,G,M)$ is
195sharp.
196
197\end{proof}
198
199\begin{lemma}\label{psharp}
200If $G$ is a $p$-group then $G$ is sharp.
201\end{lemma}
202\begin{proof}
203The greatest common divisor of a set of $p$-powers equals the smallest
204so $(1,G,M)$ is sharp.
205\end{proof}
206
207A group $G$ is {\bf strongly Sylow} if it has the following property:
208Given subgroups $H_p\subset G_p$ for each prime $p$, there exists
209$H\subset G$ such that, for all $p$, a Sylow $p$-subgroup
210of $H$ is $G$-conjugate to $H_p$.  Nilpotent groups are strongly Sylow
211because they are the direct product of their Sylow $p$-subgroups.
212The group $S_3$ is strongly Sylow but not nilpotent.
213
214\begin{theorem}\label{strongsylowone}
215If $G$ is strongly Sylow then $G$ is sharp.
216\end{theorem}
217\begin{proof}
218Let $M$ be a $G$-module.
219We must show that $(1,G,M)$ is sharp, so
220let $c\in H^1(G,M)$.
221For each prime $p$, (\ref{psharp}) gives a subgroup
222$H_p$ of $G_p$ so that $\ind(\res_{G_p}(c)) = [G_p:H_p]$.
223Choose $H$ as guaranteed by the property of $G$ being strongly Sylow.
224For each prime $p$, (\ref{conjsplit}) insures
225that every $p$-Sylow subgroup of $H$ splits $c$, i.e.,
226$\ind(\res_{H_p}(c))=1$.
227By (\ref{indpsylow}), $\ind(\res_H(c))=1$, so
228by (\ref{ordind}),  $\res_H(c)=0$. Since for each prime $p$,
229$\ord_p([G:H])=\ord_p([G:H_p])=\ord_p([G_p:H_p])=\ord_p(\ind(c))$
230it follows that $[G:H]=\ind(c)$ so that $c$ is sharp.
231Note, this proof works equally well for $c\in H^q(G,M)$.
232\end{proof}
233
234Let $f:G\ra M$ be a $1$-cocycle and define
235$\ker(f) := \{\sigma\in G : f(\sigma)=0\}$.
236\begin{proposition}\label{kersubgroup}
237$\ker(f)$ is a subgroup of $G$ and $[G:\ker(f)]=\#f(G)$.
238\end{proposition}
239\begin{proof}
240If $\sigma_1,\sigma_2\in K$ then $f(\sigma_1\sigma_2)=f(\sigma_1)+\sigma_1 f(\sigma_2)=0$
241and $0=f(1)=f(\sigma_1^{-1}\sigma)=f(\sigma_1^{-1})+\sigma_1^{-1}f(\sigma_1)=f(\sigma_1^{-1}$
242so $K$ is a subgroup.  If $\tau\in G$ and $\sigma\in K$, then
243$f(\tau\sigma)=f(\tau)+\tau f(\sigma) = f(\tau)$ so $f$ gives rise to a well-defined map
244of sets $\fbar : G/K\ra M$.  If $\tau_1, \tau_2\in G$ and
245$\fbar(\tau_1 K) = \fbar(\tau_2 K)$, then $f(\tau_1)=f(\tau_2)$ (by definition
246of $\fbar$ so
247$f(\tau_1^{-1}\tau_2)=f(\tau_1^{-1})+\tau_1^{-1}f(\tau_2) 248 = f(\tau_1^{-1})+\tau_1^{-1}f(\tau_1) = f(\tau_1^{-1}\tau_1)=0$.
249Thus $\fbar$ is injective and $\#f(G) = \# f(G/K) = \#(G/K) = [G:K]$.
250\end{proof}
251
252
253\begin{example}
254{\bf Caution:} $\ker(f)$ need {\em not} be a normal subgroup of $G$. For example,
255let $G=S_3$ and $M=\Z e_1 \oplus \Z e_2 \oplus \Z e_3$ with the natural
256permutation action.  Let $f$ be the $1$-cocycle determined by $e_1$, so
257$f(\sigma)=e_{\sigma(1)} - e_1$.  Then $\ker(f) = \{\iota, (23)\}$ is
258{\em not} a normal subgroup of $G$.  Note that
259$3=[S_3:\ker(f)] = \# f(S_3) = \# \{ 0, e_2-e_1, e_3-e_1\}$,
260as predicted by the Proposition.
261\end{example}
262
263\begin{lemma}\label{kersplit}
264Fix $c\in H^1(G,M)$.  A subgroup $H$ of $G$
265splits $c$ iff $H\subset \ker(f)$ for some cocycle $f \in c$.
266\end{lemma}
267\begin{proof}
268If $H\subset\ker(f)$ then $\res_H(c)$ is represented by $f|_H$ which is
269the $0$-cocycle,  hence $H$ splits $c$.  Suppose conversely that we know only
270that $H$ splits $c$.  If $f\in c$, then $f|_H \in \res_H(c)=0$.
271Hence there is $x\in M$ so that $f(\sigma)=\sigma(x)-x$
272for all $\sigma\in H$.  The cocycle $g: \sigma\mapsto f(\sigma)-(\sigma(x)-x)$
273represents $c$ and $g|H = 0$ so $H\subset\ker(g)$.
274\end{proof}
275Thus the set of kernels of cocycles is cofinal'' in the set of subgroups
276which split $c$, and so they can be used to compute the index.
277
278\begin{proposition}\label{indeximage}
279If $c\in H^1(G,M)$ then
280$$\ind(c)=\gcd\{[G:\ker(f)] : f \in c\}=\gcd\{\# f(G) : f\in c\}.$$
281\end{proposition}
282\begin{proof}
283The first equality follows from (\ref{kersplit}) and the fact that
284$\ker(f)$ splits $c$, and the second from (\ref{kersubgroup}).
285\end{proof}
286
287
288Let $N$ be a subgroup of $\ker(G\ra \Aut M)$, fix $c\in H^1(G,M)$ and choose $f\in c$.
289\begin{lemma}\label{fstuff}
290The image $f(N)$ is a $G$-submodule of $M$ and $\ker(f)\intersect N$ is
291normal in $G$, neither depends on the choice of $f\in c$. Further,
292$f(G)$ is a union of cosets of $f(N)$.
293\end{lemma}
294\begin{proof}
295Since $N$ acts trivially, the cocycle $f:N\ra M$ is a homomorphism of
296groups and thus $f(N)$ is a subgroup of $M$.
297If $\sigma\in G$ and $x\in N$ then, using that $x$ acts trivially on $M$, we have
298$\sigma f(x) = f(\sigma \sigma^{-1}) + \sigma f(x) 299 = f(\sigma) + \sigma f(\sigma^{-1}) + \sigma f(x) 300 = f(\sigma) + \sigma f(x) + \sigma x f(\sigma^{-1}) 301 = f(\sigma x) + \sigma x f(\sigma^{-1}) 302 = f(\sigma x \sigma^{-1}) \in f(N).$
303This proves that $f(N)$ is a $G$-submodule of $M$.
304If $f(x)=0$ then $f(\sigma x\sigma^{-1})=\sigma f(x)=0$ for all $\sigma \in G$ so
305$\ker(f)\intersect N$ is normal in $G$.  Since $N$ acts trivially on $M$, coboundaries
306restrict to $0$ on $N$ so that $f(N)$ and $\ker(f)\intersect N$ do not depend on the
307choice of $f\in c$.
308Finally, $f(G)=\union_{\sigma\in G} f(N\sigma) = \union_{\sigma\in G} (f(N)+f(\sigma))$
309is a union of cosets of $f(N)$.
310\end{proof}
311
312\begin{lemma}\label{joinlemma}
313Let $f$ be a 1-cocycle and $H, H'$ subgroups of $G$ which split $f$. Then
314the join $H.H'$ also splits $f$.
315\end{lemma}
316\begin{proof}
317By (\ref{kersubgroup}) $\ker(f)$ is a subgroup of $G$.  It contains
318both $H$ and $H'$ so it contains their join.
319\end{proof}
320{\bf Warning:} It is {\em not} true in general that if a cohomology
321{\em class} $c$ is split by subgroups $H$ and $H'$ then it is split by $H.H'$.
322See (\ref{splitexample}).
323
324If $H$ is a normal subgroup of $G$ then the sequence
325$$0\ra H^1(G/H,M^{H})\xrightarrow{\inf} H^1(G,M)\xrightarrow{\res} H^1(H,M)$$
326is exact. Upon setting $H=\ker(f)\intersect N$ we can define several
327cohomology classes naturally associated to $c\in H^1(G,M)$.
328
329\begin{theorem}\label{equivalents}
330Let $c\in H^1(G,M)$, $N\subset \ker(G\ra \Aut M)$, and $f\in c$.
331Then the following are equivalent:\\
3321. $c\in H^1(G,M)$ is sharp.\\
3332. $c'\in H^1(G/(\ker(f)\intersect N), M)$ is sharp.\\
3343. $\overline{c}\in H^1(G,M/f(N))$ is sharp.\\
3354. $\overline{c}'\in H^1(G/N,M/f(N))$ is sharp.
336\end{theorem}
337\begin{proof}
338(1)$\iff$(2):
339Suppose $H\subset G$ splits $c$.  By (\ref{kersplit})
340there exists $g\in c$ so that $H\subset\ker(g)$, and by (\ref{fstuff})
341$\ker(g)$ also contains $\ker(f)\intersect N=\ker(g)\intersect N$.  Thus
342$g$, and hence $c$, is split by the join $H.(\ker(f)\intersect N)$.
343There is a one-to-one index preserving correspondence between maximal subgroups of
344$G$ splitting $c$ and maximal subgroups of $G/(\ker(f)\intersect N)$ splitting
345the class $c'\in H^1(G/(\ker(f)\intersect N), M)$ defined by $c$.\\
346(1)$\iff$(3): Since any two cocycles in $\overline{c}$ differ by the coboundary
347of an element of $M/c(N)$, and such a coboundary is the image of the
348coboundary of an element of $M$,
349the set $\{f : f\in c\}$ surjects onto the set $\{ g : g \in \overline{c}\}$.
350If $f \in c$, (\ref{fstuff}) implies that $f(G)$ is a union of cosets
351of $N'=f(N)$ and $N'$ does not depend on $f$, hence
352$\{ \#g(G) : g\in\overline{c}\} = \{\#f(G)/\#f(N) : f\in c\}$.
353Sharpness means that the set of integers above satisfies $\min = \gcd$.  This is true
354of the left hand side iff it is true of the right hand side iff it is true
355of $\{ \#f(G) : f\in c\}$.\\
356(3)$\iff$(4): If $g\in\overline{c}$ then by (\ref{fstuff})
357$g(N)=f(N)=\{0\}\subset M/f(N)$, so $N\subset \ker(g)$.  Useing (\ref{kersplit})
358we see that there is a bijection between the maximal subgroups of
359$G$ splitting $\overline{c}$ and the maximal subgroups of
360$G/N$ which split $\overline{c}'$.
361\end{proof}
362
363We can now deduce that quotients of sharp groups are sharp.
364\begin{corollary}\label{sharpquotient}
365If $G$ is sharp and $G\ra G'$ is a surjective then $G'$ is sharp.
366\end{corollary}
367\begin{proof}
368Let $M$ be a $G'$ module.  Make $M$ a $G$-module via $G\ra G'$.
369Let $c\in H^1(G',M)$, $f\in c$, and let $N:=\ker(G\ra G')\subset\ker(G\ra \Aut M)$.
370The map $G\ra G'$ gives rise to a map $\pi: H^1(G',M)\ra H^1(G,M)$: on cocycles
371it sends $f:G'\ra M$ to the map $G\ra M$ obtained by composing with $G\ra G'$.
372Since $G$ is sharp, $\pi(c)\in H^1(G,M)$ is sharp, so by (\ref{equivalents})
373the class it defines in $H^1(G/N,M/f(N))=H^1(G',M)$ is sharp. But the latter class
374is $c$, so $c$ is sharp and by (\ref{sharpshift}) we are done.
375
376\end{proof}
377
378\begin{theorem}\label{sharpaction}
379Fix a $G$-module $M$ and
380suppose the action of $G$ on $M$ factors
381through a sharp quotient $G'$.
382Then $(q,G,M)$ is sharp for every $q$.
383\end{theorem}
384\begin{proof}
385We first prove the theorem in the case $q=1$ and then
386apply dimension shifting.
387By (\ref{equivalents}), $(1,G,M)$ is sharp
388iff $(1,G',M/f(N))$ is sharp for all $1$-cocycles $f:G\ra M$.
389This is the case because $G'$ is sharp.
390
391Next suppose the theorem is true for $q-1$.
392Let $M'$ and $M^{*}$ be as in the proof of (\ref{sharpshift})
393so there is an exact sequence of $G$-modules
394$$0\ra M \ra M^{*} \ra M' \ra 0.$$
395The action of $G$ on $M^{*}$ factors through $G'$:
396If $h\in \ker (G\ra \Aut M)$ and $a\in M$, then for any $g\in G$,
397$h.\vphi_a(g) = \vphi_a(gh)=gha=ga=\vphi_a(g)$.
398Thus the action of  $G$ on the quotient $M'=M^{*}/M$ factors through $G'$.
399As in the proof of (\ref{sharpshift}), we see that
400the set of subgroups $H$ of $G$ which
401split a fixed $c\in H^q(G,M)$ is the same as the set which split
402$\delta(c)\in H^{q-1}(G,M')$.  By our inductive assumption, using
403that $G$ acts on $M$' through $G'$, we see that
404$(q-1,G,M')$ is sharp. Thus $c$ is sharp, and hence $(q,G,M)$ is
405sharp.
406\end{proof}
407
408\section{Sharp Modules}
409
410\begin{proposition}\label{sharpsubgroupenough}
411If every (compact profinite) subgroup of $\Aut M$ is sharp then $M$ is a sharp module.
412\end{proposition}
413\begin{proof}
414We must show that $(q,G,M)$ is sharp for all groups $G$ and
415$G$-module structurs on $M$.  This follows from
416(\ref{sharpaction}) since $G$ acts through a sharp subgroup
417of $\Aut M$.
418\end{proof}
419
420We call a $\Z$-module $M$ {\bf $q$-sharp} if $(q,G,M)$ is sharp for all
421groups $G$.  Unlike in the case of groups, we do not suspect that $q$-sharpness
422for some $q$ is equivalent to $q$-sharpness for all $q$.  We have
423only the following week result.
424
425\begin{proposition}
426If $M$ is $1$-sharp and $M'$ is a direct summand of $M$ then $M'$ is
427also $1$-sharp.
428\end{proposition}
429\begin{proof}
430Write $M=M'\oplus M''$.
431Suppose $G$ is a group with an action on $M'$. Extend $G$ to act
432trivially on $M''$. There is an injection
433$H^1(G,M')\xrightarrow{\iota} H^1(G,M'\oplus M'').$
434If $c\in H^1(G,M')$ and $f\in c$ then the set of
435cardinalities of images of cocycles in $\iota(c)$ is
436$\{\#(\iota(f)+\delta(a,b))(G) : (a,b)\in M\}$
437where $\delta(a,b)$ is the cocycle determined by $(a,b)$.
438Since $(\iota(f)+\delta(a,b))(\sigma) = (f(\sigma),0) + 439(\sigma(a),b)-(a,b) = (f(\sigma)+\sigma(a)-a,0)$ we
440see that this set of cardinalities is the same as
441$\{ \# g(G) : g\in c\}$. Since $M$ is sharp, $\iota(c)$ is
442sharp so by (\ref{indeximage}), $c$ is sharp.
443\end{proof}
444
445
446\begin{corollary}\label{cyclicsharp}
447A cyclic abelian group is a sharp module.
448\end{corollary}
449\begin{proof}
450The automorphism group of a cyclic group is finite cyclic, hence every subgroup
451is sharp.  Now apply (\ref{sharpsubgroupenough}).
452\end{proof}
453
454%In the case $G$ profinite we remark that the map
455%$G\ra \Aut M$ is continuous with $M$ given the discrete topology.
456%This is because, by definition, the map $G\cross M\ra M$ is continuous with
457
458\comment{
459\begin{theorem}
460$M=\Z\oplus\Z$ is a sharp module.
461\end{theorem}
462\begin{proof}
463The profinite subgroups of the discrete group $\Aut M=\sltwoz$ are
464finite, so it suffices to show that the finite subgroups of
465$\sltwoz$ are sharp.
466[... BUT IS THIS TRUE? ...]
467\end{proof}
468}
469
470Let $\Fp=\Z/p\Z$.
471\begin{theorem}\label{gltwo}
472If $G\subset \GL_2(\Fp)$ and $q\geq 1$ then
473 $(q,G,\Fp\oplus\Fp)$ is sharp.
474\end{theorem}
475\begin{proof}
476Let $M=\Fp\oplus\Fp$.
477If $p=2$, $\GL_2(\Fp)=S_3$ and we are done because $S_3$ is sharp.
478Thus assume $p>2$.
479
480Suppose $G$ contains a scalar $t\in \Fp^{\star} \backslash \{1\}$.
481By (\ref{innerpair}) the map
482$H^q(G,M)\ra H^q(G,M)$ induced by conjugation by $t$
483is the identity map. It is also multiplication by $t$ on
484the $\Fp$ vector space $H^q(G,M)$ so since $t\neq 1$, we conclude
485that $H^q(G,M)=0$.  If $p\nmid \#G$ then (\ref{orderannihilate}) implies that
486$H^q(G,M)=0$.  We thus assume $G\subset \sltwo(\Fp)$ and
487$p\mid \# G$.
488
489{\em Case 1: $G$ contains exactly 1 Sylow $p$-subgroup:}
490Let $S$ be {\em the} Sylow $p$-subgroup of $G$.  Since all Sylow
491$p$-subgroups of $G$ are conjugate, $S$ is normal.
492The cohomology group $H^2(G/S,S)$ classifies short exact sequences
493$0\ra S \ra E \ra G/S \ra 1.$   Since $p\nmid [G:S]$
494(\ref{orderannihilate}) implies that $H^2(G/S,S)=0$ so the
495short exact sequence $0\ra S\ra G \ra G/S\ra 1$ is trivial,
496hence splits.  The image of $G/S$ under this splitting is a
497subgroup $N\subset G$ whose order is coprime to $p$ and whose
498index equals $\#S=p$. Let $c\in H^1(G,M)$ be nonzero. Since
499$M=\Fp\cross\Fp$, $c$ has order $p$. Since the order of $N$
500is coprime to $p$ (\ref{orderannihilate}) implies that
501$\res_N(c)=0$. But $[G:N]=p$ so by (\ref{ordind}) the index
502of $c$ is $p$ and this index is attained by restricting to $N$
503so that $c$ is sharp.
504
505{\em Case 2: $G$ contains more than 1 Sylow $p$-subgroup:}
506The subgroup $S=\{\abcd{1}{*}{0}{1}\}$ is a Sylow $p$-subgroup of $\sltwo(\Fp)$.
507All Sylow $p$-subgroups of $\sltwo(\Fp)$ are conjugate
508so they are of this form with respect to some basis.  Thus a Sylow $p$-subgroup is
509determined by the line it fixes and we have a bijection
510$$\{\text{ Sylow p-subgroups of \sltwo(\Fp) }\} = \{\text{ lines in \Fp\cross\Fp }\}.$$
511There are $p+1$ lines in $\Fp\cross\Fp$ hence the same number of Sylow $p$-subgroups.
512Since $G$ contains more than 1 Sylow $p$-subgroup, it must contain the subgroup
513$H$ generated by all of them.   The set of Sylow $p$-subgroups is closed under
514conjugation so $H$ is a {\em normal} subgroup of $\sltwo(\Fp)$.
515
516The order $n$ of $H$ divides the order of $\sltwo(\Fp)$
517which is $p(p+1)(p-1)$.
518The union of the $p+1$ Sylow $p$-subgroups of $\sltwo(\Fp)$ has
519cardinality $p(p+1) - p = p^2$ (the identity is counted $p+1$ times,
520but there is no other overcounting). Thus $n\geq p^2$.  Since $n \mid 521p(p+1)(p-1)$ it is exactly divisible by $p$ so $n$ can't equal $p^2$
522and hence $n\geq p^2+1=p(p+1)$ so that $p(p+1)(p-1)/n \leq p-1$.
523Therefore the quotient $\sltwo(\Fp)/ H$ is a group of order $\leq p-1$.
524
525\comment{If $p=3$, $\#\sltwo(\Fp)=24$ and the cardinality of the union
526of the four Sylow $3$-subgroups is $4\times 3 - 3=9$. Since the order
527of $H$ divides $24$, it must be at least $12$ so $[G:H]\mid 2$.
528If $p>3$, the group $\psltwo(\Fp)=\sltwo(\Fp)/\{\pm 1\}$ is simple (see \cite{atlas}).
529The image of $H$ in $\psltwo(\Fp)$ is normal and nontrivial hence equals
530$\psltwo(\Fp)$. Thus $[\sltwo(\Fp):H]\mid 2$ and
531$H\subset G$ so $[\sltwo(\Fp):G]\mid 2$ and $G$ is normal in
532$\sltwo(\Fp)$. }
533
534By the remark in VII.6 of \cite{serre} there is an exact sequence
535\begin{eqnarray*}
5360&\ra &H^1(\sltwo(\Fp)/G,M^G)\xrightarrow{\inf}H^1(\sltwo(\Fp),M)
537    \xrightarrow{\res}H^1(G,M)\\&\ra& H^2(\sltwo(\Fp)/G,M^G)
538\end{eqnarray*}
539The order of $\sltwo(\Fp)/H$ is $\leq p-1$ so by
540(\ref{orderannihilate}) $H^2(\sltwo(\Fp)/H,M)=0$. Furthermore,
541$-1\in\sltwo(\Fp)$ so $H^1(\sltwo(\Fp),M)=0$ and hence
542$H^1(G,M)=0$.  We are now allowed to write down the exact
543sequence
544\begin{eqnarray*}
5450&\ra &H^2(\sltwo(\Fp)/G,M^G)\xrightarrow{\inf}H^2(\sltwo(\Fp),M)
546    \xrightarrow{\res}H^2(G,M)\\&\ra& H^3(\sltwo(\Fp)/G,M^G)
547\end{eqnarray*}
548Argueing in the same way shows that $H^2(G,M)=0$.  Continuing
549inductively we conclude that $H^q(G,M)=0$, so $(q,G,M)$ is sharp.
550
551\end{proof}
552
553\comment{
554\begin{remark} If $p\con 1\pmod 4$ then $\sltwo(\Fp)$
555is generated by its Sylow $p$-subgroups. Let $H$ be
556the subgroup of $\sltwo(\Fp)$ generated by the Sylow $p$-subgroups.
557Since $p\con 1\pmod{4}$ there is $a\in \Fp$ so that $a^2=-1$.
558Since $H$ surjects onto the simple group $\psltwo(\Fp)$, we
559see that $\sltwo(\Fp)$ is generated by $-1$ and $H$ so that
560$\sltwo(\Fp)/H$ is generated by $-1$.  Thus one of the matrices
561$\abcd{a}{0}{0}{-a}$ or $\abcd{-a}{0}{0}{a}$ lies in $H$.
562Both of these have square $-1$, so $-1$ lies in $H$ after all.
563%Subgroups of $\sltwo(\Fp)$ are not in general sharp.
564\end{remark}
565}
566
567\begin{corollary}
568$M=\Fp\oplus\Fp$ is a $1$-sharp module.
569\end{corollary}
570\begin{proof}
571Let $G$ be an arbitrary group equipped with an action on $M$, and
572fix $c\in H^1(G,M)$.  We must show that $c$ is sharp.  Let $N=\ker(G\ra\Aut M)$
573and $f$ be a cocycle representing $c$.  By (\ref{fstuff}), $f(N)$ is a $G$-submodule
574of $\Aut M$ and by (\ref{equivalents}) $c$ is sharp iff the class
575$d=\overline{c}'\in H^1(G/N,M/f(N))$ is sharp.  If $\# f(N) > 1$ the quotient
576$M/f(N)=(\Fp\oplus\Fp)/ f(N)$ is cyclic so (\ref{cyclicsharp}) implies $d$ is sharp.
577The remaining case is $\# f(N) = 1$.   Since $G/N$ is a subgroup of
578$\Aut M = \GL_2(\Fp)$ we may apply (\ref{gltwo}) to conclude that $d$ is sharp.
579\end{proof}
580
581
582\section{Examples}
583
584Let $G$ be a finite group.
585There is a natural map $\Z[G]\ra\Z$ sending $\sum a_g g$ to $\sum a_g\in\Z$.
586The {\bf augmentation ideal} $I_G\subset\Z[G]$ is the kernel of this map so there is an exact
587sequence $0\ra I_G \ra \Z[G]\ra \Z\ra 0$.  Since $\Z[G]$ is cohomologically trivial,
588$H^1(G,I_G)=\coker(H^0(G,\Z[G])\ra H^0(G,\Z)) = \Z/(\# G)\Z$.  If $H$
589is a subgroup of $G$ then $\Z[G]$ is cohomologically trivial as an $H$
590module so we have an isomorphism
591$H^1(H,I_G)\isom \coker(H^0(H,\Z[G])\ra H^0(H,\Z))\isom \Z/(\#H)\Z=H^1(H,I_H)$
592Thus the restriction map $\res_H$ agrees with the natural map
593$\Z/(\#G)\Z\ra\Z/(\#H)\Z$.
594
595\begin{proposition}
596$(1,G,I_G)$ is sharp iff the set $\{ \# H : H\subset G \}$ is
597closed under taking least common multiples of subsets (modulo $\#G$).
598\end{proposition}
599\begin{proof}
600In the following, we work module $\#G$.
601
602First assume that $\{ \# H : H\subset G \}$ is
603closed under $\lcm$.
604Let $m\in H^1(G,I_G)=\Z/\#G \Z$. Then $\res_H(m)=0$ iff $\#H \mid m$.
605The index of $m$ is $\#G / \lcm \{\#H : \# H \mid m\}$, so $m$ is
606sharp iff $\lcm\{\#H : \# H \mid m\}$ lies in $\{\#H : \# H \mid m\}$.
607Since the $\lcm$ divides $m$ and  $\{ \# H : H\subset G \}$ is closed
608under $\lcm$, this is the case.
609
610Next assume that $(1,G,I_G)$ is sharp.
611Let $a_1,\ldots, a_n$ be elements of $\{ \# H : H\subset G \}$
612and let $m=\lcm(a_1,\ldots,a_n)$.  By sharpness,
613$\lcm\{\#H : \# H \mid m\}$ lies in $\{\#H : \# H \mid m\}$.
614Since $a_1,\ldots,a_n \in \{\#H : \# H \mid m\}$, this $\lcm$ is
615$m$ so it follows
616that there is $H\subset G$ so that $m=\lcm\{\#H : \# H \mid m\}=\#H$,
617as desired.
618\end{proof}
619
620\begin{example} The subgroups of $A_4$ have orders $1,2,3,4,12$ so
621$(1,A_4,I_{A_4})$ is not sharp.
622\end{example}
623%Similarly, the subgroups of $A_6$ have orders $1,2,3,4,5,6,8,9,10,12,18,24,36,60,360.$
624
625\begin{example}
626If $M$ and $M'$ are sharp then $M\oplus M'$ need not be sharp.
627Let $n$ be the largest integer so that $\Z^n$ is sharp.  By
628(\ref{cyclicsharp}), $n\geq 1$, and since
629$\Z^{11}$ is the $\Z$-module underlieing the augmentation
630ideal $I_{A_4}$ is not sharp, $n\leq 10$.
631Then $\Z\oplus\Z^n$ is not sharp but both $\Z$ and $\Z^n$ are.
632\end{example}
633
634\begin{example}
635Even if $(1,G,M)$ and $(1,G,M')$ are both sharp, $(1,G,M\oplus M')$
636need not be sharp.  The subgroups of $S_4$ have orders $1, 2, 3, 4, 6, 6378, 12$, and $24$, which is cosed under $\lcm$ modulo $24$.
638Thus $I=I_{S_4}$, we have that $(1,S_4, I)$ is sharp. Let $c \in 639H^1(S_4,I)$ correspond to $6$ under the isomorphism
640$H^1(S_4,I)\isom \Z/24\Z$.  Let $\vphi:S_4\ra \Z/2\Z$ be the
641nontrivial (sign) homomorphism and let
642$d=(c,\vphi)\in H^1(S_4,I\oplus(\Z/2\Z))$. Then a subgroup
643which splits $d$ must split $\vphi$ hence be contained in
644$A_4$.  Thus the subgroups splitting $d$ are the subgroups
645of $A_4$ which split $c$. But
646$\res_{A_4}(c)\in H^1(A_4,I_{S_4})=\Z/12\Z$
647is not sharp, so $d$ is not sharp.
648\end{example}
649
650
651
652\begin{example}\label{splitexample}
653Let $G=S_3$ and $c=2\in \Z/6\Z = H^1(G,I_G)$. If $H$ is any of the $3$ subgroups of
654of order $2$ then $\res_H(c)=0$.  If $H$ and $H'$ are distinct subgroups of order $2$
655they both split $c$ but their join is $S_3$ which does not split $c$.
656Thus the analogue of (\ref{joinlemma}) for cohomology classes is false.
657\end{example}
658
659\begin{example}
660If $G$ fits into an exact sequence
661$0\ra G'\ra G \ra G''\ra 0$ with $G'$ and $G''$ sharp,
662must $G$ be sharp?  No, as the example
663$0\ra V_4\ra A_4\ra C_3\ra 0$ shows.
664\end{example}
665
666\section{Questions}
667If $G$ is a sharp group is every subgroup of $G$ sharp?
668Are direct sums of sharp groups sharp?
669Give a purely group theoretic characterization of the class
670of finite sharp groups.  Are there sharp groups which are not
671strongly Sylow?  Are strongly Sylow groups closed under either
672direct sums or quotients?
673
674Are either submodules or quotients of sharp modules necessarily sharp?
675What is the smallest integer $n$ so that $\Z^n$ is a sharp module?
676For a given prime $p$,
677what is the smallest integer $n$ so that $\Fp^n$ is a sharp module?
678Are any divisible abelian groups sharp modules, for example $\Q$?
679If a $\Z$-module $M$ is $q$-sharp for some $q\geq 1$ must it be
680$q$-sharp for {\em all} $q\geq 1$.
681
682\section{Arithmetic Applications}
683This section is under construction.
684
685In the context of abelian varieties, Lichtenbaum \cite{lichtenbaum} has
686studied the quotient $\ind(c)/\ord(c)$.  Cassels \cite{cassels} gave
687an elliptic curves $E/\Q$ and a class $c\in H^1(\Q,E(\Qbar))$ for
688which $\ord(c)\neq \ind(c)$, but showed that if $c\in\Sha(E)$ then
689$\ord(c)=\ind(c)$.
690
691
692\begin{proposition}
693Let $M$ be an abelian group, $K$ a number field, and let
694$\Gal(\Kbar/K)$ act on $M$ via the cyclotomic character.
695Then $(1,\Gal(\Kbar/K),M)$ is sharp.
696\end{proposition}
697
698\begin{proposition}
699Let $K$ be a number field, $a\in K$, and $m$ a positive integer.
700Factor the polynomial $X^m-a$ as a product $\prod g_i(X)$ of
701irreducible polynomials $g_i(X)\in K[X]$.  Then the minimum
702of the degrees of the $g_i$ divide the other degrees.
703\end{proposition}
704
705We were motived to study sharpness by the following consequence
706of the Riemann-Roch theorem.
707
708\begin{theorem}
709Let $E$ be an elliptic curve of a field $K$ and $G=\Gal(\Kbar/K)$.  Then
710$(1,G,E(\Kbar))$ is sharp.
711\end{theorem}
712
713We were unable to answer the analogous question for higher dimensional
714abelian varieties.
715\begin{question} Does there exist a nonsingular plane quartic $X$ over a field
716$K$ possessing no points of degree dividing $3$ over any quadratic extension of $K$,
717but possessing a degree $6$ point over $K$.
718\end{question}
719The cohomology group $H^1(K,\Jac(X))$ associated to such an $X$ would not be sharp.
720
721
722\begin{thebibliography}{HHHHHHH}
723\bibitem[A]{atiyah} Atiyah, Wall, {\em Cohomology of Groups},
724  in {\em Algebraic Number Theory}, Ed. J.W.S. Cassels, A. Fr\"{o}hlich,
726\bibitem[C]{cassels} J.W.S. Cassels,
727{\em Arithmetic on curves of genus 1. V. Two counterexamples.}
728J. London Math. Soc. {\bf 38} (1963) 244--248.
729\bibitem[LT]{langtate} S. Lang, J. Tate,
730{\em Principal homogeneous spaces over abelian varieties},
731{\bf 80} (1958) 659--684.
732\bibitem[L]{lichtenbaum} S. Lichtenbaum,
733{\em Duality theorems for curves over $p$-adic fields}. Invent. Math.
734{\bf 7} (1969) 120--136.
735\bibitem[S]{serre} J.P. Serre, {\em Local Fields}, Springer GTM {\bf 67} (1995).
736
737\end{thebibliography}
738\end{document}
739
740
741
742
743
744
745
746
747
748
749