Sharedwww / sharp.texOpen in CoCalc
Author: William A. Stein
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% sharp.tex
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\documentclass[11pt]{article}
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\include{macros}
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\title{Subgroups in which a Cohomology Class Splits\\
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{\large (Second Rough Draft -- must reorder, finish last section.)}}
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\author{William A. Stein\footnote{UC Berkeley, Department of
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Mathematics, Berkeley, CA 94720, USA.}}
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\begin{document}
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\maketitle
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\begin{abstract}
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Let $G$ be a group, $M$ a $\Z[G]$-module, and $c\in H^q(G,M)$.
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We study the indexes of subgroups $H\subset G$
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for which $c$ restricts to $0$. In particular, we investigate
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the situation when the greatest common divisor of the indexes
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is itself an index.
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\end{abstract}
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\section{Introduction}
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Let $G$ be a (profinite) group and let $M$ be a (discrete) $G$-module.
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For any integer $q\geq 0$ let
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$H^q(G,M)$ denote the $q$-th cohomology group in the sense of \cite{atiyah}.
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Thus $H^q(G,M)$ is an additive abelian group associated covariantly to
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$M$ and contravariantly to $G$.
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If $H$ is a finite index subgroup of
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$G$ there is a restriction map $\res_H:H^q(G,M)\ra H^q(H,M)$.
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In this paper we study the set of indexes of subgroups $H\subset G$
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for which a given cohomology class restricts to $0$.
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Suppose $q\geq 1$. Define the {\bf index} of $c\in H^q(G,M)$ to be
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$$\ind(c) = \gcd\{[G:H] : \res_H(c)=0\}.$$
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Say that $c$ is a {\bf sharp cohomology class} if there exists a subgroup
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$H$ such that $[G:H]=\ind(c)$ and $\res_H(c)=0$.
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In words, ``the index is attained,'' or
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``the greatest common divisor of the indexes
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is itself an index,'' or even ``the gcd equals the minimum.''
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The triple $(q,G,M)$ is called a {\bf sharp triple} if every $c\in H^q(G,M)$
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is sharp.
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A group $G$ is a {\bf sharp group} if $(q,G,M)$ is sharp for every $G$-module
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$M$ and $q\geq 1$.
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A $\Z$-module $M$ is a {\bf sharp module} if $(q,G,M)$ is sharp
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for every $q\geq 1$ and group $G$ with an action on $M$.
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In the first section we develop basic properties of the index of a
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cohomology class. Then strongly Sylow groups are
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introduced and shown to be sharp. We next derive a criterion for
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sharpness and use it to prove that quotients of sharp groups are
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sharp. Then we show that $\Fp\cross \Fp$ is often a sharp module.
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Next we give examples showing how sharpness can fail, and some
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questions we were unable to answer.
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In the last section, sharpness is applied to
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arithmetic questions.
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\section{Sharp Groups}
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Let $G$ be a profinite group, $M$ a $G$-module, and $H^q(G,M)$ the
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$q$-th cohomology group. In this section we assume that $q\geq 1$.
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A subgroup $H$ of $G$ {\bf splits} $c$ if $\res_H(c)=0$.
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Let $\ord(c)$ denote the order of $c$ as a group element.
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\begin{lemma}\label{orderannihilate}
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Let $G$ be a finite group, $M$ a $G$-module, and
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$q\geq 1$. Then $H^q(G,M)$ is annihilated by $\# G$.
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\end{lemma}
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\begin{proof}
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This is proved as Corollary 1 of Section 6 of \cite{atiyah} by using that
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$\cores\circ \res_{\{1\}} = \#G$.
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\end{proof}
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\begin{proposition}\label{ordind}
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$\ord(c)\mid \ind(c)$ and they have the same prime factors.
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\end{proposition}
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\begin{proof}
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Suppose $H\subset G$ splits $c$. There is a map $\cores_H:H^q(H,M)\ra H^q(G,M)$ such
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that $\cores_H \circ\res_H (c) = [G:H]\cdot c$. Thus $[G:H]\cdot c=0$ and hence
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the order of $c$ divides $\ind(c)$.
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Let $p$ be a prime and let
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$G_p$ be a Sylow $p$-subgroup of $G$, so the (generalized) index
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$[G:G_p]$ is coprime to $p$ and $G_p$ is a (pro) $p$-group.
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If $p$ does not divide the order of $c$ then
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$0=\res_{G_p}(c)\in H^q(G_p,M)$ so that $\ind(c)|[G:G_p]$ is
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not divisible by $p$.
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\end{proof}
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Let $t \in G$. Consider the inner automorphism $f:G\ra G$
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given by $f(s)=tst^{-1}$. This turns $M$ into a new
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$G$-module, denoted by $M^t$, and gives a homorphism
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$f^{*}:H^q(G,M)\ra H^q(G,M^t)$. The map $g:M^t\ra M$ sending $a$ to $t^{-1}a$ defines
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an isomorphism $M^t\ra M$ and induces a map
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$g_{*}:H^q(G,M^t)\ra H^q(G,M)$.
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\begin{proposition}\label{innerpair}
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$g_{*}\circ f^{*}$ is the identity map $H^q(G,M)\ra H^q(G,M)$.
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\end{proposition}
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\begin{proof}
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This is Proposition 3 of \cite{atiyah}. The proof uses dimension shifting.
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(There is a typo in
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(4.2) of \cite{atiyah}: the first $A$ should be replaced by $A^{t}$.)
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\end{proof}
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\begin{lemma}\label{conjsplit}
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If $H$ splits $c$ then every $G$-conjugate of $H$ splits $c$.
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\end{lemma}
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\begin{proof}
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Let $t \in G$, let $f:G\ra G$ be the inner automorphism $f(s)=tst^{-1}$ and
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let $g$ be the map $M\ra M$ sending $a$ to $t^{-1}a$.
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As in section 4 of \cite{atiyah} we use functoriality of $H^q(\,,\,)$ to
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obtain a commuting diagram
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$$\begin{matrix}
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H^q(G,M) & \xrightarrow{\res_H} & H^q(H,M)\\
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\downarrow f^{*} & & \downarrow f^{*} \\
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H^q(G,M^{t}) & \xrightarrow{\res_{f^{-1}(H)}} & H^q(f^{-1}(H),M^t)\\
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\downarrow g_* & & \downarrow g_*\\
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H^q(G,M) & \xrightarrow{\res_{f^{-1}(H)}} & H^q(f^{-1}(H),M)
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\end{matrix}$$
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By (\ref{innerpair}),
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the vertical map $g_{*}\circ f^{*}$ on the left
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is the {\em identity} map.
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Thus if $\res_H(c)=0$ then so must
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$\res_{f^{-1}(H)}(c)=0$, which proves the Lemma.
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\end{proof}
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The following lemma shows that if, for each prime $p$,
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we let $G_p$ be a Sylow $p$-subgroup of $G$, then
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$\ind(c) = \prod_p\ind(\res_{G_p}(c))$.
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For $n$ an integer and $p$ a prime, let $\prt_p(n)=p^{\ord_p(n)}$.
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\begin{lemma}\label{indpsylow}
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Let $G_p$ be a Sylow $p$-subgroup of $G$.
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Then $\prt_p(\ind(c))=\ind(\res_{G_p}(c))$
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\end{lemma}
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\begin{proof}
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Suppose $H\subset G$ splits $c$ and let $H_p\subset H$ be a Sylow $p$-subgroup.
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Then $\ord_p([G:H])=\ord_p([G:H][H:H_p])=\ord_p([G:H_p])$. By Sylow's theorems,
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there exists $t\in G$ so that $t H_pt^{-1} \subset G_p$. By (\ref{conjsplit}),
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$tH_p t^{-1}$ also splits $c$. Thus
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$\ind(\res_{G_p}(c))| [G_p:t^{-1}H_p t]$
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so, since
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$\ord_p([G:t^{-1}H_p t]) = \ord_p([G:H_p])=\ord_p([G:H])$,
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it follows that
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$\ind(\res_{G_p}(c)) | \ind(c)$.
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On the other hand, $\ord_p(\ind(c))\leq \ord_p(\res_{G_p}(c))$
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because we can compute an upper bound on $\ord_p(\ind(c))$
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using subgroups of $G_p$.
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\end{proof}
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\begin{proposition}
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For any subgroup $H$ of $G$,
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$\ind(\res_H(c))|\ind(c)$.
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\end{proposition}
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\par\noindent
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{\bf Remark: } If $H'\subset G$ splits $c$ so does $H\intersect H'$. There is
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a natural injection of sets $H/(H\intersect H')\hookrightarrow G/H'$ so
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$[H:H\intersect H'] \leq [G:H']$. But, it need not be the case that
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$[H:H\intersect H'] \mid [G:H']$, for example, let $H$ and $H'$
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be distinct order $3$ subgroups of $S_4$. Then
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$3=[H:H\intersect H']$ whereas $[G:H']=\frac{24}{3}=8$.
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\begin{proof}
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It suffices to show the divisibility for each prime $p$. Let $H_p$ be a
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Sylow $p$-subgroup of $H$, and use the Sylow theorems to extend $H_p$ to
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a Sylow $p$-subgroup $G_p$ of $G$. If $H'$ is a subgroup of $G_p$ which
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splits $c$, then $H'\intersect H_p$ splits $\res_{H}(c)$. Since
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$[H_p:H'\intersect H_p] \leq [G_p : H']$ and each is a power of $p$,
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$[H_p:H'\intersect H_p] \mid [G_p : H']$. Thus
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$$\prt_p(\ind(\res_{H}(c)))=\ind(\res_{H_p}(c))\mid \ind(\res_{G_p}(c))=\prt_p(\ind(c)).$$
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\end{proof}
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%\begin{definition} A group $G$ is {\bf sharp} if $(q,G,M)$ is sharp
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%for every $G$-module $M$ and $q\geq 1$.
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%\end{definition}
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Using the technique of dimension shifting we now show that is is only necessary to
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require sharpness of $(1,G,M)$ in the definition.
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\begin{proposition}\label{sharpshift}
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If $(1,G,M)$ is sharp for every $M$ then $G$ is sharp.
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\end{proposition}
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\begin{proof}
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We proceed by induction. Let $M$ be a $G$-module and $q>1$.
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Assume that $(q-1,G,M')$ is sharp for every $G$-module $M'$.
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Consider the $G$-module $M^{*}=\Hom(\Z[G],M)$. There is a natural injection
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$M\ra M^{*}$ which maps $a\in A$ to $\vphi_a$, where $\vphi_a$ is defined
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by $\vphi_a(g) = ga$. Hence we have an exact sequence of $G$-modules
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$$0\ra M \ra M^{*} \ra M' \ra 0.$$
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Since $M^{*}$ is co-induced, it follows that
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$\delta:H^{q-1}(G,M')\ra H^q(G,M)$ is
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an isomorphism. Since a co-induced $G$-module is also a coinduced $H$-module
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for any subgroup $H$ of $G$ we obtain commuting diagrams
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$$\begin{matrix}H^{q-1}(G,M') & \ra & H^q(G,M)\\
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\downarrow \res_H & & \downarrow \res_H\\
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H^{q-1}(H,M') & \ra & H^q(H,M)
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\end{matrix}$$
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In particular, the set of subgroups $H$ of $G$ which
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split $c\in H^q(G,M)$ is the same as the set which split
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$\delta(c)\in H^{q-1}(G,M')$. By our inductive assumption,
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$(q-1,G,M')$ is sharp so $c$ is sharp, and hence $(q,G,M)$ is
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sharp.
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\end{proof}
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\begin{lemma}\label{psharp}
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If $G$ is a $p$-group then $G$ is sharp.
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\end{lemma}
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\begin{proof}
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The greatest common divisor of a set of $p$-powers equals the smallest
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so $(1,G,M)$ is sharp.
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\end{proof}
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A group $G$ is {\bf strongly Sylow} if it has the following property:
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Given subgroups $H_p\subset G_p$ for each prime $p$, there exists
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$H\subset G$ such that, for all $p$, a Sylow $p$-subgroup
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of $H$ is $G$-conjugate to $H_p$. Nilpotent groups are strongly Sylow
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because they are the direct product of their Sylow $p$-subgroups.
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The group $S_3$ is strongly Sylow but not nilpotent.
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\begin{theorem}\label{strongsylowone}
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If $G$ is strongly Sylow then $G$ is sharp.
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\end{theorem}
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\begin{proof}
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Let $M$ be a $G$-module.
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We must show that $(1,G,M)$ is sharp, so
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let $c\in H^1(G,M)$.
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For each prime $p$, (\ref{psharp}) gives a subgroup
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$H_p$ of $G_p$ so that $\ind(\res_{G_p}(c)) = [G_p:H_p]$.
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Choose $H$ as guaranteed by the property of $G$ being strongly Sylow.
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For each prime $p$, (\ref{conjsplit}) insures
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that every $p$-Sylow subgroup of $H$ splits $c$, i.e.,
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$\ind(\res_{H_p}(c))=1$.
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By (\ref{indpsylow}), $\ind(\res_H(c))=1$, so
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by (\ref{ordind}), $\res_H(c)=0$. Since for each prime $p$,
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$\ord_p([G:H])=\ord_p([G:H_p])=\ord_p([G_p:H_p])=\ord_p(\ind(c))$
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it follows that $[G:H]=\ind(c)$ so that $c$ is sharp.
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Note, this proof works equally well for $c\in H^q(G,M)$.
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\end{proof}
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Let $f:G\ra M$ be a $1$-cocycle and define
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$\ker(f) := \{\sigma\in G : f(\sigma)=0\}$.
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\begin{proposition}\label{kersubgroup}
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$\ker(f)$ is a subgroup of $G$ and $[G:\ker(f)]=\#f(G)$.
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\end{proposition}
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\begin{proof}
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If $\sigma_1,\sigma_2\in K$ then $f(\sigma_1\sigma_2)=f(\sigma_1)+\sigma_1 f(\sigma_2)=0$
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and $0=f(1)=f(\sigma_1^{-1}\sigma)=f(\sigma_1^{-1})+\sigma_1^{-1}f(\sigma_1)=f(\sigma_1^{-1}$
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so $K$ is a subgroup. If $\tau\in G$ and $\sigma\in K$, then
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$f(\tau\sigma)=f(\tau)+\tau f(\sigma) = f(\tau)$ so $f$ gives rise to a well-defined map
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of sets $\fbar : G/K\ra M$. If $\tau_1, \tau_2\in G$ and
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$\fbar(\tau_1 K) = \fbar(\tau_2 K)$, then $f(\tau_1)=f(\tau_2)$ (by definition
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of $\fbar$ so
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$f(\tau_1^{-1}\tau_2)=f(\tau_1^{-1})+\tau_1^{-1}f(\tau_2)
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= f(\tau_1^{-1})+\tau_1^{-1}f(\tau_1) = f(\tau_1^{-1}\tau_1)=0$.
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Thus $\fbar$ is injective and $\#f(G) = \# f(G/K) = \#(G/K) = [G:K]$.
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\end{proof}
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\begin{example}
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{\bf Caution:} $\ker(f)$ need {\em not} be a normal subgroup of $G$. For example,
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let $G=S_3$ and $M=\Z e_1 \oplus \Z e_2 \oplus \Z e_3$ with the natural
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permutation action. Let $f$ be the $1$-cocycle determined by $e_1$, so
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$f(\sigma)=e_{\sigma(1)} - e_1$. Then $\ker(f) = \{\iota, (23)\}$ is
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{\em not} a normal subgroup of $G$. Note that
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$3=[S_3:\ker(f)] = \# f(S_3) = \# \{ 0, e_2-e_1, e_3-e_1\}$,
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as predicted by the Proposition.
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\end{example}
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\begin{lemma}\label{kersplit}
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Fix $c\in H^1(G,M)$. A subgroup $H$ of $G$
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splits $c$ iff $H\subset \ker(f)$ for some cocycle $f \in c$.
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\end{lemma}
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\begin{proof}
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If $H\subset\ker(f)$ then $\res_H(c)$ is represented by $f|_H$ which is
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the $0$-cocycle, hence $H$ splits $c$. Suppose conversely that we know only
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that $H$ splits $c$. If $f\in c$, then $f|_H \in \res_H(c)=0$.
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Hence there is $x\in M$ so that $f(\sigma)=\sigma(x)-x$
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for all $\sigma\in H$. The cocycle $g: \sigma\mapsto f(\sigma)-(\sigma(x)-x)$
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represents $c$ and $g|H = 0$ so $H\subset\ker(g)$.
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\end{proof}
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Thus the set of kernels of cocycles is ``cofinal'' in the set of subgroups
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which split $c$, and so they can be used to compute the index.
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\begin{proposition}\label{indeximage}
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If $c\in H^1(G,M)$ then
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$$\ind(c)=\gcd\{[G:\ker(f)] : f \in c\}=\gcd\{\# f(G) : f\in c\}.$$
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\end{proposition}
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\begin{proof}
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The first equality follows from (\ref{kersplit}) and the fact that
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$\ker(f)$ splits $c$, and the second from (\ref{kersubgroup}).
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\end{proof}
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Let $N$ be a subgroup of $\ker(G\ra \Aut M)$, fix $c\in H^1(G,M)$ and choose $f\in c$.
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\begin{lemma}\label{fstuff}
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The image $f(N)$ is a $G$-submodule of $M$ and $\ker(f)\intersect N$ is
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normal in $G$, neither depends on the choice of $f\in c$. Further,
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$f(G)$ is a union of cosets of $f(N)$.
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\end{lemma}
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\begin{proof}
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Since $N$ acts trivially, the cocycle $f:N\ra M$ is a homomorphism of
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groups and thus $f(N)$ is a subgroup of $M$.
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If $\sigma\in G$ and $x\in N$ then, using that $x$ acts trivially on $M$, we have
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$\sigma f(x) = f(\sigma \sigma^{-1}) + \sigma f(x)
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= f(\sigma) + \sigma f(\sigma^{-1}) + \sigma f(x)
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= f(\sigma) + \sigma f(x) + \sigma x f(\sigma^{-1})
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= f(\sigma x) + \sigma x f(\sigma^{-1})
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= f(\sigma x \sigma^{-1}) \in f(N).$
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This proves that $f(N)$ is a $G$-submodule of $M$.
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If $f(x)=0$ then $f(\sigma x\sigma^{-1})=\sigma f(x)=0$ for all $\sigma \in G$ so
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$\ker(f)\intersect N$ is normal in $G$. Since $N$ acts trivially on $M$, coboundaries
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restrict to $0$ on $N$ so that $f(N)$ and $\ker(f)\intersect N$ do not depend on the
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choice of $f\in c$.
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Finally, $f(G)=\union_{\sigma\in G} f(N\sigma) = \union_{\sigma\in G} (f(N)+f(\sigma))$
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is a union of cosets of $f(N)$.
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\end{proof}
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\begin{lemma}\label{joinlemma}
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Let $f$ be a 1-cocycle and $H, H'$ subgroups of $G$ which split $f$. Then
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the join $H.H'$ also splits $f$.
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\end{lemma}
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\begin{proof}
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By (\ref{kersubgroup}) $\ker(f)$ is a subgroup of $G$. It contains
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both $H$ and $H'$ so it contains their join.
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\end{proof}
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{\bf Warning:} It is {\em not} true in general that if a cohomology
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{\em class} $c$ is split by subgroups $H$ and $H'$ then it is split by $H.H'$.
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See (\ref{splitexample}).
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If $H$ is a normal subgroup of $G$ then the sequence
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$$0\ra H^1(G/H,M^{H})\xrightarrow{\inf} H^1(G,M)\xrightarrow{\res} H^1(H,M)$$
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is exact. Upon setting $H=\ker(f)\intersect N$ we can define several
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cohomology classes naturally associated to $c\in H^1(G,M)$.
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\begin{theorem}\label{equivalents}
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Let $c\in H^1(G,M)$, $N\subset \ker(G\ra \Aut M)$, and $f\in c$.
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Then the following are equivalent:\\
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1. $c\in H^1(G,M)$ is sharp.\\
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2. $c'\in H^1(G/(\ker(f)\intersect N), M)$ is sharp.\\
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3. $\overline{c}\in H^1(G,M/f(N))$ is sharp.\\
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4. $\overline{c}'\in H^1(G/N,M/f(N))$ is sharp.
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\end{theorem}
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\begin{proof}
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(1)$\iff$(2):
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Suppose $H\subset G$ splits $c$. By (\ref{kersplit})
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there exists $g\in c$ so that $H\subset\ker(g)$, and by (\ref{fstuff})
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$\ker(g)$ also contains $\ker(f)\intersect N=\ker(g)\intersect N$. Thus
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$g$, and hence $c$, is split by the join $H.(\ker(f)\intersect N)$.
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There is a one-to-one index preserving correspondence between maximal subgroups of
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$G$ splitting $c$ and maximal subgroups of $G/(\ker(f)\intersect N)$ splitting
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the class $c'\in H^1(G/(\ker(f)\intersect N), M)$ defined by $c$.\\
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(1)$\iff$(3): Since any two cocycles in $\overline{c}$ differ by the coboundary
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of an element of $M/c(N)$, and such a coboundary is the image of the
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coboundary of an element of $M$,
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the set $\{f : f\in c\}$ surjects onto the set $\{ g : g \in \overline{c}\}$.
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If $f \in c$, (\ref{fstuff}) implies that $f(G)$ is a union of cosets
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of $N'=f(N)$ and $N'$ does not depend on $f$, hence
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$\{ \#g(G) : g\in\overline{c}\} = \{\#f(G)/\#f(N) : f\in c\}$.
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Sharpness means that the set of integers above satisfies $\min = \gcd$. This is true
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of the left hand side iff it is true of the right hand side iff it is true
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of $\{ \#f(G) : f\in c\}$.\\
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(3)$\iff$(4): If $g\in\overline{c}$ then by (\ref{fstuff})
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$g(N)=f(N)=\{0\}\subset M/f(N)$, so $N\subset \ker(g)$. Useing (\ref{kersplit})
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we see that there is a bijection between the maximal subgroups of
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$G$ splitting $\overline{c}$ and the maximal subgroups of
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$G/N$ which split $\overline{c}'$.
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\end{proof}
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We can now deduce that quotients of sharp groups are sharp.
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\begin{corollary}\label{sharpquotient}
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If $G$ is sharp and $G\ra G'$ is a surjective then $G'$ is sharp.
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\end{corollary}
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\begin{proof}
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Let $M$ be a $G'$ module. Make $M$ a $G$-module via $G\ra G'$.
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Let $c\in H^1(G',M)$, $f\in c$, and let $N:=\ker(G\ra G')\subset\ker(G\ra \Aut M)$.
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The map $G\ra G'$ gives rise to a map $\pi: H^1(G',M)\ra H^1(G,M)$: on cocycles
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it sends $f:G'\ra M$ to the map $G\ra M$ obtained by composing with $G\ra G'$.
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Since $G$ is sharp, $\pi(c)\in H^1(G,M)$ is sharp, so by (\ref{equivalents})
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the class it defines in $H^1(G/N,M/f(N))=H^1(G',M)$ is sharp. But the latter class
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is $c$, so $c$ is sharp and by (\ref{sharpshift}) we are done.
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\end{proof}
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\begin{theorem}\label{sharpaction}
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Fix a $G$-module $M$ and
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suppose the action of $G$ on $M$ factors
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through a sharp quotient $G'$.
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Then $(q,G,M)$ is sharp for every $q$.
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\end{theorem}
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\begin{proof}
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We first prove the theorem in the case $q=1$ and then
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apply dimension shifting.
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By (\ref{equivalents}), $(1,G,M)$ is sharp
388
iff $(1,G',M/f(N))$ is sharp for all $1$-cocycles $f:G\ra M$.
389
This is the case because $G'$ is sharp.
390
391
Next suppose the theorem is true for $q-1$.
392
Let $M'$ and $M^{*}$ be as in the proof of (\ref{sharpshift})
393
so there is an exact sequence of $G$-modules
394
$$0\ra M \ra M^{*} \ra M' \ra 0.$$
395
The action of $G$ on $M^{*}$ factors through $G'$:
396
If $h\in \ker (G\ra \Aut M)$ and $a\in M$, then for any $g\in G$,
397
$h.\vphi_a(g) = \vphi_a(gh)=gha=ga=\vphi_a(g)$.
398
Thus the action of $G$ on the quotient $M'=M^{*}/M$ factors through $G'$.
399
As in the proof of (\ref{sharpshift}), we see that
400
the set of subgroups $H$ of $G$ which
401
split 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
403
that $G$ 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
405
sharp.
406
\end{proof}
407
408
\section{Sharp Modules}
409
410
\begin{proposition}\label{sharpsubgroupenough}
411
If every (compact profinite) subgroup of $\Aut M$ is sharp then $M$ is a sharp module.
412
\end{proposition}
413
\begin{proof}
414
We 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
417
of $\Aut M$.
418
\end{proof}
419
420
We call a $\Z$-module $M$ {\bf $q$-sharp} if $(q,G,M)$ is sharp for all
421
groups $G$. Unlike in the case of groups, we do not suspect that $q$-sharpness
422
for some $q$ is equivalent to $q$-sharpness for all $q$. We have
423
only the following week result.
424
425
\begin{proposition}
426
If $M$ is $1$-sharp and $M'$ is a direct summand of $M$ then $M'$ is
427
also $1$-sharp.
428
\end{proposition}
429
\begin{proof}
430
Write $M=M'\oplus M''$.
431
Suppose $G$ is a group with an action on $M'$. Extend $G$ to act
432
trivially on $M''$. There is an injection
433
$H^1(G,M')\xrightarrow{\iota} H^1(G,M'\oplus M'').$
434
If $c\in H^1(G,M')$ and $f\in c$ then the set of
435
cardinalities of images of cocycles in $\iota(c)$ is
436
$\{\#(\iota(f)+\delta(a,b))(G) : (a,b)\in M\}$
437
where $\delta(a,b)$ is the cocycle determined by $(a,b)$.
438
Since $(\iota(f)+\delta(a,b))(\sigma) = (f(\sigma),0) +
439
(\sigma(a),b)-(a,b) = (f(\sigma)+\sigma(a)-a,0)$ we
440
see that this set of cardinalities is the same as
441
$\{ \# g(G) : g\in c\}$. Since $M$ is sharp, $\iota(c)$ is
442
sharp so by (\ref{indeximage}), $c$ is sharp.
443
\end{proof}
444
445
446
\begin{corollary}\label{cyclicsharp}
447
A cyclic abelian group is a sharp module.
448
\end{corollary}
449
\begin{proof}
450
The automorphism group of a cyclic group is finite cyclic, hence every subgroup
451
is 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}
463
The profinite subgroups of the discrete group $\Aut M=\sltwoz$ are
464
finite, so it suffices to show that the finite subgroups of
465
$\sltwoz$ are sharp.
466
[... BUT IS THIS TRUE? ...]
467
\end{proof}
468
}
469
470
Let $\Fp=\Z/p\Z$.
471
\begin{theorem}\label{gltwo}
472
If $G\subset \GL_2(\Fp)$ and $q\geq 1$ then
473
$(q,G,\Fp\oplus\Fp)$ is sharp.
474
\end{theorem}
475
\begin{proof}
476
Let $M=\Fp\oplus\Fp$.
477
If $p=2$, $\GL_2(\Fp)=S_3$ and we are done because $S_3$ is sharp.
478
Thus assume $p>2$.
479
480
Suppose $G$ contains a scalar $t\in \Fp^{\star} \backslash \{1\}$.
481
By (\ref{innerpair}) the map
482
$H^q(G,M)\ra H^q(G,M)$ induced by conjugation by $t$
483
is the identity map. It is also multiplication by $t$ on
484
the $\Fp$ vector space $H^q(G,M)$ so since $t\neq 1$, we conclude
485
that $H^q(G,M)=0$. 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:}
490
Let $S$ be {\em the} Sylow $p$-subgroup of $G$. Since all Sylow
491
$p$-subgroups of $G$ are conjugate, $S$ is normal.
492
The 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
495
short exact sequence $0\ra S\ra G \ra G/S\ra 1$ is trivial,
496
hence splits. The image of $G/S$ under this splitting is a
497
subgroup $N\subset G$ whose order is coprime to $p$ and whose
498
index 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 $N$
500
is coprime to $p$ (\ref{orderannihilate}) implies that
501
$\res_N(c)=0$. But $[G:N]=p$ so by (\ref{ordind}) the index
502
of $c$ is $p$ and this index is attained by restricting to $N$
503
so that $c$ is sharp.
504
505
{\em Case 2: $G$ contains more than 1 Sylow $p$-subgroup:}
506
The subgroup $S=\{\abcd{1}{*}{0}{1}\}$ is a Sylow $p$-subgroup of $\sltwo(\Fp)$.
507
All Sylow $p$-subgroups of $\sltwo(\Fp)$ are conjugate
508
so they are of this form with respect to some basis. Thus a Sylow $p$-subgroup is
509
determined by the line it fixes and we have a bijection
510
$$\{\text{ Sylow $p$-subgroups of $\sltwo(\Fp)$ }\} = \{\text{ lines in $\Fp\cross\Fp$ }\}.$$
511
There are $p+1$ lines in $\Fp\cross\Fp$ hence the same number of Sylow $p$-subgroups.
512
Since $G$ 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
514
conjugation so $H$ is a {\em normal} subgroup of $\sltwo(\Fp)$.
515
516
The order $n$ of $H$ divides the order of $\sltwo(\Fp)$
517
which is $p(p+1)(p-1)$.
518
The union of the $p+1$ Sylow $p$-subgroups of $\sltwo(\Fp)$ has
519
cardinality $p(p+1) - p = p^2$ (the identity is counted $p+1$ times,
520
but there is no other overcounting). Thus $n\geq p^2$. Since $n \mid
521
p(p+1)(p-1)$ it is exactly divisible by $p$ so $n$ can't equal $p^2$
522
and hence $n\geq p^2+1=p(p+1)$ so that $p(p+1)(p-1)/n \leq p-1$.
523
Therefore 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
526
of the four Sylow $3$-subgroups is $4\times 3 - 3=9$. Since the order
527
of $H$ divides $24$, it must be at least $12$ so $[G:H]\mid 2$.
528
If $p>3$, the group $\psltwo(\Fp)=\sltwo(\Fp)/\{\pm 1\}$ is simple (see \cite{atlas}).
529
The 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
534
By the remark in VII.6 of \cite{serre} there is an exact sequence
535
\begin{eqnarray*}
536
0&\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*}
539
The 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
543
sequence
544
\begin{eqnarray*}
545
0&\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*}
548
Argueing in the same way shows that $H^2(G,M)=0$. Continuing
549
inductively 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 $\sltwo(\Fp)$
555
is generated by its Sylow $p$-subgroups. Let $H$ be
556
the subgroup of $\sltwo(\Fp)$ generated by the Sylow $p$-subgroups.
557
Since $p\con 1\pmod{4}$ there is $a\in \Fp$ so that $a^2=-1$.
558
Since $H$ surjects onto the simple group $\psltwo(\Fp)$, we
559
see 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 $H$.
562
Both 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}
571
Let $G$ be an arbitrary group equipped with an action on $M$, and
572
fix $c\in H^1(G,M)$. We must show that $c$ is sharp. Let $N=\ker(G\ra\Aut M)$
573
and $f$ be a cocycle representing $c$. By (\ref{fstuff}), $f(N)$ is a $G$-submodule
574
of $\Aut M$ 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.
577
The 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
584
Let $G$ be a finite group.
585
There is a natural map $\Z[G]\ra\Z$ sending $\sum a_g g$ to $\sum a_g\in\Z$.
586
The {\bf augmentation ideal} $I_G\subset\Z[G]$ is the kernel of this map so there is an exact
587
sequence $0\ra I_G \ra \Z[G]\ra \Z\ra 0$. 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 $H$
589
is a subgroup of $G$ then $\Z[G]$ is cohomologically trivial as an $H$
590
module so we have an isomorphism
591
$H^1(H,I_G)\isom \coker(H^0(H,\Z[G])\ra H^0(H,\Z))\isom \Z/(\#H)\Z=H^1(H,I_H)$
592
Thus 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
597
closed under taking least common multiples of subsets (modulo $\#G$).
598
\end{proposition}
599
\begin{proof}
600
In the following, we work module $\#G$.
601
602
First assume that $\{ \# H : H\subset G \}$ is
603
closed under $\lcm$.
604
Let $m\in H^1(G,I_G)=\Z/\#G \Z$. Then $\res_H(m)=0$ iff $\#H \mid m$.
605
The index of $m$ is $\#G / \lcm \{\#H : \# H \mid m\}$, so $m$ is
606
sharp iff $\lcm\{\#H : \# H \mid m\}$ lies in $\{\#H : \# H \mid m\}$.
607
Since the $\lcm$ divides $m$ and $\{ \# H : H\subset G \}$ is closed
608
under $\lcm$, this is the case.
609
610
Next assume that $(1,G,I_G)$ is sharp.
611
Let $a_1,\ldots, a_n$ be elements of $\{ \# H : H\subset G \}$
612
and let $m=\lcm(a_1,\ldots,a_n)$. By sharpness,
613
$\lcm\{\#H : \# H \mid m\}$ lies in $\{\#H : \# H \mid m\}$.
614
Since $a_1,\ldots,a_n \in \{\#H : \# H \mid m\}$, this $\lcm$ is
615
$m$ so it follows
616
that there is $H\subset G$ so that $m=\lcm\{\#H : \# H \mid m\}=\#H$,
617
as 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}
626
If $M$ and $M'$ are sharp then $M\oplus M'$ need not be sharp.
627
Let $n$ 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
630
ideal $I_{A_4}$ is not sharp, $n\leq 10$.
631
Then $\Z\oplus\Z^n$ is not sharp but both $\Z$ and $\Z^n$ are.
632
\end{example}
633
634
\begin{example}
635
Even if $(1,G,M)$ and $(1,G,M')$ are both sharp, $(1,G,M\oplus M')$
636
need not be sharp. The subgroups of $S_4$ have orders $1, 2, 3, 4, 6,
637
8, 12$, and $24$, which is cosed under $\lcm$ modulo $24$.
638
Thus $I=I_{S_4}$, we have that $(1,S_4, I)$ is sharp. Let $c \in
639
H^1(S_4,I)$ 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
641
nontrivial (sign) homomorphism and let
642
$d=(c,\vphi)\in H^1(S_4,I\oplus(\Z/2\Z))$. Then a subgroup
643
which splits $d$ must split $\vphi$ hence be contained in
644
$A_4$. Thus the subgroups splitting $d$ are the subgroups
645
of $A_4$ which split $c$. But
646
$\res_{A_4}(c)\in H^1(A_4,I_{S_4})=\Z/12\Z$
647
is not sharp, so $d$ is not sharp.
648
\end{example}
649
650
651
652
\begin{example}\label{splitexample}
653
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
654
of order $2$ then $\res_H(c)=0$. If $H$ and $H'$ are distinct subgroups of order $2$
655
they both split $c$ but their join is $S_3$ which does not split $c$.
656
Thus the analogue of (\ref{joinlemma}) for cohomology classes is false.
657
\end{example}
658
659
\begin{example}
660
If $G$ fits into an exact sequence
661
$0\ra G'\ra G \ra G''\ra 0$ with $G'$ and $G''$ sharp,
662
must $G$ 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}
667
If $G$ is a sharp group is every subgroup of $G$ sharp?
668
Are direct sums of sharp groups sharp?
669
Give a purely group theoretic characterization of the class
670
of finite sharp groups. Are there sharp groups which are not
671
strongly Sylow? Are strongly Sylow groups closed under either
672
direct sums or quotients?
673
674
Are either submodules or quotients of sharp modules necessarily sharp?
675
What is the smallest integer $n$ so that $\Z^n$ is a sharp module?
676
For a given prime $p$,
677
what is the smallest integer $n$ so that $\Fp^n$ is a sharp module?
678
Are any divisible abelian groups sharp modules, for example $\Q$?
679
If 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}
683
This section is under construction.
684
685
In the context of abelian varieties, Lichtenbaum \cite{lichtenbaum} has
686
studied the quotient $\ind(c)/\ord(c)$. Cassels \cite{cassels} gave
687
an elliptic curves $E/\Q$ and a class $c\in H^1(\Q,E(\Qbar))$ for
688
which $\ord(c)\neq \ind(c)$, but showed that if $c\in\Sha(E)$ then
689
$\ord(c)=\ind(c)$.
690
691
692
\begin{proposition}
693
Let $M$ be an abelian group, $K$ a number field, and let
694
$\Gal(\Kbar/K)$ act on $M$ via the cyclotomic character.
695
Then $(1,\Gal(\Kbar/K),M)$ is sharp.
696
\end{proposition}
697
698
\begin{proposition}
699
Let $K$ be a number field, $a\in K$, and $m$ a positive integer.
700
Factor the polynomial $X^m-a$ as a product $\prod g_i(X)$ of
701
irreducible polynomials $g_i(X)\in K[X]$. Then the minimum
702
of the degrees of the $g_i$ divide the other degrees.
703
\end{proposition}
704
705
We were motived to study sharpness by the following consequence
706
of the Riemann-Roch theorem.
707
708
\begin{theorem}
709
Let $E$ 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
713
We were unable to answer the analogous question for higher dimensional
714
abelian 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 $K$,
717
but possessing a degree $6$ point over $K$.
718
\end{question}
719
The 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,
725
Academic Press (1967).
726
\bibitem[C]{cassels} J.W.S. Cassels,
727
{\em Arithmetic on curves of genus 1. V. Two counterexamples.}
728
J. 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}
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