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\title{Lecture 4-29: Restriction, corestriction, and $H^1$}
\date{April 29, 2024}
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We exhibit some important maps relating the cohomology of a group $G$ and that of its subgroups $H$; these lead to an important constraint on the cohomology of $G$.  For every $G$-module $A$ we write down explicit equations describing the first cohomology group $H^1(G,A)$ and relate this group to group extensions of $A$ by $E$  
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We begin with a definition.  Given finite groups $G,G'$, modules $A,A'$ over $G,G'$, and group homomorphisms $\phi:G'\rightarrow G,\psi:A\rightarrow A'$, we say that $\phi$ and $\psi$ are {\sl compatible} if $\psi(\phi(g')a) = g'\psi(a)$ for $g'\in 
G',a\in A$.  If this holds then $\phi$ and $\psi$ induce homomorphisms $\lambda_n:C^n(G,A)\rightarrow C^n(G',A')$ for all $n$, which commute with the coboundary maps, so that they map cocycles to cocycles and coboundaries to coboundaries.  Hence we get a homomorphism $\lambda_n:H^n(G,A)\rightarrow H^n(G',A')$.  As a simple example, any $G$-homomorphism $A\rightarrow A'$ is compatible in this sense with the identity map on $G$, so we get a natural map from $H^n(G,A)$ to $H^n(G,A')$.
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In particular, if $A$ is a $G$-module, then $A$ is also a module over any subgroup $H$ of $G$; since the inclusion of $H$ into $G$ and the identity map from $A$ to itself are compatible in this sense, we get a \alert{restriction homomorphism} Res$:H^n(G,A)\rightarrow H^n(H,A)$ for $n\ge0$, which on the cochain level just restricts a map in $C^n(G,A)$ to $H^n$ (DF, p 806).  If $H$ is a normal subgroup of $G$ and $A$ is a $G$-module, then the $H$-fixed subgroup $A^H$ of $A$ is a module for the quotient group $G/H$.  The projection $G\rightarrow G/H$ and the inclusion of $A^H$ into $A$ are then compatible, so that we get an \alert{inflation homomorphism} Inf$:H^n(G/H,A^H)\rightarrow H^n(G,A)$ for $n\ge0$.  
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Now let $H$ be a subgroup of $G$ of index $m$ and $A$ a $G$-module.  Let $g_1,\ldots,g_m$ be representatives for the left cosets of $H$ in $G$.  Define a map $\psi:M_H^G(A)\rightarrow A$ via $f\mapsto\sum\limits_{i=1}^m g_i\cdot f(g_i^{-1})$.  If we change any coset representative $g_i$ to $g_ih$ then $(g_ih)f((g_ih)^{-1} = g_ihf(h^{-1}g_i^{-1}) = g_ihh^{-1}f(g_i^{-1}) = g_if(g_i^{-1})$. so the map $\psi$ is independent of the choice of coset representatives; it is also easily seen to be a $G$-homomorphism.  By Shapiro's Lemma we have an isomorphism $H^n(G,M_H^G(A))\cong H^n(H,A)$, so by composing the two homomorphisms we get the \alert{corestriction homomorphism} Cor$:H^n(H,A)\rightarrow H^n(G,A)$ (DF, p. 806).  For a cocycle $f\in\hom_{\mathbb ZH}(P_n,A)$ representing a cohomology class $c\in H^n(H,A)$ (arising from a projective resolution $\{P_n\}$ of $\mathbb Z$ over $H$) we have that Cor$(f)\in\hom_{\mathbb ZG}(P_n,A)$ is defined by Cor$(f)(p) = \sum\limits_{i=1}^m g_if(g_i^{-1}p)$.
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Then we have
\begin{alertblock}{Proposition 26, DF, p. 807}
If $H$ is a subgroup of $G$ of index $m$, then the composite Cor$\circ$Res is multiplication by $m$.
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This follows at once from the formula for Cor$(f)(p)$ given above, since if $f$ is a $G$-homomorphism, then all terms in the sum for Cor$(f)(p)$ are equal to Cor$(f)(p)$.  As an immediate consequence (taking $H=1$) we get that \alert{given a finite group $G$ of order $m$ we have $mH^n(G,A)=0$ for all $n\ge1$, so that $H^n(G,A)$ is a torsion abelian group}.  Moreover, if $A$ is finite abelian with $|A|$ and $m$ relatively prime, then it follows from this and a previous result that $H^n(G,A)=0$ for all $n\ge1$.
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Turning now to the first cohomology group $H^1(G,A)$, it follows from the definitions of $Z^1(G,A)$ and $B^1(G,A)$ that $f:G\rightarrow A$ is a cocycle if and only if $f(gh) = f(g)+gf(h)$ for all $g,h\in G$ (using additive notation for $A$).  We call such $f$ \alert{crossed homomorphisms}); note that if $G$ acts trivially on $A$, then crossed homomorphisms are just ordinary group homomorphisms from  $G$ into $A$.    Plugging in $g=1$, we see that this condition forces $f(1)=0$.  A 1-cochain $f$ is a coboundary  if and only if there is $a\in A$ with $f(g)=ga-a$; crossed homomorphisms with this property are called \alert{principal}.  
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As an example, let $G$ be the Galois group of a finite Galois extension $K$ of $F$; take $A$ to be the multiplicative group $K^*$.  A simple calculation (see DF, p. 814) shows that $H^1(G,K^*)=0$.  (The same result holds even for infinite Galois extensions, but it does {\sl not} extend to the higher cohomology groups $H^n(G,K^*)$).  In particular, if $G$ is cyclic with generator $\sigma$, then in view of a previous computation of $H^1$ for cyclic groups this computation shows that $\alpha\in K^*$ has norm 1 (that is, the product of its conjugates under $G$ is 1) if and only if $\alpha=\sigma(\beta)/\beta$ for some $\beta\in K^*$.  If the field $F$ has $m$ distinct $m$th roots of 1, where $m=|G|$, then a primitive $m$th root $\epsilon_m$ of 1 in $F$ has norm 1, whence there is $\beta\in K^*$ with $\sigma(\beta)=\beta\epsilon_m$.  It easily follows that $K$ is an $m$th root extension of $F$ (it is generated by $\beta$ and $\beta^m\in F$).
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This is an important step in the proof that in characteristic 0, a polynomial with solvable Galois group is solvable by radicals.  The corresponding fact in characteristic $p>0$ is that a Galois extension $K$ of degree $p$ is an extension by some $\beta$ with $\beta^p-\beta\in F$; here one looks at the cohomology group $H^1(G,K)$ of $G$ with coefficients in the {\sl additive} group $K$.  In this case a polynomial with solvable Galois group need {\sl not} be solvable by radicals alone.
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Our main application of $H^1(G,A)$ is to group extensions; the cohomological theory of these is quite similar to that of module extensions.  Given a group $G$ and a $G$-module $A$, an \alert{extension of $G$ by $A$} is a short exact sequence $1\rightarrow A\rightarrow E\rightarrow G\rightarrow 1$ of groups.  (Note that exactness forces the image of $A$ in $E$ to be a normal subgroup.)  We identify two extensions $1\rightarrow A\rightarrow E\rightarrow G\rightarrow 1$ and $1\rightarrow A\rightarrow E'\rightarrow G\rightarrow 1$ whenever there is an isomorphism $\pi:E\rightarrow E'$ restricting to the identity map on the copy of $A$ sitting inside $E$ and commuting with the projection maps $\pi:E\rightarrow G,\pi':E'\rightarrow G$.  For now we consider only \alert{split} extensions, that is, ones for which there is a homomorphism $\sigma:G\rightarrow E$ such that the composite $\pi\sigma$ is the identity on $G$.  Such a $\sigma$ is called a \alert{splitting} and its image a \alert{complement to $A$ in $E$}.
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In this case we identify $E$ with the semidirect product $A\rtimes G$, consisting of all ordered pairs $(a,g)\in A\times G$ with multiplication given by $(a,g)(b,h)=(a+gb,gh)$ (recalling once again that the group operation is written multiplicatively in $E$ and $G$ but additively in $A$).  Now consider the set of all splittings $\sigma$.  Any such sends $g\in G$ to $(f(g),g)$ for some $f:G\rightarrow A$; since $(f(g),g)(f(h),h)=(f(g)+gf(h),gh)$ the condition that $\sigma$ be a homomorphism is equivalent to requiring that $f\in Z^1(G,A)$.  Composing $\sigma$ with conjugation by $(a,1)$ for any $a\in A$ then amounts to replacing $f$ by $f+a-ga$, that is, altering $f$ by a 1-coboundary.
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Reformulating the above discussion slightly, we obtain
\begin{alertblock}{Theorem, DF, Proposition 33, p. 820}
Given a semidirect product $E=A\rtimes G$ (with fixed action of $G$ on $A$) there is a bijection between $E$-conjugacy classes of complements to $A$ in $E$ and $H^1(G,A)$.
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\begin{proof}
A complement to $A$ in $E$ necessarily takes the form $\{(f(g),g);g\in G\}$ for some $f:G\rightarrow A$; we have seen that such a function must be a 1-cocycle.  As such a complement is stable under conjugation by any of its elements, its $E$-conjugacy classes are the same as its $A$-conjugacy classes, which we have shown to be in bijection to $H^1(G,A)$.
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For example, let $G=\mathbb Z/2\mathbb Z$ act on $A=\mathbb Z/4\mathbb Z$ by inversion (the unique nontrivial element of $G$ sends every element of $A$ to its inverse).  The semidirect product $E=A\rtimes G$ is then the dihedral group $D$ of order 8; the subgroup $A$ consists of the rotations in $D$.  The complements of $A$ in $E$ are then the cyclic subgroups generated by reflections in $D$.  There are two conjugacy classes of such reflections; accordingly, $H^1(G,A)$ is cyclic of order 2.  This agrees with the earlier computation of the cohomology groups of a cyclic group.  This example is discussed on p. 820.
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As an immediate consequence we get 
\begin{alertblock}{Corollary, DF, p. 821}
Given a semidirect product $E=A\rtimes G$ with $A$ abelian and the order of $G$ relatively prime to that of $A$, all complements of $A$ in $E$ are $A$-conjugate.
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This follows since we saw above that $H^n(G,A)=0$ for $n\ge0$ in this situation.
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