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\begin{document}

\begin{center}
\bf
Math 254B, UC Berkeley, Spring 2002 (Kedlaya) \\
The Cohomology of Finite Groups II: Concrete Nonsense
\end{center}

\head{Reference} Milne, II.1.

Last time, we associated to a finite group $G$ and a (right) $G$-module
$M$ a sequence of abelian groups $H^i(G, M)$, called the cohomology
groups of $M$. (Also called the \emph{Galois cohomology} groups because in
number theory, $G$ will invariably be the Galois group of some extension
of number fields, and $A$ will be some object manufactured from this
extension.) What we didn't do is make the construction at all usable in
practice! This time we will remedy this.

Recall the last point from last time: if 
\[
0 \to M \to M_0 \to M_1 \to \cdots
\]
is an acyclic resolution of $M$ (i.e., the sequence is exact, and 
$H^i(G, M_j) = 0$ for $i > 0$ and all $j$), then
\[
H^i(G, M) = \ker(M_i^G \to M_{i+1}^G)/\im(M_{i-1}^G \to M_i^G).
\]
Thus to compute cohomology, we are going to need an ample supply of
acyclic $G$-modules. As we did for injectives, we'll get these by inducing
up from the trivial group. Note that if $G$ is the trivial group, 
\emph{every} $G$-module is acyclic: if $0 \to M \to I_0 \to I_1 \cdots$ is
an injective resolution, taking $G$-invariants has no effect, so
$0 \to I_0 \to I_1 \to \cdots$ is still exact except at $I_0$
(where we omitted $M$).

We say $M$ is an \emph{induced} $G$-module if it has the form $\Ind^G_{\{e\}}
N$
for some abelian group (i.e., module for the trivial group) $N$, i.e.,
can be written as $N \otimes_\ZZ \ZZ[G]$.
\begin{lemma}[Shapiro's Lemma]
  If $H$ is a subgroup of $G$ and $N$ is an $H$-module, then
there is a canonical isomorphism $H^i(G, \Ind^G_H(N)) \to H^i(H, N)$.
In particular, $N$ is an acyclic $H$-module if and only if
$\Ind^G_H(N)$ is an acyclic $G$-module.
\end{lemma}
\begin{proof}
The key points are:
\begin{enumerate}
\item[(a)] $\Ind^G_H(N)^G = N^H$, so there is an isomorphism for $i=0$;
\item[(b)] the functor $\Ind^G_H$ from $H$-modules to $G$-modules
is exact (that is, $\ZZ[G]$ is flat over $\ZZ[H]$, which is easy to see
because it in fact is free over $\ZZ[H]$);
\item[(c)] if $I$ is an injective $H$-module, then $\Ind^G_H(I)$ is an
injective $G$-module. (This is similar to the proof from the previous
homework that there are enough injectives; or you can use the canonical
isomorphism $\Hom_G(M, \Ind^G_H(I)) = \Hom_H(M, I)$.)
\end{enumerate}
Now take an injective resolution of $N$, apply $\Ind^G_H$ to it, and the
result is an injective resolution of $\Ind^G_H(N)$.
\end{proof}
\begin{cor}
If $M$ is an induced $G$-module, then $M$ is acyclic.
\end{cor}
It is much easier to embed $M$ into an acyclic $G$-module than into an
injective $G$-module. For example, let $M_0$ be the underlying abelian group
of $M$, regarded as a module for the trivial group, and put
$N = \Ind^G(M_0) = M_0 \otimes_{\ZZ} \ZZ[G]$. Then $M$ has a natural map into
$N$:
\[
m \mapsto \sum_{g \in G} m^{g} \otimes [g^{-1}].
\]
The point is that
\[
m^h \mapsto \sum_{g \in G} m^{hg} \otimes [g^{-1}]
= \sum_{g \in G} m^g \otimes [(h^{-1}g)^{-1}]
= \left( \sum_{g \in G} m^g \otimes [g^{-1}] \right)[h],
\]
so the map is compatible with the $G$-actions.

Immediate consequence: if $M$ is finite, it can be embedded into a
finite acyclic $G$-module. Thus $H^i(G,M)$ is finite for all $i$. (But
contrary to what you might expect, for fixed $M$, the groups $H^i(G,M)$
do not necessarily become zero for $i$ large, even if $M$ is finite! 
We'll see explicit examples next time.)

Another consequence is the following result. (The case $i=1$ was a homework
problem.)
\begin{theorem}
Let $L/K$ be a finite Galois extension of fields. Then
$H^i(\Gal(L/K), L) = 0$ for $i>0$.
\end{theorem}
\begin{proof}
Put $G = \Gal(L/K)$.
The normal basis theorem (see Lang, \emph{Algebra} or Milne, Lemma
II.1.24)
states that there exists $\alpha \in L$ whose conjugates form a basis of
$L$ as a $K$-vector space. This implies that $L \cong \Ind^{G}_{\{e\}}
K$, so $L$ is an induced $G$-module and so is acyclic.
\end{proof}

Now let's see an explicit way to compute group cohomology.
Given a group $G$ and a $G$-module $M$, define the $G$-modules $N_i$
for $i \geq 0$ as functions $\phi: G^{i} \to M$, with the $G$-action
\[
(\phi^g)(g_0, \dots, g_i) = \phi(g_0g^{-1}, \dots, g_ig^{-1})^g.
\]
Notice that this module is induced: any such function is determined
by its values with $g_0=e$.

Define the map $d_i: N_i \to N_{i+1}$ by
\[
(d_i \phi)(g_0, \dots, g_{i+1})
= \sum_{j=0}^{i+1} (-1)^j \phi(g_0, \dots, \widehat{g_j}, \dots, g_{i+1}),
\]
where the hat over $g_j$ means you omit it from the list.
Then one checks that the sequence
\[
0 \to M = N_0 \to N_1 \to \dots
\]
is exact. Since the $N_i$ are induced, this is an acyclic resolution:
thus the cohomology of the complex
\[
0 \to N_0^G \to N_1^G \to \cdots
\]
coincides with the cohomology groups $H^i(G, M)$. And now we have
something we can actually compute! (Terminology: the elements of
$N_i^G$ in the kernel of $d_i$ are called (homogeneous) $i$-cochains;
the ones in the image of $d_{i-1}$ are called $i$-coboundaries.)

For example, we can give a very simple description of $H^1(G,M)$.
Namely, a 1-cochain $\phi: G^2 \to M$ is determined by $\rho(g) = \phi(e, g)$
which by $G$-invariance satisfies the relation
\begin{align*}
0 &= (d_1\phi)(e, h, gh) \\
&= \phi(h, gh) - \phi(e, gh) + \phi(e, h) \\
&= (\phi^h)(h,gh) - \rho(gh) + \rho(h) \\
&= \phi(e, g)^h - \rho(gh) + \rho(h) \\
&= \rho(g)^h + \rho(h) - \rho(gh).
\end{align*}
It is the coboundary of a 0-cochain $\psi: G \to M$ if and only if
\[
\rho(g) = \phi(e,g) = \psi(e) - \psi(g) = \psi(e) - \psi(e)^g.
\]
That is, $H^1(G,M)$ is equal to crossed homomorphisms modulo principal
crossed homomorphisms, consistent with the definition
we gave in the section on Kummer theory.

We can also give an explicit interpretation of $H^2(G,M)$ (see
Milne, example II.1.19). It classifies
short exact sequences
\[
1 \to M \to E \to G \to 1
\]
of (not necessarily abelian) groups on which $G$ has a fixed action on $M$.
(The action is given as follows: given $g \in G$ and $m \in M$, choose $h
\in E$ lifting $G$; then $h^{-1}mh$ maps to the identity in $G$, so comes
from $M$, and we call it $m^g$ since it depends only on $g$.)
Namely, given the sequence, choose a map $s: G \to E$ (not a homomorphism)
such that $s(g)$ maps to $g$ under the map $E \to G$. Then the map
$\phi: G^3 \to M$ given by
\[
\phi(a,b,c) = s(a)^{-1} s(ba^{-1})^{-1} s(cb^{-1})^{-1} s(ca^{-1}) s(a)
\]
is a 2-cocycle, and any two choices of $s$ give maps that differ by
a 2-coboundary. (Truth in advertising disclosure: I may have messed up the
formula. Verify before using.)

What ``classifies'' means precisely is that two sequences give the same
element of $H^2(G,M)$ if and only if one can find an arrow $E \to E'$ making
the following diagram commute:
\[
\xymatrix{
1 \ar[r] & M \ar^{\id}[d] \ar[r] & E \ar[d] \ar[r] & G \ar^{\id}[d] \ar[r] &
1 \\
1 \ar[r] & M \ar[r] & E' \ar[r] & G \ar[r] & 1
}
\]
Note that two sequences may not be isomorphic under this definition even
if $E$ and $E'$ are abstractly isomorphic as groups. For example, if
$G = M = \ZZ/p\ZZ$ and the action is trivial, then $H^2(G, M) = \ZZ/p\ZZ$
even though there are only two possible groups $E$, namely $\ZZ/p^2\ZZ$
and $\ZZ/p\ZZ \times \ZZ/p\ZZ$.

\head{Exercises}

\begin{enumerate}
\item
The set $H^2(G,M)$ has the structure of an abelian group. Describe the
corresponding structure on short exact sequences $0 \to M \to E \to G \to 0$.
\item
Let $G = S_3$ (the symmetric group on three letters), let $M = \ZZ^3$
with the natural $G$-action permuting the factors, and let $N = M^G$.
Compute $H^i(G, M/N)$ for $i=1,2$ however you want. (You can explicitly
compute cochains, use the alternate interpretations given above,
or use the exact sequence $0 \to N \to M \to M/N \to 0$.) Optional: use
more than one method and make sure that you get the same answer.
\item
(Artin-Schreier)
Let $L/K$ be a $\ZZ/p\ZZ$-extension of fields of characteristic $p>0$.
Prove that $L = K(\alpha)$ for some $\alpha$ such that
$\alpha^p - \alpha \in K$. (Hint: consider the short exact sequence
$0 \to \FF_p \to L \to L/\FF_p \to 0$ in which the map $L \to L/\FF_p$
is given by $x \mapsto x^p - x$.)
\end{enumerate}

\end{document}