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I have a question from Lang's Algebra (chapter twenty ex 6d). I think the vagueness confuses me as I am not even sure where to start

If $H$ is a normal subgroup of $G$, we have the cohomology groups $H^i (H,A)$.

The question asks to describe how we can get an operation of $G$ on the cohomology function $H_G$ "by conjugation" and functorality.

Any advice or direction on what this operation would be or how to look for it would be greatly appreciated.

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up vote 5 down vote accepted

Let $C^i(H,A)$ be the group of $i$-cochains, i.e., functions $H^i\rightarrow A$, which define the complex that computes the cohomology $H^i(H,A)$. For $g\in G$, you have the automorphism $\varphi_g:h\mapsto ghg^{-1}:H\rightarrow H$ because $H$ is normal in $G$, and you also have the abelian group automorphism $\psi_g:a\mapsto g^{-1}a:A\rightarrow A$, which is not usually $G$-equivariant, but is compatible with $\varphi_g$ in the sense that $\psi_g(\varphi_g(h)a)=h\psi_g(a)$. These maps give rise to maps $f\mapsto\psi_g\circ f\circ\varphi_g^i:C^i(H,A)\rightarrow C^i(H,A)$ which are compatible with the coboundary maps (here $\varphi_g^i:H^i\rightarrow H^i$ is the $i$-fold product of the map $\varphi_g$ with itself). So these maps descend to maps on cohomology $g^*:H^i(H,A)\rightarrow H^i(H,A)$ for all $i$. Then you get an action of $G$ on $H^i(H,A)$ by $g\cdot\kappa:=g^*(\kappa)$ (I may have the signs switched for a left action, i.e., you might need $\varphi_{g^{-1}}$ and $\psi_{g^{-1}}$ instead, but that works the same way). For example, on $H^0(H,A)=A^H$, this action is just given by $a\mapsto g^{-1}a$, so in particular, $H$ acts trivially on $H^0(H,A)$. Using an inductive argument with long exact sequences, it can be shown that $H$ acts trivially on $H^i(H,A)$ for all $i$, so you really have an action of $G/H$ on $H^i(H,A)$. This is important for defining the Hochschild-Serre spectral sequence $H^p(G/H,H^q(H,A))\Rightarrow H^{p+q}(G,A)$.

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That makes sense. What does being "compatible" imply or I suppose "how" does it imply? Thank you – user45150 Nov 15 '12 at 1:40
I'm just using the term ``compatible" to mean the equation I wrote down. The importance of this is that you can use it to show that the maps $\alpha^i:C^i(H,A)\rightarrow C^i(H,A)$ defined in my answer give rise to a map of complexes $C^\bullet(H,A)\rightarrow C^\bullet(H,A)$, i.e., that $\partial^{i-1}\circ\alpha^{i-1}=\alpha^i\circ\partial^{i-1}$, where $\partial^{i-1}:C^{i-1}(H,A)\rightarrow C^i(H,A)$ is the coboundary map. – Keenan Kidwell Nov 15 '12 at 1:52
This construction is a special case of a morphism of pairs $(G,A)\rightarrow(G^\prime,A^\prime)$, where $G,G^\prime$ are groups and $A$ (resp. $A^\prime$) is a $G$-module (resp. $G^\prime$-module). This is defined as the data of a map $\varphi:G^\prime\rightarrow G$ and $\psi:A\rightarrow A^\prime$ such that $\psi(\varphi(g^\prime)a)=g^\prime\psi(a)$ (this is exactly the compatibility condition, although in the special case of my answer, $G^\prime=H=G$ and $A=A^\prime$). Any morphism of pairs gives rise to a map of complexes $C^\bullet(G,A)\rightarrow C^\bullet(G^\prime,A^\prime)$ and thus – Keenan Kidwell Nov 15 '12 at 1:55
maps on cohomology $H^\bullet(G,A)\rightarrow H^\bullet(G^\prime,A^\prime)$. – Keenan Kidwell Nov 15 '12 at 1:55
Thank you. This was very helpful and much appreciated. – user45150 Nov 15 '12 at 17:26

This is explained succinctly in Ken Brown's classic text Cohomology of Groups, pg48.

You have your projective resolution $F\to\mathbb{Z}$ for your group in question, and you just need to check that the chain map $\tau:F\to F$ given by $\tau(x)=g\cdot x$ is $G$-equivariant and gives you your action (fixed $g\in G$). Then it induces a well-defined map $\tau:F_G\to F_G$ which induces the desired map on homology.

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If $M$ is a $G$-module, then $x\mapsto gx:M\rightarrow M$ is not usually $G$-equivariant if $G$ isn't abelian, right? – Keenan Kidwell Nov 15 '12 at 1:28
No, what is implied by $G$-equivariance here is $\tau(g_1x)=c(g_1)\tau(x)$ for a homomorphism $c:G\to G$, which in our case is conjugation, $c(g_1)=gg_1g^{-1}$. – Chris Gerig Nov 15 '12 at 1:37
In particular, $G$ acts trivially on $H_*(G)$, and for $H\triangleleft G$ we get an induced $G/H$-action on $H_*(H)$. – Chris Gerig Nov 15 '12 at 1:38

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