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Let $G=S_3$ and let $H$ be the Sylow $3$-subgroup in $G$. If $\mathbb{Z}$ is the trivial module, then it can be shown that

$$H^n(H,\mathbb{Z})=\begin{cases}\mathbb{Z}&n=0\\0&n\text{ odd}\\\mathbb{Z}_3&n\text{ even}\end{cases}$$

Since $H$ is normal in $G$, $G$ acts on $H^n(H,\mathbb{Z})$ as follows. Let $g\in G$ and define $c_g:H\to H$ by $c_g(h)=ghg^{-1}$. Since $H^*(-,M)$ is contravariant, we obtain an isomorphism $c_g^*:H^n(H,\mathbb{Z})\to H^n(H,\mathbb{Z})$. Then, for $z\in H^n(H,\mathbb{Z})$, define $g\cdot z=(c_g^*)^{-1}(z)$.

There is another way to define this action on cochains. If $F\to\mathbb{Z}$ is a projective resolution over $\mathbb{Z}G$, and $f\in\operatorname{Hom}_H(F,\mathbb{Z})$, then the $G$ action on cohomology is induced by the action $(g\cdot f)(x)=gf(g^{-1}x)=f(g^{-1}x)$ (notice that the action on $\mathbb{Z}$ is trivial).

I'm trying to compute explicitly the action of $G$ on $H^n(H,\mathbb{Z})$. This question addresses my ultimate goal, which is to compute the integral cohomology of $G$, but the answer given skips over what I've asked here (it answers my question referencing some mysterious exercise AE.9, which I cannot find in Brown).

Can someone show me how $G$ acts on $H^n(H,\mathbb{Z})$, using either (or both) of the definitions given above?

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Jared
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  • Oh crap sorry, so when I wrote up this solutions manual to the book, I added "additional exercises" that I solved at the end. Here I referenced: – Chris Gerig Oct 24 '13 at 07:31
  • The cyclic group $C_m$ is a normal subgroup of the dihedral group $D_m=C_m\rtimes C_2$ (of symmetries of the regular $m$-gon). There is a $C_2$-action on $C_m=\langle \sigma\rangle$ given by $\sigma\mapsto \sigma^{-1}$. Determine the action of $C_2$ on the homology $H_{2i-1}(C_m,\mathbb{Z})$, noting that there is an element $g\in D_m$ such that $g\sigma g^{-1}=\sigma^{-1}$. – Chris Gerig Oct 24 '13 at 07:32
  • @ChrisGerig, the question here is about cohomology, so what you wrote is not an answer to his question. – Mariano Suárez-Álvarez Oct 25 '13 at 06:49
  • I was only clearing up his comment on the missing "Exercise AE.9", and we use $H_{2i-1}(C_m)=H^{2i}(C_m)$ there. – Chris Gerig Oct 26 '13 at 00:17
  • @ChrisGerig, where do you get that ismomorphism from and how do you know it is equivariant for the action of $S_3$? – Mariano Suárez-Álvarez Oct 26 '13 at 05:19
  • Good point, I glossed over this when I originally worked on the problem. I think it is due to the naturality of the UCT. – Chris Gerig Oct 26 '13 at 07:31
  • @ChrisGerig, That works. Notice, though, that nothing is really gained by doing this computation on homology and then passing to cohomology using the UCT, as it is just as easy to do the computation in cohomology directly! – Mariano Suárez-Álvarez Oct 27 '13 at 08:42

1 Answers1

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A non-zero element of $H^2(H,\mathbb Z)$ is the class $\alpha$ of the non-split extension of $H$ by $\mathbb Z$ $$0\to\mathbb Z\to\mathbb Z\xrightarrow f\mathbb Z/3\to0$$ with $f(1)=1+3\mathbb Z$. Using esssentially the first description of the action (that is, a pullback construction), you can check that the transpositions of $S_3$ act on this extension by turning it into the extension with map $1\mapsto 2+3\mathbb Z$.

This means that transpositions act as $-1$ on $H^2$.

As noted above, you get the action on the whole of cohomology because $H^\bullet$ is (almost) a polynomial algebra generated by the class $\alpha$ of the above extension. Indeed, we have $H^\bullet(H,\mathbb Z)=\mathbb Z[\alpha]/(3\alpha)$ as a ring and the action of $S_3$ respects the ring structure.

Mariano Suárez-Álvarez
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  • Here, I ussed the interpretation of $H^2(G,\mathbb Z)$ as group extensions of $G$ by $\mathbb Z$, which is not one of the ingredients you wanted to use... but one very fast learns that in computing cohomology you need to use everything you have at hand :-) – Mariano Suárez-Álvarez Oct 24 '13 at 06:00
  • Thanks Mariano. I do know that classes of $H^2(G,\mathbb{Z})$ represent extensions of $G$ by $\mathbb{Z}$, but I've never used this result to compute cohomology. Thanks for this great example. It should prove very helpful to me. – Jared Oct 24 '13 at 07:05
  • I'm still studying this answer. I see what you mean by using a pullback construction to find another extension, $E'$, induced by the action of a transposition, but I'm having a difficult time seeing that $E'\cong \mathbb{Z}$. In general, is it true that all equivalence classes of extensions $E$ of $\mathbb{Z}_n$ by $\mathbb{Z}$ are just $\mathbb{Z}$ together with one of $n$ mappings $\mathbb{Z}\to\mathbb{Z}_n$? I know that there should be $n$ equivalence classes, but I can't see what they are. Shouldn't one of them be the direct sum? – Jared Oct 26 '13 at 05:04
  • If you pull back along an isomorphism, the groups will not change! – Mariano Suárez-Álvarez Oct 26 '13 at 05:20
  • To make explicit the $n$ extensions in the example you mention, find representing cocycles for each of them (you do know how to do this using a resolution) and then go through the proof that H^2 corresponds to extensions. (If you want help with this, ask a new question; these little comments are not a great place to do such calculations! ;-) ) – Mariano Suárez-Álvarez Oct 26 '13 at 05:22