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As far as I know, it holds $H^2(S_3, C_4)\cong C_2$ for the trivial operation of $C_4$ as a $S_3$-module. I have tried getting a $2$-cocycle (which is not a $2$-coboundary) by its defining equation:

Let $\gamma: S_3 \times S_3 \longrightarrow C_4$ be $2$-cocycle, i.e. it holds (because of the trivial operation)

$\gamma(h,k)+\gamma(l,hk)=\gamma(lh,k)+\gamma(l,h) \quad \forall h,k,l\in S_3$.

I do not know much about group cohomology, therefore I'd be happy to see some suggestions on how to get such functions for small finite groups like these, where the first group is not cyclic.

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    $\begingroup$ You should try to understand how such isomophism arises and then you'll find your cocycle! $\endgroup$ – Pedro Tamaroff Jun 15 '16 at 18:22
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By the universal coefficient theorem we have a short exact sequence

$$0 \to \text{Ext}^1(H_1(S_3), C_4)) \to H^2(S_3, C_4) \to \text{Hom}(H_2(S_3), C_4)) \to 0$$

where any homology group with no coefficients specified has coefficients in $\mathbb{Z}$. The Schur multiplier $H_2(S_3)$ turns out to vanish, so only the first term matters and we get

$$H^2(S_3, C_4) \cong \text{Ext}^1(H_1(S_3), C_4)).$$

We have $H_1(S_3) \cong S_3/[S_3, S_3] \cong C_2$ with the quotient map $S_3 \to C_2$ being given by the sign homomorphism, and

$$\text{Ext}^1(C_2, C_4) \cong C_2$$

so we get the desired result with the additional useful piece of information that the nontrivial class in $H^2(S_3, C_4)$ factors through the quotient map $S_3 \to C_2$; that is, it comes from the nontrivial class in $H^2(C_2, C_4)$. This in turn must be the class which classifies the nontrivial central extension

$$0 \to C_4 \to C_8 \to C_2 \to 0$$

so from here we can write down a representing $2$-cocycle by picking a section of the quotient map $C_8 \to C_2$. We'll pick the section $s : C_2 \to C_8$ that sends $0 \to 0$ and $1 \to 1$ (thinking of $C_n$ as the integers $\bmod n$). The corresponding $2$-cocycle measures the extent to which $s$ fails to be a homomorphism. Explicitly, we have (written additively)

$$c(0, 0) = s(0 + 0) - s(0) - s(0) = 0$$ $$c(0, 1) = s(0 + 1) - s(0) - s(1) = 0$$ $$c(1, 0) = s(1 + 0) - s(1) - s(0) = 0$$ $$c(1, 1) = s(1 + 1) - s(1) - s(1) = -2.$$

Here $-2$ should be interpreted as $-1 \in C_4$ since the inclusion $C_4 \to C_8$ is multiplication by $2$.

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  • $\begingroup$ I am sorry, there is no counterexample. Sorry for my confusion :( I'm still trying to find out the cocycle like my $\gamma$. could you give a hint on that? $\endgroup$ – Nogard Jun 16 '16 at 9:41
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    $\begingroup$ @Nogard: it's the pullback of the cocycle above along the sign homomorphism. $\endgroup$ – Qiaochu Yuan Jun 16 '16 at 9:43

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