# Computation of group homology $H_2 ((\mathbb{Z}/3\mathbb{Z}) \rtimes (\mathbb{Z}/4\mathbb{Z}),\mathbb{Z})$

In my research I need to compute the group homology of the dicyclic group Dic3, which is a semi-direct product of $\mathbb{Z}/3\mathbb{Z}$ and $\mathbb{Z}/4\mathbb{Z}$, and let's denote it by $G$, we have a short exact sequence, $$0 \rightarrow \mathbb{Z}/3\mathbb{Z} \rightarrow G \rightarrow \mathbb{Z}/4\mathbb{Z} \rightarrow 0$$ My idea is to use Lyndon–Hochschild–Serre spectral sequence $$H_p(\mathbb{Z}/4\mathbb{Z}, H_q(\mathbb{Z}/3\mathbb{Z},\mathbb{Z})) \Rightarrow H_{p+q}(G,\mathbb{Z})$$ We already know that $$H_2(\mathbb{Z}/4\mathbb{Z}, \mathbb{Z})=H_2(\mathbb{Z}/3\mathbb{Z}, \mathbb{Z})=0$$ so we deduce that $$H_2(G,\mathbb{Z})=H_1(\mathbb{Z}/4\mathbb{Z}, H_1(\mathbb{Z}/3\mathbb{Z},\mathbb{Z}))$$ We also know $$H_1(\mathbb{Z}/3\mathbb{Z},\mathbb{Z})=\mathbb{Z}/3\mathbb{Z}$$ Question 1: Does $\mathbb{Z}/4\mathbb{Z}$ acts trivially on $H_1(\mathbb{Z}/3\mathbb{Z},\mathbb{Z})$?

If so, we could conclude that $H_2(G,\mathbb{Z})=0$.

Question 2: If the action of $\mathbb{Z}/4\mathbb{Z}$ on $H_1(\mathbb{Z}/3\mathbb{Z},\mathbb{Z})$ is not trivial, do we still have $$H_1(\mathbb{Z}/4\mathbb{Z}, H_1(\mathbb{Z}/3\mathbb{Z},\mathbb{Z})) =0$$

• $H_1(-,\mathbf{Z})$ is the abelianization, so here the action identifies to the original one. So Question 1 has a negative answer. – YCor May 8 '18 at 12:23
• @YCor, Thank you. What about Question 2? There is only one non-trivial action of $\mathbb{Z}/4\mathbb{Z}$ on $\mathbb{Z}/3\mathbb{Z}$. Are there any references which might be helpful? – Wenzhe May 8 '18 at 12:31
• But $G$ has cyclic Sylow subgroups, from which we can conclude that $H_2(G,{\mathbb Z})=0$. – Derek Holt May 8 '18 at 12:32
• You could start with en.wikipedia.org/wiki/Schur_multiplier or groupprops.subwiki.org/wiki/Schur_multiplier – Derek Holt May 8 '18 at 12:47
• Yes that's right. In fact $H_i(F,V)$ and $H^i(F,V)$ are zero for all nonzero $i$. This is in VI.8 of K.S. Brown's book on Cohomology of Groups, for example. – Derek Holt May 8 '18 at 13:43

In the spectral sequence, notice that by the remarks of YCor and Derek Holt, it is almost trivial: since the orders of $Z/4$ and $Z/3$ are prime to each other, all homology groups of the form $H_p(Z/4,H_q(Z/3,Z))$ are zero when $p\neq 0$ or $q\neq 0$. The only remaining groups are on the bounadry'' of the spectral sequence, and all differentials are trivial. It thus holds that $H_n(G,Z) = H_n(Z/3,Z)^{Z/4}\oplus H_n(Z/4)$. To understand the action of $Z/4$, you can simply check it on the level of resolutions. Write the standard resolution for $Z$ over $Z/3$. You will then find out that $H^n(Z/3,Z)$ is zero for $n$ even, and is $Z/3$ for $n$ odd. The action of a generator of $Z/4$ is then by $-1$ if $n=4k+1$ and is trivial if $n=4k+3$. In particular, $H_2$ is trivial (but this way you can also calculate all the homology groups).