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,
\begin{equation}
0 \rightarrow \mathbb{Z}/3\mathbb{Z} \rightarrow G \rightarrow \mathbb{Z}/4\mathbb{Z} \rightarrow 0
\end{equation}
My idea is to use Lyndon–Hochschild–Serre spectral sequence
\begin{equation}
H\_p(\mathbb{Z}/4\mathbb{Z}, H\_q(\mathbb{Z}/3\mathbb{Z},\mathbb{Z})) \Rightarrow H\_{p+q}(G,\mathbb{Z})
\end{equation}
We already know that 
\begin{equation}
H\_2(\mathbb{Z}/4\mathbb{Z}, \mathbb{Z})=H\_2(\mathbb{Z}/3\mathbb{Z}, \mathbb{Z})=0
\end{equation}
so we deduce that 
\begin{equation}
H\_2(G,\mathbb{Z})=H\_1(\mathbb{Z}/4\mathbb{Z}, H\_1(\mathbb{Z}/3\mathbb{Z},\mathbb{Z})) 
\end{equation}
We also know 
\begin{equation}
H\_1(\mathbb{Z}/3\mathbb{Z},\mathbb{Z})=\mathbb{Z}/3\mathbb{Z}
\end{equation}
**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
$$H\_1(\mathbb{Z}/4\mathbb{Z}, H\_1(\mathbb{Z}/3\mathbb{Z},\mathbb{Z})) =0 $$ 



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


