For a group $G$ and integer $m>0$, view $\mathbb{I}_{m}$ as a trivial $G$-module. Prove that $H_{1}\left(G, \mathbb{I}_{m}\right) \cong G / G^{\prime} G^{m}$, where $G^{\prime}$ is the commutator subgroup and $G^{m}$ is the subgroup generated by all $m$ th powers.
Hint. Consider the exact sequence of $G$-trivial modules
$$
0 \rightarrow \mathbb{Z} \stackrel{m}{\longrightarrow} \mathbb{Z} \rightarrow \mathbb{I}_{m} \rightarrow 0 .
$$