$H^2(G;\mathbb Z) \cong H^1(G;\mathbb {C}^*)$. Where $G$ is a finite group and $G$ acts trivially on $\mathbb Z, \mathbb C^*$.
I have really tried hard but still I couldn't solve it , any help will be appreciated.
My attempt:
Universal Coefficient theorem of Group Cohomology states that:
$$ 0 \longrightarrow Ext^{1}_{\mathbb Z}(H_{n-1}(G;\mathbb Z),\mathbb Z) \longrightarrow H^n(G, \mathbb Z) \longrightarrow Hom_{Ab}((H_{n}(G;\mathbb Z),\mathbb Z)\longrightarrow 0 $$ is split exact.
I have tried calculating $H^2(G, \mathbb Z)$ using this and $ H^1(G, \mathbb C^*)$ but nothing has come up.
Then I tried calculating $ H^1(G,\mathbb {Q/Z})$ but still dead end. Could anyone please give me atleast a hint.
$\textbf{This is an Exercise in Weibels Homological Algebra Exer 6.1.6}$ Again for any help or hint I am grateful. Thanks!!