Let $G$ be a finite group and let $H^2(G,\mathbb{C}^*)$ be the second cohomology group which corresponds to the trivial $G$ action. Then there are known methods for computing $H^2(G,\mathbb{C}^*)$ for $G$ a direct product or semi-direct products of abelian groups. For example for $G=C_4\times C_4$ it is fairly easy to show that $H^2(G,\mathbb{C}^*)\cong C_4$ and for $G=S_3$ we have $H^2(G,\mathbb{C}^*)$ is trivial.
My problem is when $G$ acts in a non-trivial way, here e.g. by complex conjugation. Denoting this non-trivial action by $\eta$ I want to know how to compute $H^2_{\eta}(G,\mathbb{C}^*)$ for $G$ a direct product or semi-direct products of abelian groups. In particular I want to calculate $H^2_{\eta}(G,\mathbb{C}^*)$ for $G=C_4\times C_4$ and for $G=S_3$. My intuition says that the former is $C_4\times C_2$ and the latter $C_2$. The former "comes from" the commutator relation between the generators of $C_4\times C_4$ and from the $H^2_{\eta}(C_4,\mathbb{C}^*)$ and the latter is inflated from $H^2_{\eta}(C_2,\mathbb{C}^*)$. But these are more guesses than anything else.
I like to ask for your help, first if I am right and second how do I prove this.
Thanks in advance.
Consider
$$0 \to \Bbb Z \xrightarrow{\cdot 2\pi i} \Bbb C \xrightarrow{\exp} \Bbb C^* \to 1$$
Where $G$ acts via inversion on $\Bbb Z$ and via complex conjugation on $\Bbb C$ and $\Bbb C^*$. We will always keep these action in mind and not explicitly express them in notation.
This induces an exact sequence on cohomology groups
$H^2(G,\Bbb C) \to H^2(G,\Bbb C^*) \to H^3(G,\Bbb Z) \to H^3(G,\Bbb C)$
Now complex conjugation being $\Bbb R$-linear, $\Bbb C$ is a $\Bbb R$-linear group representation of $G$ here. Using some homological algebra, group cohomology for $\Bbb R[G]$-modules $V$ can be computed as $\operatorname{Ext}^n_{\Bbb R[G]}(\Bbb R,V)$ or as $\operatorname{Ext}^n_{\Bbb Z[G]}(\Bbb Z,V)$, these yield the same result. But due to Maschke's theorem, the category $\Bbb R[G]\textrm{-}\mathsf{Mod}$ is semisimple, so these Ext-groups vanish.
Thus we actually get $H^2(G,\Bbb C^*) \cong H^3(G,\Bbb Z)$. Now the cohomology of $S_3$ is known to be $4$-periodic (this is an exercise in Brown's book on group cohomology iirc), so we get $H^3(G,\Bbb Z)=\hat{H}^{-1}(G,\Bbb Z)$. What is the norm map? We get for $k \in \Bbb Z$ that $N(k)=k+k+k-k-k-k=0$. Thus $\hat{H}^{-1}(G,\Bbb Z)=H_0(S_3,\Bbb Z)=\Bbb Z/2\Bbb Z$.
To compute $H^3(C_4 \times C_4,\Bbb Z)$, we can use the Künneth formula for nontrivial actions. I'll assume for an example calculation that the first factor acts trivially and the latter via complex conjugation. We just need to compute $H^3_\eta(C_4,\Bbb Z)$ (with the $\eta$ indicating the nontrivial coefficient action). $H^3_\eta(C_4,\Bbb Z)\cong \hat{H}^{-1}_\eta(C_4,\Bbb Z) \cong H_{0,\eta}(C_4,\Bbb Z)\cong \Bbb Z/2\Bbb Z$ via a similar computation of norm and coinvariants as before. We also get $H^2_\eta(C_4,\Bbb Z)=\hat{H}^0_\eta(C_4,\Bbb Z)=\Bbb Z^{C_4}=0$. Of course the cohomology of $C_4$ with trivial action on $\Bbb Z$ is well-known.
Putting this into the Künneth formula, we obtain the following, where I drop the $\Bbb Z$ and just write $\eta$ for the twisted coefficients and no subscript for trivial coefficients $$H^3(C_4)\otimes H^0_\eta(C_4)\oplus H^2(C_4)\otimes H^1_\eta(C_4)\oplus H^1(C_4)\otimes H^2_\eta(C_4)\oplus H^0(C_4)\otimes H^3_\eta(C_4)=\\(0\otimes 0)\oplus (\Bbb Z/4\Bbb Z \otimes \Bbb Z/2\Bbb Z)\oplus (0 \otimes 0)\oplus (\Bbb Z\otimes \Bbb Z/2\Bbb Z)$$ This is just the first term of the Künneth formula, for the second one, we obtain:
$$\operatorname{Tor}(H^4(C_4),H^0_\eta(C_4)) \oplus\operatorname{Tor}(H^3(C_4),H^1_\eta(C_4)) \oplus\operatorname{Tor}(H^2(C_4),H^2_\eta(C_4)) \oplus\operatorname{Tor}(H^1(C_4),H^3_\eta(C_4)) \oplus\operatorname{Tor}(H^0(C_4),H^4_\eta(C_4))$$
Because $H^k(C_4)$ vanishes for odd $k$ and $H^k_\eta(C_4)$ vanishes for even $k$, these summands are all zero.
Thus we get, unless I am mistaken that $H^2(C_4 \times C_4,\Bbb C^*)=\Bbb Z/2\Bbb Z\times\Bbb Z/2\Bbb Z$