Let $G=\langle x,y,z,w \mid [y,x]=w^p=z, x^{p^2}=y^p=z^p=1 \rangle$ be a group, where $[u,v]=u^{-1}v^{-1}uv$, $p$ is a prime and the commutator which do not appear is 1.


Let $N=\langle y,w \rangle \cong C\_p \times C\_{p^2}$ and $T=\langle x \rangle$. I wish to calculate $H^2(N,\mathbb{C^{\star}})^T$, the $T$-stable subgroup of $H^2(N,\mathbb{C^{\star}})$. It is clear (by Knuth formula) that $H^2(N,\mathbb{C^{\star}}) \cong C\_p$. But I am unable to find $H^2(N,\mathbb{C^{\star}})^T$. 


Kindly give some hint or help.



How to claculate the $T$-stable subgroup of second cohomology group
The action is trivial, because the only action of $C\_{p^2}$ on the abelian group $\mathbb{Z}/p$ is trivial. You can also see it more directly, by thinking of a two cocycle as giving a twisted group algebra. If we call the cocycle $\alpha$, then we have the algebra $\mathbb{C}^{\alpha}N$, with basis $u\_g$ and multiplication $u\_gu\_h = \alpha(g,h)u\_{gh}$. If we write now $\zeta$ for a primitive $p$-th root of unity, then $u\_yu\_w = \zeta^iu\_wu\_i$ for some $0\leq i\leq p-1$. This $i$ gives you the isomorphism you mentioned above between $H^2(N,\mathbb{C}^{\times})$ and $C\_p$. The action of $x$ on this will then be given by $u\_{x(y)}y\_{x(w)} = \zeta^{x(i)}u\_{x(w)}u\_{x(y)}$. A direct calculation shows then that $x(i)=i$
(the advantage of the twisted group algebra consideration is that it works in more complicated situations)


