Calculating the second cohomology group for trivial group action

Question

Let $G$ be a finite group acting trivially on $\mathbb{R}^*$. How can I compute $H^2(G,\mathbb{R}^*)$? It seems that direct calculations are somewhat hopeless, but the answer should be simple anyway.

Answer 1

First, recall that $\mathbb{R}^{\times} \cong \mathbb{R} \times \mathbb{Z}_2$, and hence

$$H^2(G, \mathbb{R}^{\times}) \cong H^2(G, \mathbb{R}) \times H^2(G, \mathbb{Z}_2).$$

Second, since $\mathbb{R}$ is a $\mathbb{Q}$-vector space and $G$ is finite, $H^2(G, \mathbb{R}) = 0$. So the problem reduces to the computation of $H^2(G, \mathbb{Z}_2)$. At this point it depends on what else you know about $G$. For example:

If $G$ is perfect, or equivalently if $H_1(G, \mathbb{Z}) = G/[G, G] = 0$, then by universal coefficients

$$H^2(G, \mathbb{Z}_2) \cong \text{Hom}(H_2(G, \mathbb{Z}), \mathbb{Z}_2)$$

so the problem reduces to the computation of the Schur multiplier, about which a lot is known. In particular there is a method for doing this starting from a presentation of $G$ which is due to Hopf.

If $G \cong G_1 \times G_2$ is a product, then by the Kunneth formula

$$H^2(G, \mathbb{Z}_2) \cong \bigoplus_{i+j=2} H^i(G_1, \mathbb{Z}_2) \times H^j(G_2, \mathbb{Z}_2).$$

More generally, if $G$ has a normal subgroup $N$ with quotient $G/N$ then you can appeal to the Lyndon-Hochschild-Serre spectral sequence.