Consider a finite perfect group $G$ and a $G$-module, $U\left(1\right)$, on which $G$ acts trivially. Here $U\left(1\right)$ is the set of $1 \times 1$ unitary matrices over $\mathbb{C}$.

Are there any examples in which the second cohomology group $H_{2}\left(G,U\left(1\right)\right)$ is non-trivial ? If the answer is positive, is there an explicit expression for such a non-trivial cocycle ?

  • $\begingroup$ Mathstackexchange supports the standard Latex commands. Try to use them! $\endgroup$ – Stefano Apr 4 '14 at 9:34

Can't happen because of the universal coeff Theorem for Homology $0\to H_2(G,Z) \otimes U(1) \to H_2(G,U(1)) \to \mathrm{Tor}(H_1(G,Z),U(1)) \to0$. Since $G$ is perfect $H_1(G,Z)$ is zero, since $G$ is finite $H_2(G,Z)$ is finite, since $U(1)$ is a divisible group and $H_2(G,Z)$ is finite, $H_2(G,Z) \otimes U(1) = 0$. So $H_2(G,U(1)) = 0$.

  • 1
    $\begingroup$ you wrote H_2 but cohomology. If you mean cohomology , then by the universal coeff Theorem (for cohomology this time) the answer for H^2(G,U(1)) is that it is non zero when H_2(G,Z) is , for example for A_6. $\endgroup$ – Tomer Schlank Apr 4 '14 at 12:13

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