Show that $ H^2(G, \mathbb{Z}) \simeq \text{Hom} (G, U(1)) $ with $G = \mathbb{Z}/p\mathbb{Z}$

Question

There is "well-known" isomorphism, between "second-cohomology" of a group $G$ and "characters" which are maps $G \to U(1)$: $$ H^2(G, \mathbb{Z}) \simeq \text{Hom} (G, U(1)) $$

I was curious what this isomorphism looked like when $G = \mathbb{Z}/p\mathbb{Z}$. Then the RHS looks like maps $x \mapsto e^{2\pi i n x}$ which are like Fourier series. However, I have trouble picturing the LHS, which would be quotient (as in here):

$$ H^2(G, \mathbb{Z}) \simeq Z^2(G, \mathbb{Z})/ B^2(G, \mathbb{Z}) $$

Or possibly in terms of the Ext functor:

$$ H^2(G, \mathbb{Z}) \simeq \text{Ext}^2_{\mathbb{Z}[G]}(\mathbb{Z}, \mathbb{Z}) \simeq \mathbb{Z}[G]/(1 + \sigma + \dots + \sigma^{p-1}) $$ which looks close to "correct". There was also something about the long-exact-sequence and the short-exact-sequence: $$ 0 \to A^G \to B^G \to C^G \to H^1(G, A) \to H^1(G, B) \to H^1(G, C) \to H^2(G, A) \to \dots $$ Except that I don't know how to choose the exact sequence to get the thing going: $$ 0 \to A \to B \to C \to 0 $$ This question should have an answer within the first chapter of these notes which are also challenging to read.

Answer 1

Maybe I’m misreading the question, but one considers the short exact sequence of Abelian groups ~ $G$-modules, for any finite $G$: $$0\to\Bbb Z\hookrightarrow\Bbb R\twoheadrightarrow S^1\to0$$Because $\Bbb R$ is divisible and $G$ is finite, $H^1(G;\Bbb R)$ vanishes (no homomorphisms exist for torsion reasons) as does $\mathsf{Ext}(H_1(G;\Bbb Z),\Bbb R)$. It is known $H_\ast(G;-)$ always produces a torsion group module in the finite case, so you also know $\hom(H_2(G;\Bbb Z),\Bbb R)=0$. It follows $H^2(G;\Bbb R)$ vanishes thus the coboundary associated to the short exact sequence $\hom(G,S^1)\cong H^1(G;S^1)\to H^2(G;\Bbb Z)$ is an isomorphism.

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If you use the bar resolution and unwind how the connecting homomorphism for $\mathsf{Ext}$ works, you can use the above and the known characterisation of extensions via $H^2$ to deduce the following:

The following recipe determines an explicit isomorphism between characters of a finite group $G$ and extension classes: $$0\to\Bbb Z\to X\to G\to1$$Whose induced $G$-action on $\Bbb Z$ is trivial.

Let $\chi$ be a character of $G$. Let $\psi:G\to\left(-\frac{1}{2},\frac{1}{2}\right]\subset\Bbb R$ be the function $g\mapsto\frac{1}{2\pi i}\log\chi(g)$, where $\log$ is the principal logarithm with argument $-\pi<\arg\le\pi$.

Due to branch-jumping concerns, $\psi$ need not be a homomorphism; $\log ab\neq\log a+\log b$ in general for complex numbers. In fact, it follows from the claim that $\psi$ is a homomorphism iff. $\chi$ is the trivial character. In general the terms $\psi(g)+\psi(h)-\psi(gh)$ are always integer in range $\{-1,0,1\}$ but definitely can be nonzero.

Define the group $X$ by its underlying set $\Bbb Z\times G$ and group action: $$(n,g)(m,h):=(n+m+\psi(g)+\psi(h)-\psi(gh),gh)$$Then with the obvious projections and inclusions, we get an extension class associated to $0\to\Bbb Z\to X\to G\to1$.