Schur multiplier of finite group is finite and exponent of Schur multiplier divides group order

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Statement

Let G be a finite group. Then the Schur multiplier M(G) of G is finite, and its exponent divides the order of G.

Proof

Let n be the order of G.

We show that any 2-cocycle for the action of G on \mathbb{C}^* is cohomologous (viz, a coboundary times) a 2-cocycle with values in the n^{th} roots of unity. To see this let \alpha be a 2-cocycle \alpha: G \times G \to \mathbb{C}^*. Then:

\alpha(x,y)\alpha(xy,z) = \alpha(x,yz)\alpha(y,z)

We now take the product over all x \in G. We get:

\prod_{x \in G} \alpha(x,y)\alpha(xy,z) = \prod_{x \in G} \alpha(x,yz) \alpha(y,z)

Set t(y) = \prod_{x \in G} \alpha(x,y). The above then simplifies to:

t(y)t(z) = t(yz)\alpha(y,z)^n

For any x \in G choose g(x) \in \mathbb{C}^* such that g(x)^n = t(x)^{-1}. Then the coboundary \partial g is the map (x,y) \mapsto g(xy)/g(x)g(y) and we can easily see that \beta = (\partial g) \alpha is a 2-cocycle such that \beta(x,y)^n = 1 for all x,y \in G. This completes the proof.