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

## 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.