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.