Tensor square of finite group is finite

Statement
Suppose $$G$$ is a finite group. Then, the tensor square of $$G$$ is also a finite group.

Related facts

 * Tensor product of finite groups is finite
 * Exterior square of finite group is finite
 * Exterior product of finite groups is finite

Facts used

 * 1) uses::Exterior square of finite group is finite
 * 2) uses::Exact sequence giving kernel of mapping from tensor square to exterior square

Proof
By Fact (2), we have an exact sequence:

$$H_3(G;\mathbb{Z}) \to \Gamma(G^{\operatorname{ab}}) \to G \otimes G \to G \wedge G \to 0$$

By Fact (1), we already have that $$G \wedge G$$ is finite. In order to show that $$G \otimes G$$ is finite, it thus suffices to show that $$\Gamma(G^{\operatorname{ab}})$$ is finite. This follows from the fact that $$G^{\operatorname{ab}})$$ is finite, and the universal quadratic functor computation always gives a finite output for a finite group.