Tensor product of finite groups is finite

Statement
Suppose $$G$$ and $$H$$ are finite groups with a compatible pair of actions on each other. Consider the tensor product $$G \otimes H$$ with respect to these actions. Then, $$G \otimes H$$ is also a finite group. 

Related facts

 * Tensor product of p-groups is p-group
 * Exterior product of finite groups is finite
 * Exterior product of p-groups is p-group

Facts used

 * 1) uses::Exterior product of finite groups is finite
 * 2) uses::Order of extension group is product of order of normal subgroup and quotient group

Case where both the groups live as normal subgroups inside a big group with the action given by conjugation in the big group
-- we basically use Fact (1) and finiteness of the universal quadratic functor of any finite abelian group.

Given: A group $$Q$$. $$G,H$$ are finite normal subgroups of $$Q$$. We have natural actions of each on the other by conjugation; this is a compatible pair of actions.

To prove: The tensor product $$G \otimes H$$ with respect to the actions on each other by conjugation is finite.

Proof: By definition, there is a surjective homomorphism from the tensor product to the exterior product of groups $$G \wedge H$$:

$$G \otimes H \to G \wedge H$$

given on a generating set by:

$$g \otimes h \mapsto g \wedge h$$

By Fact (1), the group $$G \wedge H$$ is finite. Thus, by Fact (2), it suffices to show that the kernel of the homomorphism is finite.

We know, again from definition, that the kernel of the homomorphism is generated by elements of the form $$x \otimes x, x \in G \cap H$$. We note first that we only care about the coset of $$x$$ mod $$[G,H]$$, i.e., $$x \otimes x = y \otimes y$$ if $$x,y$$ are in the same coset of $$[G,H]$$. Thus, $$x \mapsto x \otimes x$$ descends to a set map from the abelian group $$(G \cap H)/[G,H]$$ to $$G \otimes H$$, and the subgroup generated by the image of this set map is the kernel we are interested in. Finally, note that the set map is quadratic, hence gives a group homomorphism from the universal quadratic functor $$\Gamma((G \cap H)/[G,H])$$ to $$G \otimes H$$. Since the universal quadratic functor of the finite group $$(G \cap H)/[G,H]$$ is finite, so is the image of the group homomorphism, which is the kernel we wanted to show is finite. We are thus done.