TITLE: Group rings over central products
QUESTION [5 upvotes]: I have a proof of the following result but I was wondering if anyone had a reference for it. I have asked on math.stackexchange here but didn't receive any replies.
Let $G$ a finite group given by the (inner) central product of two subgroups $H$ and $K$ over $M$ (I am using the definition of central product given in Gorenstein's "Finite groups"; in particular $G=HK$, $H \cap K=M \subset Z(G)$ and $H$ centralizes $K$). If R is a commutative ring, then $R[G]=R[H] \otimes_{R[M]} R[K]$.

REPLY [6 votes]: I don't have a reference, but the proof is routine so that the result should be well-known. Also note that $H, K$ don't need be finite. 
For a proof first note that if $H, K$ are subgroups of $G$ and $M$ is a subgroup of $H \cap K$ than $$\mu: R[H] \times {R[K]} \to R[G],\,(h,k) \mapsto h \cdot k$$
is bilinear such that $\mu(hm,k)=\mu(h,mk)$. Hence $\mu$ induces a homomorphism 
$$\mu: R[H] \otimes_{R[M]} R[K]\to R[G],\,h \otimes k \mapsto hk.$$
Now let $G$ be the central product of $H,K$, i.e. (1) $G=HK$, (2) elements of $H,K$ commute, (3) $M= H\cap K$ is central in $G$. 
By (1) $\mu$ is surjective and by (2) $\mu$ is an $R$-algebra homomorphism. Define a map
$$\varphi: G \to R[H] \otimes_{R[M]} R[K],\,hk \mapsto h \otimes k.$$
By (3) $\varphi$ is well-defined (i.e. doesn't depend on the choice of $h,k$). Extend $\varphi$ linearly to $R[G]$. Than by definition, $\varphi \circ \mu = \text{id}$, i.e. $\mu$ is injective and hence an isomorphism of $R$-algebras. q.e.d.