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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 $R[G]=R[H] \otimes_{R[M]} R[K]$.

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  • $\begingroup$ Can you sketch the proof out of interest? $\endgroup$
    – Ali Caglayan
    Commented Mar 17, 2018 at 19:04
  • $\begingroup$ I had a proof which used the fact that $R[H] \otimes_R R[K] \cong R[H \times H]$ and the augmentation ideal of $R[M]$. However, I think that the proof that Todd Leason provides below is probably clearer. $\endgroup$
    – David Watson
    Commented Mar 19, 2018 at 11:10
  • $\begingroup$ @David: Maybe there is also a converse, i.e. something like: "If the natural $R$-linear map $\mu$ is an isomorphism of $R$-algebras then $G$ is the central product of $H, K$" ? $\endgroup$
    – Todd Leason
    Commented Mar 19, 2018 at 12:13
  • $\begingroup$ I suspect that one wouldn't necessarily be able to recover that $M < Z(G)$. For example, let $G = (Z \times A) \rtimes (B_1 \times B_2)$ where $Z$, $A$, $B_1$ and $B_2$ are groups of order p generated by $z, a, b_1, b_2$ respectively and $[b_i, a] = z$. If $H = \langle z, a, b_1 \rangle$ and $K = \langle z, a, b_2 \rangle$, then $G$ is not the central product of $H$ and $K$ but something more general (I don't have a name for it), however I believe that $\mu$ is still an isomorphism of $R$-algebras. Perhaps you'd need to assume $M < Z(H)$ and $M < Z(K)$. $\endgroup$
    – David Watson
    Commented Mar 19, 2018 at 15:52
  • $\begingroup$ Does $M \le Z(G)$ not follow already from the condition "H centralizes K" ? For, if $H$ centralizes $K$, then $hk = kh$ for all $h \in H, k\in K$. Let $m \in M = H \cap K$. Then we can take either $h := m$ (showing that $m$ centrlizes $K$) and $k := m$ (showing that $m$ centralizes $H$). Hence $m$ centralizes $HK=G$. $\endgroup$
    – Todd Leason
    Commented Mar 20, 2018 at 6:44

1 Answer 1

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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.

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