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I recently completed an exercise showing that $$ H_1(G*H,A) \cong H_1(G,A) \oplus H_1(H,A) $$ for $A$ a trivial $G*H$-module, and also proved a similar statement for cohomology. This is exercise 6.2.5 in Weibel's Homological Algebra.

Now I'm going back and seeing if I can prove this in a different way than I originally did. Here is my attempt:

Using the fact that $$ H_1(G,A) \cong \frac{G}{[G,G]} \otimes_{\mathbb Z G} A $$ for $A$ a trivial $G$-module, we have \begin{align*} H_1(G*H,A) &\cong \frac{G*H}{[G*H,G*H]} \otimes_{\mathbb Z(G*H)} A \\ &\cong \left( \frac{G}{[G,G]} \oplus \frac{H}{[H,H]} \right) \otimes_{\mathbb Z(G*H)} A \\ &\cong \left( \frac{G}{[G,G]} \otimes_{\mathbb Z(G*H)} A \right) \oplus \left( \frac{H}{[H,H]} \otimes_{\mathbb Z(G*H)} A \right) . \end{align*} Does it follow from some kind of change of base theorem that $$ \frac{G}{[G,G]} \otimes_{\mathbb Z(G*H)} A \cong \frac{G}{[G,G]} \otimes_{\mathbb Z G} A ? $$

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You implicitly are seeing $G/[G,G]$ as a right $\mathbb Z(G*H)$-module. The action of elements of $G$ is the obvious one, and the action of those of $H$ is trivial.

Now, the tensor product $G/[G,G]\otimes_{\mathbb Z[G*H]}A$ is the quotient of $G/[G,G]\otimes_{\mathbb Z}A$ by the abelian subgroup generated by the elements of the form $$xg\otimes a-x\otimes ga,\qquad g\in G, x\in G/[G,G], a\in A \tag{$\clubsuit$}$$ and those of the form $$xh\otimes a-x\otimes ha,\qquad h\in H, x\in G/[G,G], a\in A.\tag{$\spadesuit$}$$ This is not obvious, but easy to see.

Now the elements of listed in ($\spadesuit$) are all zero! It follows that $G/[G,G]\otimes_{\mathbb Z[G*H]}A$ is actually the quotient of $G/[G,G]\otimes_{\mathbb Z}A$ by the abelian subgroup generated by the elements listed in ($\clubsuit$), which is $G/[G,G]\otimes_{\mathbb Z[G]}A$.

N.B. The general statement underlying this is the following: if $R$ is a ring and $S\subseteq R$ is a set which generates $R$ as a ring, and $M$ and $N$ are right and left $R$-modules, respectively, then $M\otimes_RN$ is the quotient of $M\otimes_{\mathbb Z}N$ by the subgroup generated by the elements $$ms\otimes n-m\otimes sn, \qquad m\in M,\quad n\in N,\quad s\in S.$$

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