Homology/Cohomology of Free Product

Question

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 $$ \frac{G}{[G,G]} \otimes_{\mathbb Z(G*H)} A \cong \frac{G}{[G,G]} \otimes_{\mathbb Z G} A ? $$

Answer 1

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 $$ms\otimes n-m\otimes sn, \qquad m\in M,\quad n\in N,\quad s\in S.$$

Answer 2

Here is another way to think about this, using topological ideas. Let $BG$ and $BH$ be the classifying spaces of the groups $G$ and $H$. Then, there is a weak equivalence $BG\lor BH\to B(G*H)$ by van-Kampen. Thus this induces an isomorphism on homology and cohomology groups $H_i(BG\lor BH;A)\xrightarrow{\sim} H_i(B(G*H);A)=H_i(G*H,A)$ and $H^i(G*H;A)\xrightarrow\sim H^i(BG\lor BH; A)$.

Furthermore, by Mayer-Vietoris there are isomorphisms $H^i(BG\lor BH;A)\cong H^i(BG;A)\oplus H^i(BH;A)=H^i(G;A)\oplus H^i(H;A)$ and $H_i(BG\lor BH;A)\cong H_i(G;A)\oplus H_i(H;A)$ for each $i>0$.