Reference Request: Abelianization of free product is the direct sum of abelianizations

Question

Abelianization of free product is the direct sum of abelianizations

I am interested in the result above, and would like to study a textbook that delves into it deeper.

However the standard algebraic topology textbooks (e.g. Hatcher/Munkres) and standard algebra books like Hungerford do not contain this result.

Is there any suitable book that I can consult? Thanks!

Answer 1

Let $\{G_i:i\in I\}$ be a collection of groups, $F$ be the free product of the $G_i$ and $A$ the direct sum of the abelianizations of the $G_i$. Free products, direct sums, and abelianizations all have universal properties. The universal property for the direct sum $A$ is that, for any abelian group $B$, we have

$$\mathrm{Hom}(A,B)=\prod_{i\in I}\mathrm{Hom}(G_i/[G_i,G_i],B)\text{.} $$

On the other hand, applying the universal property of abelianization, the free product, and finally abelianization again, gives

$$\mathrm{Hom}(F/[F,F],B)=\mathrm{Hom}(F,B)=\prod_{i\in I}\mathrm{Hom}(G_i,B)=\prod_{i\in I}\mathrm{Hom}(G_i/[G_i,G_i],B)$$.

These equalities (really canonical isomorphisms) constitute an isomorphism of functors $\mathrm{Hom}(A,-)\simeq\mathrm{Hom}(F/[F,F],-)$ from the category of abelian groups to the category of sets. Yoneda's lemma tells us that such an isomorphism is induced by a unique isomorphism $A\simeq F/[F,F]$.

Answer 2

This is Exercise 69.1 of Munkres' Topology.

The previous solution uses results not available in Munkres' textbook. Here I am giving an alternative solution that is more suitable for students following Munkres.

Our goal is to apply the "converse" part of Lemma 67.5.

Since $[G_\alpha, G_\alpha] \in [G, G]$ for $\alpha = 1,2$, we have induced homomorphisms $i_\alpha: G_\alpha/[G_\alpha,G_\alpha] \rightarrow G/[G,G]$ such that $i_\alpha \circ \pi_\alpha = \pi$, where $\pi_\alpha$ is the quotient map $G_\alpha \rightarrow G_\alpha/[G_\alpha,G_\alpha]$ and $\pi$ is the quotient map $G\rightarrow G/[G,G]$.

$$\require{AMScd} \require{cancel} \def\diaguparrow#1{\smash{\raise.6em\rlap{\ \ \scriptstyle #1} \lower.6em{\cancelto{}{\Space{2em}{1.7em}{0px}}}}} \begin{CD} && G_\alpha/[G_\alpha,G_\alpha] \\ & \diaguparrow{\pi_\alpha} @VV i_\alpha V \\ G_\alpha @>> \pi> G/[G,G] \end{CD}$$

In particular, we want to show that given any abelian group $H$ and any family of homomorphism $h_\alpha: G_\alpha/[G_\alpha,G_\alpha] \rightarrow H$, there exists a homomorphism $h: G/[G,G] \rightarrow H$ such that $h \circ i_\alpha = h_\alpha$ for each $\alpha$.

$$\require{AMScd} \require{cancel} \def\diaguparrow#1{\smash{\raise.6em\rlap{\ \ \scriptstyle #1} \lower.6em{\cancelto{}{\Space{2em}{1.7em}{0px}}}}} \begin{CD} && G/[G,G] \\ & \diaguparrow{i_\alpha} @VV h V \\ G_\alpha/[G_\alpha,G_\alpha] @>> h_\alpha> H \end{CD}$$

Note that $h_\alpha \circ \pi_\alpha$ defines a homomorphism $G_\alpha \rightarrow H$. Applying Lemma 68.1, we have $h': G \rightarrow H$ whose restriction to $G_\alpha$ equals $h_\alpha \circ \pi_\alpha$, for each $\alpha$.

$$\require{AMScd} \require{cancel} \def\diaguparrow#1{\smash{\raise.6em\rlap{\ \ \scriptstyle #1} \lower.6em{\cancelto{}{\Space{2em}{1.7em}{0px}}}}} \begin{CD} && G \\ & \diaguparrow{j_\alpha} @VV h' V \\ G_\alpha @>> h_\alpha \circ \pi_\alpha> H \end{CD}$$

Now we apply Lemma 69.3 and $h': G \rightarrow H$ induces a homomorphism $h: G/[G,G] \rightarrow H$ such that $h \circ \pi = h'$.

$$\require{AMScd} \require{cancel} \def\diaguparrow#1{\smash{\raise.6em\rlap{\ \ \scriptstyle #1} \lower.6em{\cancelto{}{\Space{2em}{1.7em}{0px}}}}} \begin{CD} && G/[G,G] \\ & \diaguparrow{\pi} @VV h V \\ G @>> h'> H \end{CD}$$

We claim that $h \circ i_\alpha = h_\alpha$ for each $\alpha$, because for any $x \in G_\alpha$, we have $$h_\alpha \circ \pi_\alpha(x) = h'(x) = h \circ \pi(x) = h \circ i_\alpha \circ \pi_\alpha(x).$$

Q.E.D.