Abelianization of free product is the direct sum of abelianizations I define $\text{Ab}(G)=G/[G,G]$ where $[G,G]$ is the commutator subgroup. I want to show that $$\text{Ab}(G_1*G_2)\cong \text{Ab}(G_1)\oplus\text{Ab}(G_2)$$
This page gives a categorical proof, but I don't know much category theory. Can someone give a purely group-theoretic proof of this (I know the universal property of abelianizations)? By the universal property, it would suffice to show that the RHS is an abelianization of $G_1*G_2$.
 A: Let $A$ be an abelian group. Then
$$\begin{eqnarray*}
\text{Hom}(G_1\ast G_2, A) & = & \text{Hom}(G_1,A)\times \text{Hom}(G_2,A) \\
& = & \text{Hom}(Ab(G_1),A)\times \text{Hom}(Ab(G_2),A) \\
& = & \text{Hom}(Ab(G_1)\oplus Ab(G_2),A)
\end{eqnarray*}$$ and there you have your universal property.

Explicitly, for any group $G$ I write $[g]_G$ for the class of $g\in G$ in $Ab(G)$. Then you get $\varphi: Ab(G_1\ast G_2)\to Ab(G_1)\oplus Ab(G_2)$ defined by $\varphi([g_1h_1g_2h_2\cdots g_rh_r]_{G_1\ast G_2}) = \left( [g_1\cdots g_r]_{G_1}, [h_1\cdots h_r]_{G_2}\right)$ with $g_i\in G_1$, $h_i\in G_2$.
A: I'll try to give another proof of the following fact (please let me know if it is correct)

If $G={\mathop{\LARGE \ast}}_{a\in J}G_a$ then $G/G'\cong \oplus_{a\in J}(G_a/G_a')$

by showing that $G/G'$ satisfies the universal property of direct products:

Let $H$ be an abelian group and $H_a, \ a\in J$  family of subgroups of $H$. If $H=\oplus_a H_a$ then for every abelian group $T$ and for every family of homomorphisms $\phi_a:H_a\to T$ there is a unique homomorphism $\phi:H\to T$ with $\phi\circ i_a=\phi_a$.

So it suffices to show that $G/G'$ satisfies the universal property of the direct sum $G_a/G_a'$.
We first want homomorphisms $\lambda_a:\frac{G_a}{G_a'}\to \frac{G}{G'}$.
$$\require{AMScd}
\require{cancel}
\def\diaguparrow#1{\smash{\raise.6em\rlap{\ \ \scriptstyle #1}
   \lower.6em{\cancelto{}{\Space{2em}{1.7em}{0px}}}}}
\begin{CD}
&& G_a/G_a' \\
& \diaguparrow{p_a} @VV\lambda_aV \\
G_a @>> p> G/G'
\end{CD}$$
Since $G_a'\subseteq \ker{p}$ from the universal property of group quotients there is $\lambda_a:\frac{G_a}{G_a'}\to \frac{G}{G'}$ such that the above diagram is commutative.
We want the triple $(G/G',G_a/G_a',\lambda_a)$ to satisfy the universal property of $\oplus_a(G_a/G_a')$.
Let $H$ be an abelian group and a family of homomorphisms $\phi_a:G_a/G_a'\to H$.

From the universal property of free products we have $\phi:{\mathop{\LARGE \ast}}_a G_a\to H$ such that $(1)$ is commutative.
The group $H$ is abelian so $G'\subseteq \ker\phi$ so from the universal property of the quotient group $G/G'$ there is $\bar{\phi}:G/G'\to H$ s.t. $(2)$ is commutative.
It is easy to see that $(\Delta)$ is commutative and $\bar{\phi}$ is unique.
