Generators give quotients of free products of groups?

Question

I have some confusion here.

I have this idea that:

If $\Gamma$ is generated by (all of) the elements of subgroups $G_1,\dots,G_k$, then $\Gamma=\langle G_1,\dots, G_k\rangle$ is a quotient of the free product $G_1\ast \cdots \ast G_k$.

Is this actually true? I am wary about mixing quotients and subgroups in this fashion.

As an example of what I think is true:

Where $\mathbb{Z}_2\ast\mathbb{Z}_2=\langle a,b\mid a^2=b^2=e\rangle$ we have that $S_3$ is a quotient $\mathbb{Z}_2\ast \mathbb{Z}_2\to S_3$, via, say, $a\mapsto (12)$ and $b\mapsto (13)$.

In other words, is it true to say that if $\Gamma$ is generated by two elements of order two, then $\Gamma$ is a quotient $\mathbb{Z}_2\ast\mathbb{Z}_2\to \Gamma$?

Consider $\Lambda:=(\mathbb{Z}_2\oplus \mathbb{Z}_2)\ast \mathbb{Z}_3\ast \mathbb{Z}_2$. Because $S_4$ can be generated by the elements of the Klein four group in $S_4$, a three cycle, and a two cycle, does it follow that $S_4$ is a quotient $\Lambda\to S_4$?

Answer 1

The answer is "yes".

Let $\iota_j\colon G_j\to G_1*\cdots*G_k$ be the canonical embedding into the free product. The free product has the following universal property (in fact, the property defines the free product up to unique isomorphism, and shows that it is the coproduct in the category of groups and group homomorphisms):

Theorem. Let $H$ be any group, and let $f_j\colon G_j\to H$ be group homomorphisms. Then there exists a unique group homomoprhism $F\colon G_1*\cdots*G_k\to H$ such that for all $j$, $f_j = F\circ \iota_j$.

So, suppose that $\Gamma$ is generated by its subgroups $G_1,\ldots,G_k$. Let $\rho_i\colon G_i\to \Gamma$ be the inclusion maps. Then there exists a (unique) homomorphism $F\colon G_1*\cdots*G_k\to \Gamma$ such that for each $j$, if $x\in G_j$, then $F(\iota_j(x)) = x$. In particular, $G_j$ is contained in the image of $F$. Moreover, since $F$ is a homomorphism, the image of $G_1*\cdots*G_k$ is a subgroup of $\Gamma$. Therefore, $$\Gamma = \langle G_1,\ldots,G_k\rangle\leq F(G_1*\cdots*G_k) \leq \Gamma$$ proving that $F$ is surjective. Thus, $\Gamma$ is (isomorphic to) a quotient of $G_1*\cdots*G_k$, as desired.