Can I conclude that my group is finitely generated, if it is a homomorphic image of a free-group on finitely many generators?

Question

Say $X$ is a finite set, $F \langle X \rangle$ is the free group on the set $X$ and $G$ be a group.

If I have a surjective homomorphism $$\varphi : F \langle X \rangle\longrightarrow G$$ then can I conclude that $G$ is finitely generated?

Answer 1

Yes, you can conclude that if $x_1,\dots ,x_n$ are elements of $X$, $\phi(x_1),\dots,\phi(x_n)$ are generators of $G$. Let $g\in G$, $g=\phi(u)$ where $u$ is an element of $F(X)$, you can write $u=x_{i_1}^{n_1}\dots x_{i_p}^{n_p}$ where $i_j\in\{1,\dots ,n\}$. This implies that $$g=\phi(u)=\phi(x_{i_1})^{n_1}\dots\phi(x_{i_p})^{n_p}.$$

Answer 2

Yes, since $F\langle X\rangle/ \operatorname{ker}\varphi \cong G$, by the first isomorphism theorem, and every quotient of a finitely generated group is finitely generated.