# Show that a group $G$ is finitely generated if and only if it is a quotient of a free group on a finite set of letters

As stated in the title, I am working on this exercise:

Show that a group $$G$$ is finitely generated if and only if it is a quotient of a free group on a finite set of letters.

I have showed $$(\Rightarrow)$$ as follows:

$$(\Rightarrow).$$ Suppose $$G$$ is finitely generated, and let $$S=\{a_{\alpha}\}$$ be the set of all the generators. Then consider the free group on the same set of generators, denoted by $$Fr(S)$$. Then there exists a homomorphism $$h:Fr(S)\longrightarrow G$$ such that $$h(a_{\alpha})=a_{\alpha}$$ for all $$\alpha$$, and thus $$h$$ is surjective.

However, I don't know how to show the converse. Suppose $$F$$ is a free group on a finite set of letters $$B$$, and suppose $$G\cong F/N$$ for some $$N$$, then I know that this gives us an isomorphism $$\phi:F/N\longrightarrow G,$$ but what should I do to get the information of the generators of $$G$$?

Thank you!

The generators of $$G$$ are simply the images of the generators of the free group. You can see this because the homomorphism $$F\to G$$ is surjective; if you can get any element in the domain, the images can get any element in the codomain. In general, a quotient of a finitely generated group is finitely generated.
• @Jacob The quotient map $F\to F/N$ is by definition surjective. That's what it means to be a quotient. By assumption, $F/N=G$, it is not simply an isomorphism. Commented Nov 14, 2019 at 17:40