Homomorphic image of a free group

Question

Given a group $G$ with a generating set $S$, such that $|S|=n$, I need to prove that $G$ is a homomorphic image of $F_n$.

Right now I'm just looking for any tips for how to even start this proof or how to start thinking about the question.

Answer 1

Hint: $F_n$ is a free object in the category of groups i.e $Hom_{Group}(F_n,G)=Hom_{Set}(\{1,...,n\},G)$.

This is equivalent to saying that the forgetful functor from the category of sets to the category of groups has a left adjoint which associates to a set $S$ the free group $F_S$.

Answer 2

Define the map

$$\phi: \begin{cases}\langle a_1,\ldots, a_n\rangle = F_n\to G = \langle g_1,\ldots, g_n\rangle \\ a_i\mapsto g_i & 1\le i\le n\end{cases}$$

Then this is the map. Since the $g_i$ generate $G$ it's surjective. That's the basic idea.

Answer 3

I think the following definition (the most easy to grab and/or standard I know) of a free group will solve the problem:

The group $\;F_n\;$ is free on $\;S:=\{s_1,...,s_n\}\;$ iff for any function $\;f:S\to G\;,\;\;G\;$ any group,

there exists a unique homomorphism $\;\phi:F_n\to G\;$ extending $\;f\;$ , meaning: such that $\;\left.\phi\right|_S=f\;$

Answer 4

Let $a_1,\dots a_n$ be generators for $F_n$, and $g_1,\dots g_n$ generators for $G$. Note that a homomorphism from $F_n$ is given by the images of $a_i$. So let's take the homomorphism given by $a_i\mapsto g_i$. Any element in $G$ can be written as product of $g_i$-s. Then the product of corresponding $a_i$-s will be in the preimage of our element.

ADDITIONAL: you can study the kernel of the map. What do elements in the kernel represent?