Quotient of a free group on a set

Question

Suppose I have a free group $F_S$ on a set $S,$ with a nonempty proper subset $T.$ If $N$ is the smallest normal subgroup of $F_S$ containing $T$, I would like to show that $F_S/N$ is also free. There's an obvious inclusion $S\setminus T \hookrightarrow F_S/N$ so I need to show that this satisfies the universal property. This seems like it would follow pretty straightforwardly from something like Yoneda's lemma, and while I'm primarily interested in seeing proofs that don't use category theory, I would also appreciate a short solution in those terms.

Answer 1

This can be proved only using group theory, along the lines you suggested. Hint: a function $S\setminus T\to G$ into some group $G$ can be extented to a function $S\to G$ by sending all elements of $T$ to $e_G$, the neutral element of $G$.

Then, using the universal property of $F_S$, this function can be extended to a group morphism $F_S\to G$. Since elements of $T$ are sent to $e_G$, so are the elements of the smallest normal subgroup $N$ containing $T$, and this morphism factorises into a morphism $F_S/N\to G$.