# Is $\langle S \rangle / \operatorname{nc}(T) \simeq \langle S \setminus T \rangle$, where $T \subseteq S$?

Let $$S$$ be a set and $$\langle S \rangle$$ denote the free group generated by the set $$S$$.

If we take a subset $$T$$ of $$S$$ and consider the quotient group $$\langle S \rangle / \operatorname{nc}(T)$$, where $$\operatorname{nc}(T)$$ denotes the normal closure of $$\langle T \rangle$$ in $$\langle S \rangle$$, is it true, in general, that it is isomorphic to $$\langle S \setminus T \rangle$$?

My intuition says it should be, but I’m having a hard time actually closing in on a proof.

My first approach was to take $$\pi: \langle S \rangle \to \langle S \rangle / \operatorname{nc}(T)$$ to be the canonical projection, and take a reduced product $$p = s_1^{\epsilon_1} ... s_n^{\epsilon_n}$$. Then:

$$\pi(p) = p\operatorname{nc}(T) = \pi(s_1^{\epsilon_1} ... s_n^{\epsilon_n}) = \pi(s_1^{\epsilon_1}) ... \pi(s_n^{\epsilon_n})$$

Now this becomes a product of equivalence classes, where every term $$\pi(s_i^{\epsilon_i})$$ with $$s_i \in T$$ will cancel, thus yielding a product in $$S \setminus T$$. Since $$\pi$$ is surjective, we know $$S \setminus T$$ generates this quotient group (modulo $$\operatorname{nc}(T)$$).

I now have a hard time proving this a free group.

I also tried a second approach. Let’s define $$f : S \to \langle S \setminus T \rangle$$ by $$s \mapsto s$$ if $$s \notin T$$ and $$s \mapsto 1$$ if $$s \in T$$. This extends into a homomorphism $$F : \langle S \rangle \to \langle S \setminus T \rangle$$ which is surjective because of the definition of $$f$$.

Thus, we get $$\langle S \rangle / \operatorname{ker}(F) \simeq \langle S \setminus T \rangle$$, but I can’t seem to figure out if this kernel is indeed $$\operatorname{nc}(T)$$. This seems easier, but I am stuck nonetheless.

Am I correct in my belief? If so, how do I actually prove this result?

I think the easiest thing is to prove the quotient has the universal property of the free group on $$S\setminus T$$.
Let $$F=\langle S\rangle$$ be the free group on $$S$$, let $$N=\langle T\rangle^F$$ be the normal closure of $$T$$, and let $$S'=S\setminus T$$. We prove that $$F/N$$ has the universal property of the free group on $$S'$$, via the map that sends $$S'$$ to $$F$$ and then to $$F/N$$ through the canonical embedding and projection, respectively.
To that end, let $$G$$ be any group, and let $$f\colon S'\to G$$ be any set map. Let $$\mathfrak{f}\colon S\to G$$ be the map given by $$\mathfrak{f}(s) = \left\{\begin{array}{ll} f(s) &\text{if }s\in S';\\ e_G &\text{otherwise.} \end{array}\right.$$ Then $$\mathfrak{f}$$ induces a homomorphism $$\mathcal{F}\colon F\to G$$ with $$\mathcal{F}(s)=\mathfrak{f}(s)$$ for all $$s\in S$$. Since $$T\subseteq \ker(\mathcal{F})$$, then $$\mathcal{F}$$ factors through $$F/N$$, so we obtain a morphism $$\phi\colon F/N\to G$$ such that if $$s\in S'$$, then $$\phi(sN) = \mathcal{F}(s)=\mathfrak{f}(s) = f(s)$$.
Finally, suppose that $$g\colon F/N\to G$$ is such that $$g(sN) = f(s)$$ for all $$s\in S'$$. Then $$g\circ\pi\colon F\to G$$, where $$\pi\colon F\to F/N$$ is the canonical projection, has $$g\circ\pi(s)=\mathcal{F}(s)$$ for all $$s\in S$$, hence $$g\circ \pi =\mathcal{F}$$. Since $$\phi\circ\pi = \mathcal{F}=g\circ \pi$$ and $$\pi$$ is surjective, it follows that $$g=\phi$$, so the function $$\phi$$ is unique.
Thus, for every group $$G$$ and every set function $$f\colon S'\to G$$, there exists a unique group morphism $$\mathcal{F}\colon F/N\to G$$ such that $$\mathcal{F}(sN) = f(s)$$ for all $$s\in S'$$. This proves that $$F/N$$ is the free group on $$\{sN\mid s\in S'\}$$, and hence is isomorphic to the free group on $$S'$$.