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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?

Thanks a lot in advance!

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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 $S'$.

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