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!


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 $S'$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.