Find rank of subgroup of $F_2 = \langle x,y\rangle$ generated by $xyx^{-1}y, yxy^{-1}x$.

Question

This is from an exercise in a group theory course I'm taking, I believe the subgroup of $F_2 = \langle x,y\rangle$ generated by $a = xyx^{-1}y$ and $ b = yxy^{-1}x$ is isomorphic to $F_2$. What I've tried is:

Let $\phi : F_2 \rightarrow \langle a,b\rangle$ be a group homomorphism mapping $x$ to $a$ and $y$ to $b$. This is clearly surjective and let's assume an element $1\neq g \in \ker\phi.\;$ If $g= s_1 \cdots s_n$ is a reduced word in $F_2$ with $s_i \in \{x,y,x^{-1},y^{-1}\}$ then

$$\phi(g) = \phi(s_1)\cdots \phi(s_n) = 1$$ and since in any way we multiply the elements $\{xyx^{-1}y, yxy^{-1}x, \left(xyx^{-1}y\right)^{-1}, \left(yxy^{-1}x\right)^{-1}\}$ together nothing cancels out (unless we pair inverses together of course), this implies $\phi(s_1)\cdots \phi(s_n)$ is a reduced word in $\langle a,b\rangle$, which is a contradiction and thus $\phi$ is an isomorphism.

I kinda feel I've went somewhere wrong, also the wording on the exercise was "find the free rank of the subgroup and a freely generating set" so one would expect the generating set would be something different than $\{a,b\}$ and the rank higher than 2 but maybe it's just bad wording. So have I gone anywhere wrong? Also I'm not yet familiar enough with algebraic topology to understand such solutions so try to keep it group theoretic.

Answer 1

You are wrong. This subgroup is a proper free subgroup of rank $2$. Indeed, modulo $[F_2,F_2]$ it is generated by $x^2, y^2$, so modulo the derived subgroup of $F_2$ it is a proper free Abelian subgroup of rank $2$. Thus your subgroup is proper, free as every subgroup of a free group, and its rank is at least $2$. Hence its rank is exactly $2$ since it is $2$-generated.

Answer 2

Now your proof is fine in outline. But your phrase "nothing cancels out" is rather vague and needs justification.

One way to fill in these kinds of vague justifications regarding cancellation is by using Stallings fold method.

Start with two rose graphs: one is $R\langle x,y\rangle$ with two oriented edges labelled $x,y$; the other is $R\langle a,b\rangle$ with two oriented edges labelled $a,b$. Let the $a$ edge be subdivided into four labelled, oriented subedges forming the word $xyx^{-1}y^{-1}$. Let the $b$ edge be similarly subdivided forming the word $yxy^{-1}x$.

Consider the induced map $f : R\langle a,b \rangle \to R \langle x,y \rangle$ taking each oriented subedge of the domain that is labelled $x$ to the (unique) oriented edge of the range labelled $x$, and similarly for $y$. The group in question that you need to analyze is $f_*(\pi_1(R\langle a,b \rangle))$.

Now when you said "nothing cancels out", that's wasn't quite true, because the terminal $y^{-1}$ subedge on $a$ cancels out with the initial $y$ subedge on $b$: those two subedges can be concatenated to form a $y^{-1}y$ subpath in $R\langle a,b \rangle$ whose image under $f$ cancels.

Using Stallings method, the thing to do is to identify those two subedges by a "fold", producing a quotient map $q : R\langle a,b \rangle \to R'$ where the graph $R'$ has two vertices $p,q$, one orineted edge from $p$ to $q$ labelled $x$, one oriented edge from $q$ to $p$ labelled $x$, and one oriented edge from $p$ to $q$ with three oriented, labelled subedges forming the word $y x y^{-1}$. The map $f$ factors as $$R \langle a,b \rangle \xrightarrow{h} R' \xrightarrow{g} R\langle x,y \rangle $$ The map $g$ is locally one-to-one.

Now one applies a theorem: A locally one-to-one map from one graph to another induces an injection on the fundamental group.

Also, the map $h$ is a homotopy equivalence and so $h_*$ is an isomorphism. It follows that $g_*$ is injective, and it clearly has the same image as $f_* = g_* \circ h_*$.

Hence your group is the same as the $g_*(\pi_1(R'))$. Since $\pi_1(R')$ is clearly generated by the classes of the two loops labelled by the original two words, you are done. But you do also obtain a slightly simpler free generating set, namely $xyx^{-1}y$ and $x^2$.