# Show that any automorphism of the free group maps the normal commutator group to the normal commutator group

Consider the free group $F = \langle x, y \rangle$. Let $N = \langle\langle xyx^{-1}y^{-1}\rangle\rangle$ be the smallest normal subgroup of $F$ generated by $xyx^{-1}y^{-1}$. Show that any automorphism of $F$ sends $N$ to $N$.

My attempt: I know that any automorphism from $F$ to $F$ would have to be defined by either $\phi(x) =x, \phi(y) = y$ or $\phi(x) = \phi(y), \phi(y) = x$, so from this it is intuitively clear that $\phi(N) = N$, but I don't know how to formally write it.

Also, I know $F/N$ is abelian, and that isomorphism preserve the abelian property, so I think I might somehow be able to use this but I can't figure out how. Any help would be appreciated!

• The commutator subgroup $[G,G]$ of any group $G$ is characteristic, so any automorphism of $G$ maps $[G,G]$ to itself. – Derek Holt Sep 2 '17 at 10:41
• "I know that any automorphism from $F$ to $F$ would have to be defined by either $\phi(x) =x, \phi(y) = y$ or $\phi(x) = \phi(y), \phi(y) = x$"... What about the automorphism defined by $\phi(x)=xy$, $\phi(y)=xyx$? – Lee Mosher Sep 3 '17 at 0:00

This has nothing to do with $F$ being free. Since $F$ is generated by the two element $x, y \in F$ it follows that the commutator subgroup $[F,F]$ is the smallest normal subgroup which contains the commutator $[x,y]$:
Since $[F,F]$ is a normal subgroup which contains $[x,y]$ we find that $\langle\!\langle [x,y] \rangle\!\rangle \subseteq [F,F]$. On the other hand we find that $F/\langle\!\langle [x,y] \rangle\!\rangle$ is generated by the two commuting elements $\overline{x}$ and $\overline{y}$ and is therefore abelian, which shows that $[F,F] \subseteq \langle\!\langle [x,y] \rangle\!\rangle$.
Thus we have found that $N = \langle\!\langle [x,y] \rangle\!\rangle = [F,F]$. The commutator subgroup $[F,F]$ is invariant under every homomorphism $\phi \colon F \to F$ since $$\phi([F,F]) = [\operatorname{im} \phi, \operatorname{im} \phi] \subseteq [F,F].$$