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Let $F_n$ be free of rank $n$, and $g\neq e$ an element of $F_n$. The $\langle g\rangle$ is a non trivial cyclic subgroup of $F_n$. We can define $N=\{u\in F_n:u\langle g\rangle = \langle g \rangle u\}$ to be the normalizer of $\langle g \rangle $. So I want to show that $N$ is a cyclic subgroup of $F_n$.

If $u\in N$, then $u\langle g \rangle =\langle g \rangle u$, so in particular there is $g^m$ such that $ug=g^mu$, which implies $ug^{-1}=g^{-m}u$, so that $ug^k=g^{mk}u$ for any $k\in \mathbb{Z}$. So in general one has $$u^l g^k=g^{m^lk}u^l$$

The reason I'm trying to understand this relation is because the hint to the problem suggests that considering $\langle u,g\rangle$, and in particular remarks that it is free if $u\in N$.

The problem is,

  1. I don't see how $\langle u,g\rangle$ is free. I think I have to show that any word from $\{u,g,u^{-1},g^{-1}\}$ does not reduce to identity, but it's not clear to me unless $u=g^k$ for some $k$

  2. I don't see how $\langle u,g\rangle$ being free helps me, especially when I need to show something about $N$.

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    Your argument in Problem 1 shows that $\langle u,g \rangle$ is not free of rank $2$, so it must be free of rank $1$ and hence cyclic - say $\langle u,g \rangle = \langle t \rangle$. So $u=t^k$ for some $k \in {\mathbb Z}$. Choose such a $t$ with $|k|$ maximal, and you must have $N = \langle t \rangle$. – 2017-02-16

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