The centralizer of an element x in free group is cyclic

If $F$ is free group and $1 \neq x \in F$, then $C_F(x)$ is cyclic.

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The subgroup $\langle x,C_F(x)\rangle$ is a subgroup of a free group, hence is free. The only free groups with nontrivial center are the free groups of ranks $0$ or $1$, and $x\neq 1$ is clearly central in $\langle x, C_F(x)\rangle$. Thus, $\langle x, C_F(x)\rangle$ is cyclic, hence its subgroup $C_f(x)$ is cyclic.

Here is a sketch of the proof:

If $F$ is a free group, then there exists a $S \subset F$ such that every element of $F$ can be written uniquely as a product of elements of $S$. Suppose $x \neq 1$. Then $x = (s_1…s_n)^k$ for some $n \geq 1$, $s_i \in S$, and $k$ chosen to be largest. Show that $C_F(x)$ is the cyclic group generated by $s_1…s_n$. Use the fact that everything in $F$ can be written uniquely as product of elements of $S$.