The problem says that Let F be the free group on $x_1,x_2, \ldots ,x_n$. Show that the normalizer/centralizer of any element $\neq 1$ is a cyclic group.
The problem seems much harder to me as none of my attempts are anywhere working. Can somebody give me some hints.
On
Another hint: if you don't know that subgroups of free groups are free but do know about the normal form for elements of a free group (as a product of powers of the generators), note that $x$ centralizes $x$, so if the centralizer of $x$ is cyclic, generated by $y$ say, you must have $x = y^n$ for some $n$. Now turn this round and show that if you write $x$ in its normal form as a product of powers of the generators $x_i$, the shortest substring $y$ of the normal form such that $x = y^n$ for some $n$ generates the centraliser of $x$.
Hint: Let $F$ be a free group. If $a,b \in F$ commute, then $\langle a,b \rangle$ is an abelian subgroup; but it is also free, as a subgroup of a free group. Therefore, $\langle a,b \rangle$ is cyclic.