Let $\mathbb{F}_n$ be a free group on $n$ generators. Let $H$ be a subgroup of $\mathbb{F}_n$. When is $$C_{\mathbb{F}_n}(H)=\{g \in \mathbb{F}_n: gh=hg, \forall h \in H\}$$ the centraliser of $H$ in $\mathbb{F}_n$ trivial?
For each $h \in H$, $C_{\mathbb{F}_n}(h)$ is cyclic and contains $C_{\mathbb{F}_n}(H)$. Moreover $$C_{\mathbb{F}_n}(H)=\cap_{h \in H}C_{\mathbb{F}_n}(h)$$
I believe that intersection is $\{e\}$, unless $H$ is itself a cyclic group, in which case all of them are equal.
Please correct me on this, if I am wrong.
Thanks for the help!!
Yes, your conjecture is correct.
Observe that any subgroup of a cyclic group is itself cyclic. Also, note that each centralizer $C_{\mathbb F_n}(h)$ for $h\neq e$ is cyclic.
As your expression for $C_{\mathbb F_n}(H)$ is an intersection of centralizers of elements, we can combine these facts to say that $C_{\mathbb F_n}(H)$ is itself cyclic, so long as $H$ is nontrivial.
Now, let $K$ be the centralizer of $H$. Either $K$ is trivial, in which case we are done, or we can look at the centralizer $H'$ of $K$. Note that $H\subseteq H'$, however we have the tools to show that $H'$ is cyclic, thus so is $H$.