G is a group and the centralizer of a subgroup $H$ is
$Z_G(H) = \{g ∈ G;\ ∀h ∈ H\ gh = hg\}$
$S$ is any subset of $Z_G(H)$
I am trying to prove (or disprove) that $H = Z_G(S)$
It seems obvious that $∀h ∈ H,\ h ∈ Z_G(S)$
However, I cannot prove the opposite direction. Is it possible that there is an element of $G$ outside of $H$ that commutes with every element in $S$? Thank you for reading!
The other direction is obviously false.
For example, let H be $\{e\}$, which is the trivial subgroup of $G$. $Z_G(H)$ would be the whole group $G$.
Let $S$ be $Z_G(H)= G$ itself. If your assertion was true, $G$ would have a trivial center. Since there is no restriction on $G$, let $G$ be any abelian group would contradict it.