Let $G$ be a group, $N$ a normal subgroup of $G$, $C_G(X)$ the centralizer of $X$ in $G$ and $G'$ the commutator subgroup of $G$. Is it true that $C_G(N) = C_G(N \cap G')$?
If true, how do I prove it? If not, I'd like to know a counterexample or even better a characterization of the groups for which equality holds.
Best Answer
No. Take a prime $p$, a non-abelian group $G$ of order $p^3$, and $N$ a subgroup of $G$ of order $p^2$. We have that $G'$ has order $p$, and $G'\le N$.
Then $C_{G}(N) = N$, while $C_{G}(N \cap G') = C_{G}(G') = G$.