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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.

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    $\begingroup$ A counter example is $G = S_3$ and the normal subgroup $G$. In this case, $C_G(G)$ is the center which is trivial. But $C_G(G') = G'$ ($G'$ is abelian, so the centralizer certainly contains it, and it is easy to check that it contains nothing else as $G'$ cannot be central). $\endgroup$ Commented Jul 11, 2013 at 12:31
  • $\begingroup$ thanks, I should have checked $S_3$. Funnily it holds for $S_4$ $\endgroup$
    – vuur
    Commented Jul 11, 2013 at 12:46
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    $\begingroup$ I think this rarely holds. S4 and SL(2,3) work, but just because they have so few normal subgroups, they form a chain, and $G'$ is centerless. For groups of order less than 72, it is equivalent to having $G$ abelian or $G'$ nonabelian, but again mostly because there are few normal subgroups to worry about. $\endgroup$ Commented Jul 11, 2013 at 15:16
  • $\begingroup$ @JackSchmidt, there are examples in all non-abelian $p$-groups $G$, I believe. Since $[G', Z_2(G)] = 1$, it is enough to take $N$ as a normal subgroup which is not centralized by $Z_2(G)$. For instance, take the normal closure of $\langle a \rangle$, where $a \notin C_G(Z_2(G)) < G$. $\endgroup$ Commented Jul 11, 2013 at 22:23

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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 $C_{G}(N \cap G') = C_{G}(G') = G$.

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  • $\begingroup$ More generally, as I wrote in comment to OP above, there are examples in all non-abelian $p$-groups $G$. Since $[G', Z_2(G)] = 1$, it is enough to take $N$ as a normal subgroup which is not centralized by $Z_2(G)$. For instance, take the normal closure of $\langle a \rangle$, where $a \notin C_G(Z_2(G)) < G$. $\endgroup$ Commented Jul 11, 2013 at 22:27

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