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Let $p$ be a prime number and $P$ a $p$-group.

(1) If $A$ is a maximal Abelian subgroup, what are nice examples where it isn't self-centralizing?

(2) What if $A$ happens to be normal as well?

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    $\begingroup$ A maximal abelian subgroup is always self-centralising: $A$ and an element from the centraliser generate an abelian subgroup. $\endgroup$ – Torsten Ekedahl Jul 26 '11 at 9:31
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    $\begingroup$ I don't understand your question. If $A$ is maximal abelian subgroup, it must be self-centralizing. Otherwise you can add to $A$ an element from $Cent_G(A)$ and get a bigger abelian subgroup. $\endgroup$ – Yiftach Barnea Jul 26 '11 at 9:34
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    $\begingroup$ Let me add to the comments of Torsten and Yiftach that the converse is also true, i.e. in every (either finite or infinite) group the self-centralizing subgroups are exactly the maximal abelian subgroups. $\endgroup$ – Ralph Jul 26 '11 at 11:56
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    $\begingroup$ This is not a research level question: I think it is better to close it. $\endgroup$ – Salvatore Siciliano Jul 26 '11 at 13:39
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    $\begingroup$ @Salvo: Thanks, for confirming. @Alex: Let $H \le G$ with $H=C_G(H)$ where $C_G(H) := \{g \in G \mid \forall h \in H: gh=hg\}$. Since $H\le C_G(H)$, $H$ is abelian. Assume $H$ is not max. abelian. Then there is an abelian $A \varsupsetneqq H$. It follows $A \le C_G(H) =H$ which is impossible. Thus $H$ is max. abelian. qed. $\endgroup$ – Ralph Jul 26 '11 at 16:34

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