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"If $G$ is a finite non-abelian group, then there exists an element $a\in G$ whose normalizer is abelian. Here $N(a)= \{g \in G : ga=ag \}$ is normalizer of a."

I have verified the above fact for the symmetric group $S_{3}$. I was trying it for quite long, but couldn't get anything. Any help or hint would be helpful. Thanks in advance.

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  • $\begingroup$ Maybe this holds for finite simple groups though. $\endgroup$ – Moishe Kohan Aug 4 '17 at 20:57
  • $\begingroup$ @MoisheCohen That does seem like it might be the case. I have had GAP running all day trying to find an example of a simple group where it fails, but so far it has not found any. $\endgroup$ – Tobias Kildetoft Aug 7 '17 at 15:15
  • $\begingroup$ @TobiasKildetoft This guess is motivated by simple complex Lie groups where it is true (one takes any regular semisimple element). $\endgroup$ – Moishe Kohan Aug 8 '17 at 12:25
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The claim is not true. There is a counterexample of order $32$ which is a semidirect product of $D_8\times C_2$ with $C_2$. I found this example by asking GAP to go through the groups and check. The group in question is the one with ID [32,49].

Note that what you have written up is the definition of the centralizer of that element, not the normalizer. Normalizers are usually defined for subgroups, but one can also define them for elements as the normalizer of the subgroup generated by them.

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  • $\begingroup$ Why does minimality imply that every proper subgroup of $G$ is abelian? On the face of it, minimality only implies that for every proper subgroup $H< G$ the centralizer $Z_H(a)$ of every $a\in H$ is abelian. $\endgroup$ – Moishe Kohan Aug 4 '17 at 7:16
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    $\begingroup$ @MoisheCohen You are right, I need to think about this some more. If I can't fix it shortly, I will delete the answer for now. $\endgroup$ – Tobias Kildetoft Aug 4 '17 at 7:18
  • $\begingroup$ @MoisheCohen Ok, I have now fixed it and undeleted. It turned out that the claim was not true after all. $\endgroup$ – Tobias Kildetoft Aug 4 '17 at 7:50
  • $\begingroup$ How does $C_2$ act on the product? Is it by an inner automorphism of $D_8$ and trivially on the $C_2$ factor? $\endgroup$ – Moishe Kohan Aug 4 '17 at 7:59
  • $\begingroup$ @MoisheCohen I did not actually check (I don't recall how to figure this out in a nice way). It is the group with ID [32,49] (I will add this to the answer as well). $\endgroup$ – Tobias Kildetoft Aug 4 '17 at 8:02

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