# For a non-abelian group, there exists an non-trivial element whose normalizer is abelian subgroup.

"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.

• Maybe this holds for finite simple groups though. – Moishe Kohan Aug 4 '17 at 20:57
• @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. – Tobias Kildetoft Aug 7 '17 at 15:15
• @TobiasKildetoft This guess is motivated by simple complex Lie groups where it is true (one takes any regular semisimple element). – Moishe Kohan Aug 8 '17 at 12:25

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].
• 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. – Moishe Kohan Aug 4 '17 at 7:16
• How does $C_2$ act on the product? Is it by an inner automorphism of $D_8$ and trivially on the $C_2$ factor? – Moishe Kohan Aug 4 '17 at 7:59