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.

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.