To show that normalizer of an element 'a' in a group G is a normal subgroup of G.

Question

I couldn't solve the normalcy criteria. Any help will be highly appreciated.

Answer 1

This is not true. For counterexample, take $G=S_3$, the symmetric group on $\{1, 2, 3\}$. Put $\tau = (1 2)$ and $\sigma = (1 2 3)$. Then $N_G(\tau)=\{ \alpha \in S_3 : \alpha \tau = \tau \alpha \}=\{id, \tau\}$. But $ \sigma^{-1} \tau \sigma = \sigma \tau = (1 3) \notin N_G(\tau) $.

If you interested in finite groups whose normalizer of an arbitrary element is normal, see this post.