How to compute the normalizer $N(H)$? 
Let  $H$ be  a   subgroup   generated  by $(12)$  in $S_3$. Compute the  normalizer  $N(H)$ of $H$.

 My attempt:
$\sigma$ $\in N(H) $ then $ \sigma H \sigma^{-1}=H=(12)$
implies $\sigma (12)= (12)\sigma$ 
Now applying  both sides  to $3$
$\sigma (3) (12) =(12) \sigma(3)$
$(12)$ cancel both  each other
$\sigma (3)  = \sigma(3)$
I thinks  I'm  using  wrong logic.
Any  hints/suggestion  on how to compute the normalizer $N(H)$?

If you know this theorem:
Theorem: Let $H$ be a Sylow -$p$ subgroup of $G$. Then the number of Sylow $p$-subgroup of $G$ is equal to $\frac{o(G)}{o(N(H))}$.
Since $S_3$ has $3$ Sylow $2$-subgroup. Thus, by above theorem $$3=\frac{o(G)=6}{o(N(H))}$$
$\Rightarrow o(N(H))=2$
Thus $N(H)=H$.
