How to compute the normalizer $N(H)$?

Question

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 $N(H)$?

Answer 1

Hint: For any $\sigma\in S_n$, we have $$\sigma(12)\sigma^{-1}=(\sigma(1)\sigma(2)),$$ where $\sigma(i)$ is $\sigma$ applied to $i$, with $i$ from the set $S_n$ acts on.

Answer 2

Shaun's answer works for general $S_n$. For an answer that depends more heavily on working in $S_3$, note that $N(H) \leq S_3$ and has an element of order $2$, so $\vert N(H) \vert = 2 \text{ or } 6$. But $S_3$ has three Sylow-$2$ subgroups, so $H$ is not normal in $S_3$; thus $N(H)=H$.

Answer 3

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 $N(H)=H$.