In this example, why are the only possibilities for $N_{S_3}(H)$ either $H$ or $S_3$? I am reading Dummit and Foote's Abstract Algebra: 3rd edition.
The following is the beginning sentence of the example on page 91.

Let $H=\langle~ (1~2)~\rangle\leq S_3$. Since $H$ is of prime index $3$ in $S_3$, by Lagrange's Theorem, the only possibilities for $N_{S_3}(H)$ are $H$ or $S_3$.

Could someone explain how the fact $|S_3:H|=3$ and the Lagrange's Theorem imply that the only possibilities for $N_{S_3}(H)$ are $H$ or $S_3$?

For any group $G$ and subgroup $H$,
$$H \subseteq N_{G}(H) \subseteq G.$$
That is, the normalizer of $H$ in $G$ includes at least the elements $H$. This is because for any $h \in H$, $hH = Hh = H$.
Also, $N_G(H)$ is a subgroup (closed under multiplication and inverses).
By Lagrange's theorem, $N_G(H)$ must divide the order of the group $G$, but the order of $H$ divides the order of $N_G(H)$ as well. So $|N_G(H)|$ must be some number divisible by $|H|$ and dividing $|S_3|$.
So we have that $N_{S_3}(H)$ has order dividing $|S_3| = 6$, and divisible by $|H| = 2$. The only numbers divisible by $2$ and dividing $6$ are $2$ and $6$ itself, so $N_{S_3}(H) = H$ or $N_{S_3}(H) = S_3$.
Edit to clarify: The reasoning "there are no numbers divisible by $2$ and dividing $6$" is the same thing as saying that $|G : H| = 3$ is prime. If $n$ is the size of $G$ and $H$ has size $\frac{n}{p}$ where $p$ is prime, then there will be no numbers dividing $n$ and divisible by $\frac{n}{p}$ except for $n$ and $\frac{n}{p}$ themselves (the divisors of $n$ which are divisible by $\frac{n}{p}$ are exactly corresponding with the divisors of $p$, and $p$ is prime). Another example may help here: there only divisors of $88$ that also are divisible by $8$ are $8$ and $88$, because $\frac{88}{8} = 11$ is prime.
So in general, if $|G : H|$ is prime, either $N_G(H) = H$ or $N_G(H) = G$, by the same reasoning as above.
