Calculating the normalizer of a sylow p-subgroup It seems that its explained pretty well online how to find a p-sylow subgroup, but Im having a hard time finding an explanation of how to find a p-sylow subgroups normalizer. 
Take a specific 2-sylow subgroup of $S_4$: It will have order $2^3=8$ and could look something like $D_4$ in $S_4$ (symmetries of square): {e,(1234),(1432),(12)(34),(14)(23)...}. What is the best method for finding the normalizer of this group? 
 A: A $2$-Sylow subgroup in $S_4$ is a maximal subgroup, meaning there is no other proper subgroup of $S_4$ containing it (this can be seen by the fact that it has prime index, namely $3$). Thus it is either its own normalizer or normal. Which is it?
A: Addressing one of the additional cases raised in the comments: let us find the normalizer of a $5$-Sylow subgroup of $S_5$.
First, let $n_5$ denote the number of $5$-Sylow subgroups of $S_5$. From Sylow's theorem, we have $n_5 \equiv 1$ (mod $5$) and $n_5$ must divide $4! = 24$. The possibilities are $n_5 = 1,6$. In this case, a $5$-Sylow subgroup has order $5$, so it is generated by any $5$-cycle. Since there are more than four $5$-cycles, there must be more than one $5$-Sylow subgroup, so $n_5 = 6$.
Now we can use the fact that $|G:N_G(S)| = n_5 = 6$, where $S$ is any $5$-Sylow subgroup. So the normalizer of $S$ has index $6$, hence order $20$. Certainly $S \leq N_G(S)$. What else is in $N_G(S)$? Without loss of generality (relabeling if necessary), we may assume that $S = \langle(12345)\rangle$.
Note that there are four generators of $S$, namely:
$$(12345)$$
$$(12345)^2 = (13524)$$
$$(12345)^3 = (14253)$$
$$(12345)^4 = (15432)$$
Any element $g \in S_5$ for which $g(12345)g^{-1}$ is one of the generators of $S$ will normalize $S$. We can easily see that, for instance,
$$g(12345)g^{-1} = (13524)\quad\text{ if }\quad g = (2354)$$
This shows that $N_G(S)$ contains the cyclic subgroup $H = \langle(2354)\rangle$. Therefore, $N_G(S)$ contains $SH$. Moreover, $|SH| = |S||H|/|S \cap H| = (5)(4)/(1) = 20$, so $N_G(S) = SH$.
