Let $H$ be the subgroup $\{(),(12)(34)\}$ of $S_4$. Find the normalizer $N(H)$ and the quotient $N(H)$.
What I have so far: in this case, $N(H)=Z((12)(34))$. The order of the conjugacy class of $(12)(34)$ is $3$, so the order of its centralizer (which is $N(H)$) is $8$. The quotient has order $4$ so it's either $C_2\times C_2$ or $C_4$.
To find what elements lie in $N(H)$, do I need to check whether each of the $24$ elements commute with $(12)(34)$ (until I find $8$ elements), or is there an easy way to see what elements lie in $N(H)$? How to understand whether the quotient is $C_4$ or $C_2\times C_2$?
Probably the easiest way is to explicitly find $N(H)$. To do that you can ue the fact that $g(12)(34)g^{-1} = (g(1)g(2))(g(3)g(4))$, where $g(1)$ is the action of the permutation $g$ to $1$. It's not hard to notice that if $g \in N(H)$ then it's uniquely determined by its action on $1$ and $3$.
For example if $g(1) = 1$, then we must have $g(2) = 2$. If we choose $g(3) = 4$, then we must have $g(3) = 4$. This choice would yield $g=(34)$. In particular as we have $8$ choiches (choosing g's action on $1$ immediately determines the action on $2$, leaving 2 choices for the action on $3$) for the action to $1$ and $3$ any choice would work. Such a compuation would yield:
$$N(H) = \{e,(34),(12),(12)(34),(13)(24),(1324),(14)(23), (1423)\}$$
For the second part you can notice that every element of the quotent group is of order $2$ or $1$, which should help you conclude that $N(H)/H \cong C_2 \times C_2$. Indeed this is obvious for the elements consisting of two cycles. Also:
$$((1324)H)^2 = (1324)^2H = (12)(34)H = H$$
$$((1423)H)^2 = (1423)^2H = (12)(34)H = H$$