Let $H$ be the cyclic subgroup of $S_4$ generated by the cycle $(1234)$.
What is the order of the normalizer $N$ of $H$ in $S_4$? Give generators for $N$.
How do I go about solving a problem like this? We have $H = \{(1), (1234), (13)(24), (4321)\}$. I've seen a formula for $[S_4 : N]$ as $\frac{s}{r}$, where $s$ is the number of elements in $S_4$ in the same conjugacy class of $(1234)$, and $r$ is the number of elements in $H$ with the same cycle type as $(1234)$. So $s = 6$, and $r = 2$, so $\frac{s}{r} = 3$, and $|N| = \frac{24}{3} = 8$. What is a good way to find the generators of $N$?
My guess would be $N$ would be generated by $(1234)$ and any transposition, say $(12)$ (assuming $N \cong D_8$).
Best Answer
Hints:
By Lagrange, and since $\;H\le N\;$, it must be that $\;|N|\in\{4,8,12,24\}\;$
Show that $\;|N|=12\implies N=A_4\;$ , and thus it must be $\;|N|=8\;$ since $\;(12)(34)\in N\setminus H\;$
Now prove
$$(12)(34)(1234)(12)(34)=(1234)^{-1}\implies N\cong D_8\;$$