Prove that $S= \langle(1,2,3,4) \rangle$ has 3 conjugates in $S_4$ Some things I know:


*

*$S = \{ (1),(1,3)(2,4), (1,2,3,4),(1,4,3,2)\}$

*$(2,4) \in N_G(S)$

*Number of conjugates = $[G: N_G(S)]$


This seems like such a easy question but it made me realised that I do not know how to go about thinking about (and finding) the right cosets of $N_G(S)$. If I slowly work it out, I'm sure I can find all the elements of $N_G(S)$ and from there calculate the index, but is there a better way of finding out the answer?
 A: Conjugation in $S_n$ (i.e. relabelling the points) preserves the cycle structure. So all conjugate subgroups must be of the form $<(a,b,c,d)>$, because one of the points is $1$ it really is $<(1,a,b,c)>$, and a conjugating permutation would be one mapping $2\mapsto a$, $3\mapsto b$, $4\mapsto c$ while fixing $1$. Of the ($3!=6$) possible maps, two will give you the same subgroup, because $(1,a,b,c)^{-1}=(1,c,b,a)$, and so both elements lie in the same subgroup (as $(1,2,3,4),(1,4,3,2)$ in your example. Now it is not hard to find the three possibilities (e.g. besides $S$ itself $<(1,2,4,3)>$ and $<(1,3,2,4)>$).
A: Consider the conjugates of the element $(1,2,3,4)$: they are all and only the elements of $S_4$ with the same cyclic structure. Hence they are
$$
(1,2,3,4),\;\;(1,4,3,2)\\
(1,2,4,3),\;\;(1,3,4,2)\\
(1,4,2,3),\;\;(1,3,2,4)\\
$$
Now consider the corrispondent cyclic groups: the first couple of element I wrote is one the inverse of the other, hence they generate the same subgroup, and no other element with this structure can be in this subgroup.
Same argument for the remaining two couple.
Thus we found three different subgroups conjugate to $S=\langle(1,2,3,4)\rangle$.
But since every conjugate of it must be generated in this way, there are no more conjugate subgroups so we are finished.
