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


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

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

*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?

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 $<(1,3,2,4)>$).
