I would like to know how is possible to calculate the order of the normalizer of $H=\langle s\rangle$ in $S_n$ where $s$ is an assigned permutation of $S_n$.
I know that finding the order of the centralizer is relatively easy using the following "propositions":
1) Two permutations are conjugate IFF they have the same cycle structure.
2) $\frac{|G|}{|cl(x)|}=|C_{G}(x)|$.
but for the normalizer what do I have to do? there is a standard method?
I thought to use the theorem $N/C$ and in particular the fact that $\frac{|N_{G}(H)|}{|C_{G}(H)|}$ must divide $|Aut(H)|$ but sometimes this process is not enough to reach a solution.
For instance if $s=(1,2,3,4)(5,6,7,8)(9,10)(11,12)$ in $S_{12}$ then $|cl(s)|=\frac{12!}{2^8}$ and $|C_{G}(H)|=2^8$. Now $|Aut(H)|=2$ so by theorem $N/C$ i only can conclude $|N_{S_{12}}(H)|=2^8$ or $=2^9$.
I really hope you can help me I've my algebra exam in less then a week!!
P.S I'm sorry for my English I hope everything is at least understandable
Best Answer
I believe [N:C] = |Aut(H)| always (for H a cyclic subgroup of the symmetric group).
If H is generated by a single cycle, then it is true: whenever H is a regular permutation group, then the normalizer contains the automorphism group fairly literally as a permutation group.
If H is generated by a product of disjoint cycles, then the product of the normalizers of the cycles contains the automorphism group too.
For instance, the normalizer of $H = \langle (1,2,3,4)(5,6,7,8)(9,10)(11,12) \rangle$ contains the (orbit) normalizers $\langle (1,2,3,4), (1,3) \rangle$, $\langle (5,6,7,8), (5,7) \rangle$, $\langle (9,10) \rangle$, and $\langle (11,12) \rangle$. In particular, the normalizer of H induces every automorphism of H, just by working on each orbit.