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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

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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.

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Thank you Mr.Schmidt your answer have been really useful! – Lorenzo Rossi Jun 20 '11 at 22:07
I think instead of $(1,3)(2,4)$ you want $(1,3)$, and similarly for $(5,7)(6,8)$. – user641 Dec 20 '11 at 6:34
@SteveD: thanks – Jack Schmidt Dec 20 '11 at 14:30

If $\sigma = (1\ 2\ 3\ 4)(5\ 6\ 7\ 8)(9\ 10)(11\ 12)$ then $\sigma^2 = (1\ 3)(2\ 4)(5\ 7)(6\ 8)$, $\sigma^3 = (1\ 4\ 3\ 2)(5\ 8\ 7\ 6)(9\ 10)(11\ 12)$ and $\sigma^4 = id$.

We're looking for the number of $\tau \in S_{12}$ that satisfy $\tau\sigma\tau^{-1}\in\langle\sigma\rangle$.

Conjugation does not change the cycle structure of a permutation. This means that the only reachable powers are $\sigma$ and $\sigma^3$. Since we already know the order of $Z(\sigma)$ we only need to find the number of $\tau \in S_{12}$ that satisfy $\tau\sigma\tau^{-1}=\sigma^3$

We observe that once we fix the image of an element in a cycle we fixed the image of every other element in that cycle.

$1$ can be sent in any of $8$ possible elements of a $4$-cycle of $\sigma^3$. $5$ can be sent in any of the $4$ possible elements of the $4$-cycle where we didn't send $1$. Reasoning analogously we find that $9$ has $4$ possible images and $11$ has $2$ possible images.

This gives $8\cdot4\cdot4\cdot2 = 2^{8}$ new elements of $N(\sigma)$, for a total of $2^9$ elements.

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Thanks Jacopo!!!!! – Lorenzo Rossi Jun 20 '11 at 22:07

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