Given a permutation $\sigma\in S_{n}$, is there a way to know the order of the centraliser $C_{S_{n}}\left(\sigma\right)=\left\{ \pi\in S_{n},\,\pi\sigma=\sigma\pi\right\}$ , i.e what is $\left|C_{S_{n}}\left(\sigma\right)\right|$?

I would appreciate a proof if the answer is yes.

Also, if the answer above is yes, is there also a way to calcualte the order of the centraliser of a given subset of $S_{n}$, or at least for a pair of permutations?


3 Answers 3


Let $n_1,n_2,\ldots,n_k$ be the distinct lengths of the cycles of $\sigma$ (including 1 if there are fixed points) and suppose that there are $m_i$ cycles of length $n_i$. Then the centralizer of $\sigma$ can permute the cycles of the same length. Its order is $\prod_{i=1}^k n_i^{m_i}m_i!$.

Calculating the centralizer of a subgroup $H$ of $S_n$ is not difficult, but it is more complicated. The order of the centralizer of a single orbit is equal to the number of fixed points (in that orbit) of the stabilizer of a point in the orbit. But if $H$ has more than one orbit with equivalent actions then the equivalent orbits can be permuted by the centralizer, so the complete centralizer is a direct product of wreath products of centralizers of sets of equivalent orbits.

  • $\begingroup$ i must say i didn't exactly understand the second part. any chance of getting an example with 2 different permutation or something of that sort? Thanks in any case! $\endgroup$
    – IBS
    Nov 28, 2011 at 13:28

You let the permutations act by conjugation on the permutation and you seek the size of the stabilizer of $\sigma$. By the orbit-stabilizer theorem, it is enough to know the size of the orbit, which is the well-known size of the conjugacy class of $\sigma$.


Factor $\sigma$ into its orbits, and let $m_1, m_2, \ldots, m_i$ be the order of each of those orbits, with $m = \sum_{j=1}^i m_j$. For a permutation $\pi$ to commute with $\sigma$, for each of the orbits of $\pi$ we must have either that it is a power of an orbit of $\sigma$, or that it acts on none of the elements that $\sigma$ does. Thus we have that the centralizer group is isomorphic to

$$ \mathbb{Z}_{m_1}\oplus \mathbb{Z}_{m_2}\oplus \cdots \oplus\mathbb{Z}_{m_i} \oplus S_{n-m} $$

  • 3
    $\begingroup$ This is wrong, and I'm sorry for writing something like this thinking it was an actual answer. The point that I've found that I'm wrong is that even though a set of orbits in $\pi$ and in $\sigma$ aren't powers of one another, they might both be powers of some other permutation. Example: in $S_6$, you have the translation $\phi = (1, 2, 3, 4, 5, 6)$. Let $\sigma = \phi^2 = (1, 3, 5)(2, 4, 6)$ and $\pi = \phi^3 = (1, 4)(2, 5)(3, 6)$. Then $\sigma$ and $\pi$ commutes without any of their respective orbits commuting. Disregard my answer, please. $\endgroup$
    – Arthur
    Nov 27, 2011 at 6:09
  • $\begingroup$ So why don't you delete it, then? $\endgroup$
    – Alex M.
    Dec 12, 2015 at 17:26
  • 1
    $\begingroup$ @AlexM. I didn't delete it because wrong as it is, with my comment below it still has some minimal value. Also, it is not in anyone's way. How about you? Why did you bump a four year old question without having anything real to contribute? $\endgroup$
    – Arthur
    Dec 12, 2015 at 17:34
  • $\begingroup$ I've bumped into it while reviewing close votes: a similar question was proposed for closing as a duplicate of this one, so it was natural to see what this one was about. $\endgroup$
    – Alex M.
    Dec 12, 2015 at 17:38

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