I was trying to find the structure of centralizer of any permutation in $ Sym(n) $ .

First ı consider as an example $C_{Sym n}(123)$. So ı am looking for

the set {$\delta \in Sym(n) | (\delta(1)\delta(2)\delta(3)) = (123) $}.

This set is actually {$\delta \in Sym(n) | \delta(1)=1, \delta(2)=2, \delta(3)=3$ } $\bigcup$ {$\delta \in Sym(n) | \delta(1)=3, \delta(2)=1, \delta(3)=2$ } $\bigcup$ {$\delta \in Sym(n) | \delta(1)=2, \delta(2)=3, \delta(3)=1$ }.

(İt is come from the variations of $(123) : (123),(312),(231)$).

So we can write these sets actually like this :$Sym \{4,...,n\}\bigcup (123)Sym \{4,...,n\}\bigcup (132)Sym \{4,...,n\} = \langle (123)\rangle .Sym \{4,...,n\}$ .

After all these ı think general version such that if $\alpha = \alpha_1 ...\alpha_n$ is a permutation but $\alpha_i $ and $\alpha_j $ are disjoint cycles then $C_{Sym n} \alpha = \langle \alpha_1, ...,\alpha_n \rangle$. $ Sym \{x| \forall i \in \{ 1,...,n \} , \alpha_i (x) = x \} $. Am ı correct ?

  • 1
    $\begingroup$ That's not quite right. Cycles of equal length can be interchanged by the centralizer. For example $(1,2)(3,4)$ is centralized by $(1,3)(2,4)$. $\endgroup$
    – Derek Holt
    Jul 6, 2018 at 7:14
  • $\begingroup$ @Derek Holt Sure, like in Klein four group. I think ı can fix it by looking these kind of subsets of Sym(n) but do you know the general case ? It seems to me there must be a natural answer... $\endgroup$
    – wrabbit
    Jul 6, 2018 at 8:53

1 Answer 1


Let $\sigma \in S_n$ acting on $X$ with $|X|=n$. Write $\sigma$ as a product of disjoint cycles, and group those cycles of the same length together (including length 1), so that we get $X = X_{i_1} \cup X_{i_2} \ldots X_{i_r}$, where the $i_j$ are distinct, and $\sigma$ acts as a product of cycles all of length $i_j$ on each $X_{i_j}$. Then $C_{S_n}(\sigma)= C_1 \times \cdots \times C_r$, where $C_j$ is the centralizer of the restriction of $\sigma$ to $X_{i_j}$.

So this effectively reduces the problem to the case where $\sigma$ acts on $X$ as a product of cycles $\sigma_1 \cdots \sigma_m$ all of the same length $k$ on $X$. (e.g. $\sigma = (1,2,3)(4,5,6)(7,8,9)(10,11,12)$ with $k=3$, $m=4$, $n=12$).

The centralizer $C$ of $\sigma$ in $S_n$ in this case is (isomorphic to) the wreath product of $\langle \sigma_1 \rangle$ with $S_m$. This has a normal abelian subgroup $N$ isomorphic to $C_k^m$ generated by $\sigma_1,\ldots,\sigma_k$. Then $C/N \cong S_m$, where the $S_m$ acts by permuting the cycles.

So, in the example, there is a subgroup $S_4$ of $C$ generated by $(1,4)(2,5)(3,6)$ and $(1,4,7,10)(2,5,8,11)(3,6,9,12)$.

I have noticed this is also answered here.


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