It is known that any two disjoint cycles in $S_n$ commutes. Therefore, any $\pi\in S_n$ which is disjoint with $\sigma$ is in the centralizer of $\sigma$: $C_{S_n}(\sigma)$. Also $$ \sigma^i\pi\in C_{S_n}(\sigma), 0\leq i\leq|\sigma|. \tag{*} $$ Here are my questions:

Are these all the elements in $C_{S_n}(\sigma)$? If the answer is no, is there an method to find all the elements in $C_{S_n}(\sigma)$ other than check the elements one by one?

A quick search returns several relevant questions:

The method ($*$) is what I learned so far from these questions and their answers. An immediate example in which ($*$) does not work that I can come up with is $\sigma=(12)(34)(56)$ in $S_6$. No disjoint $\pi$ can be found in this case and $$\langle(12),(34),(56)\rangle\subset C_{S_6}(\sigma).$$


The key is to look at how conjugation affects permutations of a given cycle type:

$$g=(a_1~a_2~\cdots~a_{\lambda_1})(b_1~b_2~\cdots~b_{\lambda_2})\cdots(c_1~\cdots~c_{\lambda_r}); \\ \sigma g\sigma^{-1}=(\sigma(a_1)~\sigma(a_2)~\cdots~\sigma(a_{\lambda_1}))\,\cdots(\sigma(c_1)~\cdots~\sigma(c_{\lambda_r})).$$

The first line denotes the disjoint cycle representation of an arbitrary $g\in S_n$. Thus, conjugation may only permute cycles of the same length in the representation. Let's revamp our notation:

$$g=a_{1,1}a_{1,2}\cdots a_{1,e_1}\cdots a_{s,1}\cdots a_{s,e_s},$$

where $a_{k,1},\cdots,a_{k,e_k}$ denote the $e_k$ cycles of length $\lambda_k$ in the decomposition, and $g$'s cycle type is

$$\lambda=\big(\underbrace{\lambda_1,\cdots,\lambda_1}_{e_1},\cdots,\underbrace{\lambda_s,\cdots,\lambda_s}_{e_s}\big)\vdash n.$$

The $\lambda_i$s are distinct so that the $e_i$s describe their multiplicities in the integer partition $\lambda$ of $n$.

Fix written representations of the cycles i.e. $a_{t,\ell}=(a_{t,\ell}^{(1)}~a_{t,\ell}^{(2)}~\cdots~ a_{t,\ell}^{(\lambda_{\large t})})$ for $1\le t\le s$ and $1\le\ell\le e_t$; denote, for an arbitrary element $\tau\in S_{e_1}\times\cdots\times S_{e_s}=E$, the induced permutation $\phi(\tau)\in S_n$ that sends $a_{t,\ell}^{(f)}\mapsto a_{t,\tau_{\large t}(\ell)}^{(f)}$, again for each index $1\le t\le s$, $1\le \ell\le e_t$. Thus the inclusion $\phi(E)\subseteq C(g)$ holds. The only other ways to not affect $g$'s structure is to cycle through its disjoint cycles, but note these cyclings can all be done independently from each other. We therefore deduce

$$C(g)=\left\langle \underbrace{\phi(S_{e_1}\times\cdots\times S_{e_s})}_E,\underbrace{\prod_{t=1}^s\prod_{\ell=1}^{e_t}\langle a_{t,\ell}\rangle}_{P}\right\rangle.$$

Observe the cyclic groups generated in the products in $P$ are each trivially intersecting, so the internal product is in fact direct. The elements of $E$ don't slide past those of $P$ without a fight, however; they act on them so that we are creating an unrestricted internal wreath product.

Note then that every element of $C(g)$ can be written as $\alpha\beta$ with $\alpha\in E$ and $\beta\in P$.

We say unrestricted because the arguments here generalize to symmetric groups of arbitrary cardinality, with care taken to only allow infinite products of disjoint cycles and replace $\lambda_i$s and $e_i$s with cardinal numbers as need be. In conclusion,

$$C(g)\cong \prod_{t=1}^s (C_{\lambda_{\large t}}\wr S_{e_{\large t}}),\quad \#C(g)=\prod_{t=1}^s \lambda_t^{e_t}e_t!$$

edit: fixed notation explanation

  • $\begingroup$ I feel like there could be a simpler way of doing this which involves the combinatorial facts about partitions of a given $n$...but I haven't been able to do it yet. Tried working with some tableaux interpretation... $\endgroup$ – Tanner Strunk Dec 11 '16 at 0:14
  • $\begingroup$ I know it's 7 years later, but I am not sure about one thing here - in the centraliser of $g$ should be also permutations disjoint with $g$ - I am not sure if this is described here. Anyway, very nice answer! $\endgroup$ – Igor Sikora Nov 20 '19 at 19:29
  • 1
    $\begingroup$ @IgorSikora In my description (IIRC) I am including fixed points as $1$-cycles in the cycle type, so e.g. $e=(1)(2)(3)$ in $S_3$. $\endgroup$ – anon Dec 9 '19 at 20:57

I will outline a way in which you (or somebody else who reads this thread in the future) can seek an answer to the originally asked question.

Step 1 : If you have a $k$ cycle $\tau$ in $S_n$ then, what can you say about the number of elements in the centralizer of $\tau$ ?

Answer: $k \cdot (n-k)!$ is the number of elements in the centralizer of a $k$ cycle in $S_n$, and it is rather easy to show this. Leave a comment if you want this to be described, but I am not sure when I will find time to reply to a comment.

Step 2: Next, if you denote by $T_{\tau}$ the set $T_{\tau} = \{ \tau ^i \cdot \sigma ~ : 0 \le i \le k-1, \sigma \in S_{n-k} \},$ then how many elements are there in $T_{\tau}$ ?

Answer: $k \cdot (n-k)!$ is the number of elements in $T_{\tau}$ and this is so straightforward that you should be able to show this yourself.

Step 3: Finally notice that $T_{\tau}$ defined as above is a subset of the centralizer $C_{\tau}$ of $\tau$.

All these would tell you that $C_{\tau}$ is in fact isomorphic (as a group) to $\mathbb{Z}_k \times S_{n-k}.$


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