I need to find the centralizer of the permutation $\sigma=(1 2 3 ... n)\in S_n$.
I know that:
$C_{S_n}(\sigma)=\left\{\tau \in S_n|\text{ } \tau\sigma\tau^{-1}=\sigma\right\}$
In other words, that the centralizer is the set of all the elements that commute with $\sigma$, and I also know that if two permutations have disjoint cycles it implies that they commute, but the thing is; there are no $\tau\in S_n$ s.t. $\tau$ and $\sigma$ have disjoint cycles, since $\sigma=(1 2 3...n)$.
So can I conclude that $\sigma$ does not commute with any other $\tau$ in $S_n$ (besides $id$ of course)?
I guess my question reduces to: is the second direction of the implication mentioned above also true? meaning, if two permutation commute, does it imply that they have disjoint cycles?
If the answer is no, how else can I find the $C_{S_n}(\sigma)$?

By the way, on related subject, I noticed that if an element $g$ of a group $G$ is alone it its conjugacy class, it commutes with all elements in $G$.
What does it mean, intuitively, for an element to share its conjugacy class with another element? does it mean it "almost" commute with everyone in the group?
Is it true that the bigger the conjugacy class, the lesser its members commutes with others in the group?

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    $\begingroup$ No, the other direction does not hold. A permutation will always commute with all powers of itself. $\endgroup$ Jun 19, 2014 at 20:23
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    $\begingroup$ For your last paragraph: “The size of a conjugacy class is equal to the index of its centralizer.” is the way your truth is often expressed. $\endgroup$ Jun 19, 2014 at 20:24
  • $\begingroup$ The part in quotation marks in @JackSchmidt's comment also gives you a good way to see how many elements should be in this centralizer, assuming you are familiar with how conjugacy classes look in the symmetric groups. $\endgroup$ Jun 19, 2014 at 20:25
  • $\begingroup$ @TobiasKildetoft, I know that "two permutations conjugate iff they have the same cycle type". Is that what you meant? $\endgroup$ Jun 19, 2014 at 20:34
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    $\begingroup$ Yes, precisely. $\endgroup$ Jun 19, 2014 at 20:40

3 Answers 3


Hint: conjugacy in $S_n$ leaves cycle types in tact: $\tau^{-1}(1 2 3 \dots n)\tau=(\tau(1) \tau(2)\tau(3) \cdots \tau(n))$.

  • $\begingroup$ It will only help me figure out how many are there ($(n-1)!$), but will I be able to find them? I'm just confused about how I'm supposed to explicitly write them (the above is a question that might appear in my exam next week). $\endgroup$ Jun 19, 2014 at 20:40
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    $\begingroup$ Well if $\sigma$ is centralized by $\tau$, so $\tau^{-1}(1 2 3 \dots n)\tau=(\tau(1) \tau(2)\tau(3) \cdots \tau(n))=(1 2 3 \cdots n)$, can you see that $\tau$ must be a power of $\sigma$ (try small examples $n=2,3$)? Your observation of $(n-1)!$ is not correct ... $\endgroup$ Jun 19, 2014 at 20:49
  • $\begingroup$ Yeah, you're right, my $(n-1)!$ observation was mistakenly referred to the size of the centralizer, instead of to the size of the conjugacy class. $\endgroup$ Jun 19, 2014 at 21:06
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    $\begingroup$ OK, proving that $\sigma$ commutes with all powers of itself was easy, and thus giving $<\sigma>\subset C_{S_n}(\sigma)$, but I couldn't figure out from the hint why does it mean that any other element which isn't a power of $\sigma$ does not belong to the centralizer, meaning that $<\sigma>= C_{S_n}(\sigma)$ $\endgroup$ Jun 19, 2014 at 21:18
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    $\begingroup$ Ah great! That's your "learning point", glad you see it now! $\endgroup$ Jun 19, 2014 at 21:43

Let $S_n$ act on itself by conjugation. Let $g = (1,2,\ldots,n)$. The size of the orbit of $g$ in this action is its conjugacy class $\{ h^{-1} gh: h \in S_n\}$. The stabilizer subgroup of $g$ in this action is the set of elements $h$ in $S_n$ such that $h^{-1}gh=g$, i.e. the centralizer $C_{S_n}(g)$. By the orbit-stabilizer lemma, the size of the orbit equals the index of the stabilizer. The size of the orbit is the number of elements that have the same cycle structure as $g$, which is $(n-1)!$. Thus, the index of the centralizer is $(n-1)!$, whence the centralizer has $n! / (n-1)!=n$ elements. Thus the powers of $g$ exhaust all of $C_{S_n}(g)$.

A second proof is as follows. For the special case where $g=(1,2,\ldots,n)$, we can determine its centralizer in $S_n$ without using the orbit-stabilizer lemma. If $h^{-1}gh = g$, then $(h(1),h(2),\ldots,h(n)) = (1,2,\ldots,n)$. Now, $h(1)$ can be chosen in $n$ ways to be any of $1,2,\ldots,n$, but once $h(n)$ is chosen, the remaining elements $h(2),\ldots,h(n)$ are uniquely determined. In fact, if $h(1)=i$, then $h(2)=i+1$, and so on, and so $h$ is just a power of $g$. Thus, there are exactly $n$ different elements $h$ such that $h^{-1}gh=g$.


Note that $|\sigma| = n$ so that $\sigma^n = 1$. So $$ \langle \sigma \rangle = \{1, \sigma, \sigma^2, \ldots, \sigma^{n-1}\} $$ forms a cyclic commutative subgroup of $S_n$ with order $|\langle \sigma \rangle| = n$. For any permutation $x$ of a symmetric group $S_n$ we have the very convenient formula: $$ |C(x)| = 1^{\alpha_1}2^{\alpha_2}\cdots n^{\alpha_n}\alpha_1!\alpha_2!\ldots\alpha_n!, $$ where $\alpha_i$ is the number of cycles in $x$ of length $i$. For your permutation $\alpha_1 = \alpha_2 = \cdots \alpha_{n-1} = 0$ and $\alpha_n = n$ so $|C(\sigma)| = n$. We have $|C(\sigma)| = |\langle \sigma \rangle| = n$, implying $C(\sigma) = \langle \sigma \rangle$.

  • $\begingroup$ This is not right; for example, $|C(x)| = 2 \cdot 2! \cdot 3 \cdot 1! = 12$ for $x = ((12)(34)(567))$ in $S_{10}$, but this is not the correct answer. $\endgroup$
    – Ben123
    Mar 17 at 19:03

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