Okay, so I'm teaching myself some algebra here, and I just need to cement some of the relations.
We have that disjoint permutations commute, a permutation commutes with itself, and therefore all powers of itself, and the identity, of course, commutes with everything.
So to find the set of all permutations that commute with a given permutation, would it be suffice to include all of the above?
Commuting permutations
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Commuting permutations
Mike_Bson wrote:What's next, something representing less than nothing?
Re: Commuting permutations
You are looking for the centralizer of a given permutation. If you don't know about wreath products this will be difficult to describe, but if you want to know:
Suppose that the cycle type is [imath]1^{a_1}2^{a_2}\ldots n^{a_n}[/imath]. The centralizer is the direct product of the centralizers of the individual cycle types: i.e., we consider the centralizer of a permutation of type [imath]i^{a_i}[/imath], and then take the direct product of these things. This centralizer has the form [imath]C_i\wr S_{a_i}[/imath]. What this means is that it is generated by the [imath]a_i[/imath] cycles of length [imath]i[/imath] that you can obviously see, and all permutations sending one cycle to another.
I can give you more information, or point you to a reference if you want, but this is actually a good topic for you to calculate the answer yourself. To check you have all of the permutations, use the orbitstabilizer theorem and count the number of permutations of a given cycle type.
Suppose that the cycle type is [imath]1^{a_1}2^{a_2}\ldots n^{a_n}[/imath]. The centralizer is the direct product of the centralizers of the individual cycle types: i.e., we consider the centralizer of a permutation of type [imath]i^{a_i}[/imath], and then take the direct product of these things. This centralizer has the form [imath]C_i\wr S_{a_i}[/imath]. What this means is that it is generated by the [imath]a_i[/imath] cycles of length [imath]i[/imath] that you can obviously see, and all permutations sending one cycle to another.
I can give you more information, or point you to a reference if you want, but this is actually a good topic for you to calculate the answer yourself. To check you have all of the permutations, use the orbitstabilizer theorem and count the number of permutations of a given cycle type.
 MartianInvader
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Re: Commuting permutations
The answer is basically, given above, but to give a specific example of why the answer to your question is "no", notice that (123)(456) commutes with (14)(25)(36). Remember, two group elements commute iff conjugation by one leaves the other unchanged. And there's a very nice description of what conjugation does for permutations.
Let's have a fervent argument, mostly over semantics, where we all claim the burden of proof is on the other side!
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Re: Commuting permutations
Unfortunately, DavCrav's post is a bit above me. Let's say I want to find the centralizer of the cycle [imath]p=(1,2, \dots ,n) \in S_n[/imath]. (I think this should be a simple example, because it moves all points). I think that the centralizer of [imath]p[/imath] is only the cyclic group [imath]\langle p \rangle[/imath]. However, I am having a hard time proving this to myself. I can prove that all elements of the centralizer of [imath]p[/imath] will lack fixed points, but can't show that they must be powers of [imath]p[/imath]. Does anyone have a somewhat simpler way to explain at least this subcase?

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Re: Commuting permutations
For that specific case, here's a hint. Let p be the cycle as you've defined it, and let c be any permutation.
If x \in {1,2,...,n}, then what is p(x)?
So what is p(c(x))?
And what is c(p(x))?
If x \in {1,2,...,n}, then what is p(x)?
So what is p(c(x))?
And what is c(p(x))?
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Re: Commuting permutations
Alright, thanks lightvector; I think I have it now.
Showing that the centralizer [imath]C_{S_n}(a)=\langle a \rangle[/imath], is equivalent to showing that
the stabilizer [imath]\text{Stab}_{S_n}(a)=\langle a \rangle[/imath], for [imath]S_n[/imath] acting on itself by conjugation.
[imath]a^ka=a^{k+1}=aa^k, \forall k[/imath] implies that a commutes with its powers (this is so in any group).
So [imath]C_{S_n}(a)\geq n[/imath], and [imath]C_{S_n}(a) \supseteq \langle a \rangle[/imath].
Now I just need to show that [imath]C_{S_n}(a)= n[/imath]. So here goes:
The number of conjugates (that is, the size of the orbit of a in [imath]S_n[/imath], denote [imath]\mathcal{O}_{S_n}(a)[/imath]) is the index [imath][S_n:C_{S_n}][/imath], by the orbitstabilizer theorem. Counting the conjugates is simple: there are [imath]n![/imath] choices for the elements of the ncycle, but this counts each cycle n times, so we have [imath]n!/n[/imath] conjugates. Thus,
[math]n!=S_n=\mathcal{O}_{S_n}(a)\text{Stab}_{S_n}(a)= \frac{n!}{n} C_{S_n}(a)\Rightarrow C_{S_n}(a)=n.[/math]
So [imath]C_{S_n}(a)=\langle a \rangle[/imath].
Showing that the centralizer [imath]C_{S_n}(a)=\langle a \rangle[/imath], is equivalent to showing that
the stabilizer [imath]\text{Stab}_{S_n}(a)=\langle a \rangle[/imath], for [imath]S_n[/imath] acting on itself by conjugation.
[imath]a^ka=a^{k+1}=aa^k, \forall k[/imath] implies that a commutes with its powers (this is so in any group).
So [imath]C_{S_n}(a)\geq n[/imath], and [imath]C_{S_n}(a) \supseteq \langle a \rangle[/imath].
Now I just need to show that [imath]C_{S_n}(a)= n[/imath]. So here goes:
The number of conjugates (that is, the size of the orbit of a in [imath]S_n[/imath], denote [imath]\mathcal{O}_{S_n}(a)[/imath]) is the index [imath][S_n:C_{S_n}][/imath], by the orbitstabilizer theorem. Counting the conjugates is simple: there are [imath]n![/imath] choices for the elements of the ncycle, but this counts each cycle n times, so we have [imath]n!/n[/imath] conjugates. Thus,
[math]n!=S_n=\mathcal{O}_{S_n}(a)\text{Stab}_{S_n}(a)= \frac{n!}{n} C_{S_n}(a)\Rightarrow C_{S_n}(a)=n.[/math]
So [imath]C_{S_n}(a)=\langle a \rangle[/imath].
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