If two permutations commute they are disjoint

In summary, the conversation discusses the statement that if two permutations, \alpha and \beta, in the symmetric group S_n commute, then \beta must permute those integers that are left fixed by \alpha. The participants also question whether this statement is true and provide counterexamples to show that it is not. They also consider a related statement and give an example in S_5 to show that it is not true.
  • #1
3029298
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Homework Statement


If [tex]\alpha,\beta\in S_n[/tex] and if [tex]\alpha \beta = \beta \alpha[/tex], prove that [tex]\beta[/tex] permutes those integers which are left fixed by [tex]\alpha[/tex]. Show that [tex]\beta[/tex] must be a power of [tex]\alpha[/tex] when [tex]\alpha[/tex] is a n-cycle.

The other way round is easy to see, since if two cycles are disjoint they do not do anything with the numbers permuted by the other cycle, hence they commute. But I don't know how to start when I want to prove the statement above... can anyone hint me?
 
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  • #2
Hi 3029298! :smile:

(have an alpha: α and a beta: β :wink:)
3029298 said:
If [tex]\alpha,\beta\in S_n[/tex] and if [tex]\alpha \beta = \beta \alpha[/tex], prove that [tex]\beta[/tex] permutes those integers which are left fixed by [tex]\alpha[/tex].

But that's obviously not true …

put α = β: then α does not permute those integers which are left fixed by α. :confused:
 
  • #3
Yes, I saw that as well, but I think (for the question to make sense) alpha must permute different elements than beta. But this makes the second part of the question questionable...
 
  • #4
Hi. I'm new in the forum. I'd like to prove (or find counter example) the next statement.

If two permutations [tex]\alpha,\beta\in S_n[/tex] such that [tex]\alpha\neq\beta^i[/tex] for every [tex]i\in\mathbb{Z}[/tex] commute then they're disjoint.

Do you think it's true?? I think it is.
 
  • #5
Here is an example in [itex]S_5.[/itex] Take [itex]\alpha=(12345),[/itex] and [itex]\beta=(13524)[/itex].

They commute:
[itex](12345)(13524)=(14253),[/itex]
[itex](13524)(12345)=(14253).[/itex]

But they are not disjoint (and neither is one a multiple of the other, re Unviaje).
 
  • #6
oops I just realized that [itex](12345)^2=(13524)[/itex]. I think you are right, Unviaje. I haven't found a proof yet though.
 

Related to If two permutations commute they are disjoint

What does it mean for two permutations to commute?

Two permutations commute if they can be done in either order and result in the same outcome. Essentially, both permutations have no effect on each other.

How do you know if two permutations are disjoint?

If two permutations have no elements in common, they are considered disjoint. This means that they have no shared elements and do not affect each other when performed in any order.

What is the significance of two permutations commuting?

Two permutations commuting can be seen as a special case of commutativity in mathematics. It is a property that allows for the manipulation and rearrangement of elements without affecting the outcome, making it a useful tool in various fields such as algebra and group theory.

Can two permutations that commute be equal?

Yes, two permutations that commute can be equal. This means that they are essentially the same permutation, just written in a different order. For example, the permutations (1 2)(3 4) and (3 4)(1 2) commute and are also equal.

Are there any exceptions to the rule that two commuting permutations are disjoint?

No, there are no exceptions to this rule. If two permutations commute, they will always be disjoint. This is because the definition of commutativity requires that both permutations have no effect on each other, meaning they cannot share any elements.

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