Suggestion: Pick a small value of n and see if the statement is true.
Dummit and Foote Section 1.3 Symmetric Groups Exercise 15 is as follows:
Let p be a prime. Show that an element has order p in if and only if its cycle decomposition is a product of commuting p-cycles. Show by explicit example that this need not be the case if p is not prime.
Would appreciate help with this problem
Peter
sketch of proof:
decompose σ in Sn into disjoint (non-trivial, that is, length > 1) cycles. now if |σ| = p, can any of these cycles have order < p or > p? why do disjoint cycles necesarily commute?
for a counter-example, consider (1 2)(3 4 5) in S6.
(Answer to AlexMahone)
Good idea
Take n = 3
Consider the element = (1 2 3) which is of order 3
Then it should follow that it cycle decomposes in two commuting 3-cycles.
But problems at this early stage - how to find them?
I know (1 2 3) = (1 3) o (1 2) ... and these are not even commuting ... 3-cycles??
How do I find the required 3-cycles in this case?
Peter
(Answer to Deveno)
Let be such that | | = p where p is prime
Then | | = p | | is a p-cycle (is that correct?)
Now every permutation of a finite set can be expressed as a product of disjoint cycles and, further, disjoint cycles are commutative - they do not have an effect outside their own elements.
So is a product of commuting disjoint cycles BUT!!
these cycles would be m-cycles where m is less than p (wouldn't they???)
Surely if we express a cycle in terms of the product of a number of cycles the cycles in the product would be smaller - wouldn't they?
Peter
I was quite wrong to say in the above post that
| | = p is a p-cycle
(I think I can make the claim that if a permutation can be expressed as a p-cycle then it has oder p)
So removing the erroneous statement from the argument we now have the following:
===============================================
Let be such that | | = p where p is prime
Then we may that can be expressed as a product of commuting cycles since
(1) every permutation of a finite set can be expressed as a product of disjoint cycles
(2) disjoint cycles are commutative
But! Where to go from here??
It must be the case (as Deveno implies in a post above) that these commuting cycles are p-cycles - but WHY?
I have the intuition that each of the cycles must have order p - but ...
Can anyone help me further with this proof
(Thanks to Alexmahone and Deveno for help so far)
Peter
OK, thanks for help.
Will try to put together the proof that:
has order p in the cycle decomposition of is a product of commuting p-cycles
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So let be such that | | = p where p is prime
Then can be expressed as a product of disjoint, and hence commuting cycles
Thus = ... where the are commuting cycles of lengths, say, , , ...
Then | | = p = lcm( , , ... )
Thus | p, | p, ... | p
Thus = 1 or p with at least one = p since p prime
[If p was not prime a number of the cycles could have lengths less than p]
Thus cycle decomposition of is a product of commuting p-cycles
Is that OK?
If so I now have to show that if the cycle decomposition of is a product of commuting p-cycles then | | = p
- but of course, reading through the posts again, Alexmahone has shown how to do this (thanks again!)
Peter