Commuting p-cycles in Sn

• Oct 15th 2011, 08:38 PM
Bernhard
Commuting p-cycles in Sn
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 $\displaystyle S_n$ 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
• Oct 15th 2011, 08:52 PM
alexmahone
Re: Commuting p-cycles in Sn
Suggestion: Pick a small value of n and see if the statement is true.
• Oct 15th 2011, 09:00 PM
Deveno
Re: Commuting p-cycles in Sn
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.
• Oct 15th 2011, 09:46 PM
Bernhard
Re: Commuting p-cycles in Sn

Good idea

Take n = 3

Consider the element $\displaystyle \rho$ = (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
• Oct 15th 2011, 10:12 PM
alexmahone
Re: Commuting p-cycles in Sn
Quote:

Originally Posted by Bernhard

Good idea

Take n = 3

Consider the element $\displaystyle \rho$ = (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

In this case, (1 2 3) itself is the commutating 3-cycle.

Take n = 4.

Consider the element $\displaystyle \sigma = \left(\begin{array}{cccc}1 & 2 & 3 & 4 \\ 3 & 4 & 1 & 2\end{array}\right)$, which has order 2.

$\displaystyle \sigma=(1\ 3)(2\ 4)$, which is the product of commutating 2-cycles.
• Oct 15th 2011, 10:16 PM
Bernhard
Re: Commuting p-cycles in Sn

Let $\displaystyle \sigma$$\displaystyle \in$$\displaystyle S_n$ be such that |$\displaystyle \sigma$| = p where p is prime

Then |$\displaystyle \sigma$| = p $\displaystyle \Longrightarrow$ |$\displaystyle \sigma$| is a p-cycle (is that correct?)

Now every permutation $\displaystyle \sigma$ 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 $\displaystyle \sigma$ 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
• Oct 15th 2011, 10:22 PM
Bernhard
Re: Commuting p-cycles in Sn
Thanks

I assumed (wrongly) that the 'product' had to involve at least two cycles - hence was thrown!

Peter
• Oct 15th 2011, 11:03 PM
Bernhard
Re: Commuting p-cycles in Sn
I was quite wrong to say in the above post that

|$\displaystyle \sigma$| = p $\displaystyle \Longrightarrow$ $\displaystyle \sigma$ 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 $\displaystyle \sigma$ $\displaystyle \in$$\displaystyle S_n be such that |\displaystyle \sigma| = p where p is prime Then we may that \displaystyle \sigma can be expressed as a product of commuting cycles since (1) every permutation of a finite set can be expressed as a product of disjoint cycles (2) disjoint cycles are commutative But! Where to go from here?? It must be the case (as Deveno implies in a post above) that these commuting cycles are p-cycles - but WHY? I have the intuition that each of the cycles must have order p - but ... Can anyone help me further with this proof (Thanks to Alexmahone and Deveno for help so far) Peter • Oct 15th 2011, 11:15 PM alexmahone Re: Commuting p-cycles in Sn Quote: Originally Posted by Bernhard (I think I can make the claim that if a permutation can be expressed as a p-cycle then it has oder p) Assume that \displaystyle \sigma=abc... where \displaystyle a,\ b,\ c,\ ... are commutative p-cycles. So, \displaystyle \sigma^p=(abc...)^p=a^pb^pc^p...=e Now try proving the converse. • Oct 15th 2011, 11:41 PM Deveno Re: Commuting p-cycles in Sn the "missing bit" is this lemma, which you need to prove: if we factor \displaystyle \sigma as a product of r distinct cycles of lengths \displaystyle k_1,k_2,\dots,k_r then \displaystyle |\sigma| = lcm(k_1,k_2,\dots,k_r) . • Oct 16th 2011, 12:50 AM Bernhard Re: Commuting p-cycles in Sn OK, thanks for help. Will try to put together the proof that: \displaystyle \sigma has order p in \displaystyle S_n \displaystyle \Longrightarrow the cycle decomposition of \displaystyle \sigma is a product of commuting p-cycles ---------------------------------------------------------------------------------------------------- So let \displaystyle \sigma \displaystyle \in \displaystyle S_n be such that |\displaystyle \sigma| = p where p is prime Then \displaystyle \sigma can be expressed as a product of disjoint, and hence commuting cycles Thus \displaystyle \sigma = \displaystyle a_1$$\displaystyle a_2$ ... $\displaystyle a_r$ where the $\displaystyle a_i$ are commuting cycles of lengths, say, $\displaystyle k_1$, $\displaystyle k_2$, ... $\displaystyle k_r$

Then |$\displaystyle \sigma$| = p = lcm($\displaystyle k_1$, $\displaystyle k_2$, ... $\displaystyle k_r$)

Thus $\displaystyle k_1$ | p, $\displaystyle k_2$ | p, ... $\displaystyle k_r$ | p

Thus $\displaystyle k_i$ = 1 or p with at least one $\displaystyle k_1$ = p since p prime

[If p was not prime a number of the cycles could have lengths less than p]

Thus cycle decomposition of $\displaystyle \sigma$ is a product of commuting p-cycles

Is that OK?

If so I now have to show that if the cycle decomposition of $\displaystyle \sigma$ is a product of commuting p-cycles then |$\displaystyle \sigma$| = p

- but of course, reading through the posts again, Alexmahone has shown how to do this (thanks again!)

Peter