Re: Commuting p-cycles in Sn

Suggestion: Pick a small value of n and see if the statement is true.

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.

Re: Commuting p-cycles in Sn

(Answer to AlexMahone)

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

Re: Commuting p-cycles in Sn

Quote:

Originally Posted by

**Bernhard** (Answer to AlexMahone)

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.

Re: Commuting p-cycles in Sn

(Answer to Deveno)

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

Re: Commuting p-cycles in Sn

Thanks

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

Peter

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

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.

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) $.

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