Commuting p-cycles in Sn

**Bernhard** · October 15th 2011, 08:38 PM

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 $S_n$ 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

**alexmahone** · October 15th 2011, 08:52 PM

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

**Deveno** · October 15th 2011, 09:00 PM

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.

**Bernhard** · October 15th 2011, 09:46 PM

(Answer to AlexMahone)

Good idea

Take n = 3

Consider the element $\rho$ = (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

**alexmahone** · October 15th 2011, 10:12 PM

Originally Posted by Bernhard

(Answer to AlexMahone)

Good idea

Take n = 3

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

$\sigma=(1\ 3)(2\ 4)$ , which is the product of commutating 2-cycles.

**Bernhard** · October 15th 2011, 10:16 PM

(Answer to Deveno)

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

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

Now every permutation $\sigma$ 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 $\sigma$ 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

**Bernhard** · October 15th 2011, 10:22 PM

Thanks

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

Peter

**Bernhard** · October 15th 2011, 11:03 PM

I was quite wrong to say in the above post that

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

Then we may that $\sigma$ 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

**alexmahone** · October 15th 2011, 11:15 PM

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 $\sigma=abc...$ where $a,\ b,\ c,\ ...$ are commutative p-cycles.

So, $\sigma^p=(abc...)^p=a^pb^pc^p...=e$

Now try proving the converse.

**Deveno** · October 15th 2011, 11:41 PM

the "missing bit" is this lemma, which you need to prove:

if we factor $\sigma$ as a product of r distinct cycles of lengths $k_1,k_2,\dots,k_r$

then $|\sigma| = lcm(k_1,k_2,\dots,k_r)$ .

**Bernhard** · October 16th 2011, 12:50 AM

OK, thanks for help.

Will try to put together the proof that:

$\sigma$ has order p in $S_n$ $\Longrightarrow$ the cycle decomposition of $\sigma$ is a product of commuting p-cycles
----------------------------------------------------------------------------------------------------

So let $\sigma$ $\in$ $S_n$ be such that | $\sigma$ | = p where p is prime

Then $\sigma$ can be expressed as a product of disjoint, and hence commuting cycles

Thus $\sigma$ = $a_1$ $a_2$ ... $a_r$ where the $a_i$ are commuting cycles of lengths, say, $k_1$ , $k_2$ , ... $k_r$

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

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

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

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

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

Is that OK?

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

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

Peter

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