Prove that a product of disjoint m-cycles is a power of a cycle

[Bachelier]

let π be a product of disjoint m-cycles. Prove that π is a power of a cycle?

So this is like asking show that π = β^x for some cycle β and pos. integer x. right?

I don't know how to proceed on this except for the fact that the order of π is m.

any hints please

 

[mathwonk]

what have you tried? remember, when you have no clue what will work, any idea at all is progress.

 

[Bachelier]

what have you tried? remember, when you have no clue what will work, any idea at all is progress.

man I have no idea.

I know I can write π = (...)(...)(...)(...)(...)(...)(...)(...)(...)(...)

probably need to consider when m is even and odd.

I can break down π to a product of transpositions.

But the end result is too abstract I can get my head around it.

 

[micromass]

First, note that

$$\sigma (i_1~i_2~i_3~...~i_n) \sigma^{-1}=(\sigma(i_1)~\sigma(i_2)~\sigma(i_3)~...~\sigma(i_n))$$

This allows you to bring everything back to the cycle (1 2 3 ... n).

Now, take powers of (1 2 3 ... n) and see what types of disjoint cycle decompostions you meet.

 

[Bachelier]

let ∏= ∏α_i for all 1≤i≤n with α_i = (a_i1 a_i2 a_i3 ...a_im)

consider θ = (a₁₁ a₂₁...a_n1 a₁₂ a₂₂...a_1m a_2m a_3m...a_nm)

then applying θ n times will give us the original ∏.

Hence ∏ = θⁿ