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Bachelier
Nov16-11, 07:22 PM
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 = βx 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
Nov16-11, 07:52 PM
what have you tried? remember, when you have no clue what will work, any idea at all is progress.

Bachelier
Nov16-11, 10:47 PM
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
Nov17-11, 11:55 AM
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
Nov22-11, 03:01 AM
let ∏= ∏αi for all 1≤i≤n with αi = (ai1 ai2 ai3 ...aim)

consider θ = (a11 a21......an1 a12 a22......a1m a2m a3m...anm)

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

Hence ∏ = θn