# Proof about cycle with odd length

## Homework Statement

In the following problems let ##\alpha## be a cycle of length ##s##, and say
##\alpha = (a_1a_2 . . . a_s)##.

5) If ##s## is odd, ##\alpha## is the square of some cycle of length s. (Find it. Hint: Show ##\alpha = \alpha^{s+1}##)

## The Attempt at a Solution

I know ##(12345)(12345) = (12345)## and ##(1234)(1234) = (13)(24)##. And I think I can prove this is true for any integer n, such that if a cycle ##\alpha## has an odd length, then squaring it produces another cycle of odd length, and ##\alpha## has an even length, s, then its square is the product of two disjoint cycles, each of length s/2. I guess i've pretty much restated the problem... i'm stuck.

For the hint.. I know ##\alpha## of length ##s## has ##\alpha## distinct powers but I'm not sure how to prove it.

Sorry this is so bare bones, thank you for any help

andrewkirk
Homework Helper
Gold Member
I know ##(12345)(12345) = (12345)##
That doesn't look right. I get ##(12345)(12345)=(13524)##.
The 'square root' of (12345) is (13524) because (13524)(13524)=(12345).

The hint is a good one. Have you tried proving it?
What is the position in the cycle to which the ##k##th element ##a_k## is mapped by one application of ##\alpha## (in other words, find an expression in terms of ##k## for the index ##j## such that ##\alpha a_k=a_j##).

That doesn't look right. I get ##(12345)(12345)=(13524)##.
The 'square root' of (12345) is (13524) because (13524)(13524)=(12345).

The hint is a good one. Have you tried proving it?
What is the position in the cycle to which the ##k##th element ##a_k## is mapped by one application of ##\alpha## (in other words, find an expression in terms of ##k## for the index ##j## such that ##\alpha a_k=a_j##).

Thanks for the help and correction, I agree that (12345)12345) = (13524)

So we can write ##\alpha^na_i = a_j## where ##j = (i + n) \mod s##
Therefore ##\alpha a_i = a_{((i + 1) \text{mod s})}##
and ##\alpha^{s+1}a_i = a_{((s+1+i) \text{mod s})}##
Since (i + 1) = (s + 1 + i) (mod s)
we have ##\alpha a_i = \alpha^{s+1}a_i##
##\alpha a_ia_i^{-1} = \alpha^{s+1}a_ia^{-1}##
##\alpha e = \alpha^{s+1}e##
##\alpha = \alpha^{s+1}##
[]

##\alpha^sa_i = a_{i + s} = a_i##
##\alpha^{s+1}a_i = a_{i + s + 1} = a_{i + 1}##

I'm not sure how to apply this to the original problem but ill keep thinking.. I also thought of this which I think is a proof in response to the "Find it" part of the hint.

Proof: Suppose ##\alpha## is a cycle of length ##s## where ##s## is an odd integer. Then we can write ##\alpha## as ##(a_1a_3...a_sa_2a_4...a_{s-1}).## But ##(a_1a_3...a_sa_2a_4...a_{s-1}) = (a_1a_2...a_s)(a_1a_2...a_s).## Let ##\beta = (a_1a_2..a_s).## Note that ##\beta## is a cycle of length s. Then ##\alpha = \beta^2##. This concludes the proof.

andrewkirk
Homework Helper
Gold Member
Give that ##\alpha=\alpha^{s+1}## and ##s## is odd, how can you write ##\alpha## as the square of an integer power of itself?

Give that ##\alpha=\alpha^{s+1}## and ##s## is odd, how can you write ##\alpha## as the square of an integer power of itself?

s is odd, so we can write s = 2k + 1, where k is an integer.
We note that ##\alpha^n## is a cycle of length s for all integers n.
Then ##\alpha = \alpha^{s+1} = \alpha^{2(k+1)} = \alpha^{k+1}\alpha^{k+1}##.
so ##\alpha## is the square of ##\alpha^{k+1}##.