Proof about cycle with odd length

[fishturtle1]

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 $\alpha = \alpha^{s+1}$)

Homework Equations

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]

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 $\alpha a_k=a_j$).

Then what about two applications of $\alpha$. What about $s$ applications? What about $s+1$?

 

[fishturtle1]

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 $\alpha a_k=a_j$).

Then what about two applications of $\alpha$. What about $s$ applications? What about $s+1$?

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]

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

 

[fishturtle1]

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 $\alpha^{k+1}$.