For some reason I'm finding permutation cycles to be strange and hard to deal with.
Let $\alpha$ and $\beta$ be cycles of odd length (not disjoint). Prove that if $\alpha^2 = \beta^2$, then $\alpha = \beta$.
I don't see what the odd lengths of $\alpha$ and $\beta$ have to do with anything. I see that they are even permutations. I see that $\alpha^2 = \beta^2 \implies \alpha^2\beta^{-2} = \varepsilon$. I don't think that helps. I don't know. The book went over the proof that a permutation can't be both odd and even, but I don't see anything in the proof that helps.
If $\alpha = (a_1, a_2, \dots, a_s)$, then $\alpha^2 = (a_1, a_3, \dots, a_s, a_2, \dots, a_{s-1})$. If $\beta = (b_1, b_2, \dots, b_r)$, then $\beta^2 = (b_1, b_3, \dots, b_r, b_2, \dots, b_{r-1})$. I don't see how to match up the elements of $\alpha$ and $\beta$ if I'm using different letters, but I can't use the same letters because I don't know which ones are the same or different. And maybe they are the same, but what if one of the cycles is rotated and starts on a different element? How do you deal with things like this?
Edit: Is the following proof sound?
The only way for $\alpha^2$ to be equal to $\beta^2$ while $\alpha \neq \beta$ is for $\alpha^2$ and $\beta^2$ to both be equal to $\varepsilon$. But the only way for that to happen is if the exponent, 2, is equal to a multiple of the length of the cycle. But since the lengths of $\alpha$ and $\beta$ are odd, their lengths have to be at least 3, since if they were length 1, they would be the same as length 0. So $\alpha^2$ and $\beta^2$ cannot be equal to $\varepsilon$, so $\alpha$ must be equal to $\beta$.