Cycles of odd length: $\alpha^2=\beta^2 \implies \alpha=\beta$

Question

Let $\alpha$ and $\beta$ be cycles of odd length (not disjoint). Prove that if $\alpha^2=\beta^2$, then $\alpha=\beta$.

I need advice on how to approach this. I recognized that $\alpha,\beta$ are even permutations (because they have odd length). I'm not sure if this is relevant to the problem.

I'm also trying to figure out why this wouldn't work for even length cycles.

Any hints on how to begin would be appreciated. Thanks.

Answer 1

Suppose that $\alpha$ and $\beta$ both have length $2k + 1$ for some $k \in \mathbb N$. Then observe that: \begin{align*} \alpha &= \alpha\varepsilon \\ &= \alpha^1\alpha^{2k+1} \\ &= \alpha^{2k+2} \\ &= (\alpha^2)^{k+1} \\ &= (\beta^2)^{k+1} \\ &= \beta^{2k+2} \\ &= \beta^1\beta^{2k+1} \\ &= \beta\varepsilon \\ &= \beta \end{align*}

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To see why this doesn't work for even length cycles, consider $\alpha = (1,2)$ and $\beta = (1,3)$.