Let $\alpha$ be a $2$-cycle and $\beta$ be a $t$-cycle in $S_n$. Prove that $\alpha\beta\alpha$ is a $t$-cycle $\bf Claim:$ Let $\alpha$ be a $2$-cycle and $\beta$ be a $t$-cycle in $S_n$.
Prove that $\alpha\beta\alpha$ is a $t$-cycle.
If $\alpha$ and $\beta$ are disjoint, they commute and thus the product $\alpha\beta\alpha$ is equal to $\beta$, a $t$-cycle, but how would the proof proceed if $\alpha$ and $\beta$ are not disjoint?
 A: Here is a different method to the one proposed in the comments.
Edit: As Dan Shved points out in his comment, the argument below shows that IF $\alpha\beta\alpha$ is a cycle, then it is a $t$-cycle; that is, it doesn't show why $\alpha\beta\alpha$ must be a cycle.

A $t$-cycle is a cycle $\beta$ such that the smallest $k > 0$ with $\beta^k = e$ is $k = t$, where $e$ the identity.
As $\alpha$ is a $2$-cycle, we have $$(\alpha\beta\alpha)^k = (\alpha\beta\alpha)(\alpha\beta\alpha)\dots(\alpha\beta\alpha) = \alpha\beta(\alpha\alpha)\beta(\alpha\alpha)\dots(\alpha\alpha)\beta\alpha = \alpha\beta\beta\dots\beta\alpha = \alpha\beta^k\alpha.$$
Now note that 
\begin{align*}
& \alpha\beta^k\alpha = e\\
\Leftrightarrow & \alpha\alpha\beta^k\alpha = \alpha\\
\Leftrightarrow & \beta^k\alpha = \alpha\\
\Leftrightarrow & \beta^k\alpha\alpha = \alpha\alpha\\
\Leftrightarrow & \beta^k = e.
\end{align*}
So the smallest value of $k > 0$ such that $(\alpha\beta\alpha)^k = e$ is also the smallest value of $k > 0$ such that $\beta^k = e$, which we know is $t$. Therefore $\alpha\beta\alpha$ is a $t$-cycle.
