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In S6, let $\alpha=(135)(156)(135)$ how do I compute $\alpha^{24}$? I'm given the hint that I first have to express alpha as a product of disjoint cycles which I computed as (15)(36) and then I have to find $\alpha^2$.

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Hint: $\alpha^{24}=(\alpha^2)^{12}$

Alternative hint: Don't even rewrite $\alpha$ as product of disjoint cycle. Just note that it is a permutation of $\{1,3,5,6\}$ only and that $|S_4|=24$.

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HINT: What is the order of this permutation? The lowest common multiple of the lengths in its disjoint cycle notation.

Now what is the identity permutation multiplied by itself 12 times?

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