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I tried this example in $S_3$ and saw that this held but I couldn't figure out how to prove it for all $S_n$. Any help would be appreciated.

Martin Sleziak
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user139985
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1 Answers1

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First, you should show that if $\sigma$ is a cycle, then $\alpha\sigma\alpha^{-1}$ is also a cycle of the same length. In fact, you can actually write out what this permutation has to be.

Take an arbitrary permutation $\sigma = \sigma_1 \sigma_2 \dots \sigma_n$, written in disjoint cycle notation. Then $\alpha\sigma\alpha^{-1} = \alpha \sigma_1 \sigma_2 \dots \sigma_n \alpha^{-1}$. Can you find a way to rewrite this that takes advantage of the previous fact?

Thad Janisse
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  • I don't understand how to show that they have to be the same length though – user139985 Oct 12 '15 at 05:19
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    Well, where does $\alpha \sigma \alpha^{-1}$ send $i$? Well, we can write $i = \alpha (j)$ for some $j$ by bijectivity. Then $\alpha \sigma \alpha^{-1}(i) = \alpha \sigma(j)$. So if $\sigma = (a_1 a_2 \dots a_n)$, what does conjugacy do to it? – Thad Janisse Oct 12 '15 at 05:22