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This question might be duplicate because of a representation theory question. I don't know representation theory enough so I didn't tried to check that section. Please notify.

I heard and experienced that if $s_1,s_2 \in Sym(n)$, then there exists a $k \in Sym(n)$ such that $k^{-1}s_1k = s_2 \iff s_1$ and $s_2$ has same cycle type.

I tried to proof this statement but in the middle of it someone told me it has really short proof so I started to trying to find it but I couldn't. I hope you can help me and find that short proof.

Thanks for any help.

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Do you know how easily to calculate the conjugate of a cycle? – Tobias Kildetoft Aug 9 '13 at 21:42
@TobiasKildetoft: With transpositions, right? – Konformist Liberal Aug 9 '13 at 21:43
No, I mean do you know what you get if you compute $g(1 2 3 4 5)g^{-1}$ for example (in terms of $g$). – Tobias Kildetoft Aug 9 '13 at 21:45
@TobiasKildetoft: No. – Konformist Liberal Aug 9 '13 at 21:46
@KonformistLiberal try and solve Tobias' question. Once you have a solution, you can solve the conjugacy question on your own! – Olivier Bégassat Aug 9 '13 at 21:47
up vote 1 down vote accepted

As is discussed in the comments, the following fact should be very helpful:

If $\sigma,\tau\in S_n$, with $\tau=(a_1\ldots a_m)$ an $m$-cycle, then we have the formula:


For the proof, see what $\sigma\circ\tau\circ\sigma^{-1}$ does to symbols of the form $\sigma (a_i)$, and also symbols of the form $\sigma (b)$ where $b\ne a_i$.

Next, use the above formula on the cycle decomposition of a generic element $g=\tau_1\circ\ldots\circ\tau_k$ to see that conjugation by $\sigma$ preserves the cycle structure.

The other direction of the proof will also make use of the conjugation formula.

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