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Let $G$ be finite, and group action on $X\subseteq G$: $g\cdot x:=g^{-1}xg$. Let $G=S_n$, and $X=S_n.$ Show that $[x]_R$ consists of all elements of $S_n$ that are of the same cycle-type as $x$.

I know that a group action defines an equivalence relation $R$ on $X$ s.t. $\exists\,g\in G,\,g\cdot x=y.$ But I am not sure how should I approach this problem.

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You have the action of $S_n$ on itself by conjugation. The equivalence classes, or orbits, under an action by conjugation are called conjugacy classes. What you need to show is that two elements in $S_n$ are conjugate if and only if they share the same cycle type. You can find a proof of that on page 126 of Dummit and Foote under Proposition 11, for one.

Just to illustrate, the elements $\sigma = (1,2)(3,4)(5,6,7,8)$ and $(3,4)(5,6)(8,1,2,9)$ have the same cycle type and are thus conjugate in, say, $S_9$. To see that, consider the element $\tau = (1,3,5,7)(2,4,6,8)$, which is not unique in this regard. A straightforward calculation shows $$ \tau \sigma \tau^{-1} = (3,4)(5,6)(8,1,2,9) \enspace. $$

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