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In Dummit and Foote on p.126, they use the following assertion to complete a proof of a proposition:

Since there is a bijection between the conjugacy classes of $S_n$ and the permissible cycle types and each cycle type for a permutation in $S_n$ is a partition of $n$, the number of conjugacy classes of $S_n$ equals the number of partitions of $n$.

I have emphasized the part I am not seeing. How can I easily see that the above bijection indeed exists?

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Given any cycle $a =(a_1a_2\cdots a_k)$ in $S_n$ and any element $\sigma$ in $S_n$, conjugating $a$ by $\sigma$ gives $(\sigma(a_1)\sigma(a_2)\cdots\sigma(a_k))$. Now, since $S_n$ acts $n$-transitively on the $n$ elements $a_1,\dots,a_n$ we can pick $\sigma$ such that we get any other cycle of the same length as $a$ by conjugating by $\sigma$. Since any element can be written uniquely as a product of disjoint cycles and those cycles commute, we now get that any two cycles of the same type are conjugate. On the other hand, conjugation preserves the cycle type by the same argument, so this establishes the desired bijection.

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Now I see. I had understood that part earlier, but failed to see that my emphasized part is just a restatement. Thanks, –  Mike Jan 9 '13 at 19:12

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