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I was reading this answer The size of a conjugacy class in the symmetric group but I didn't understand point 1). Why does this contribute a factor of $2!\times2!$? After all, the 4 cycles can be moved around freely: the cycle type $(3,3,2,2)$ is an unordered list. Shouldn't the factor be $4!$?

QuartelQuartz
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  • The conjugate of a $3$-cycle is a $3$-cycle, so no, there are only $2!$ choices for which $3$-cycle; and similarly for the $2$-cycles. – ancient mathematician Aug 13 '21 at 09:34
  • @ancientmathematician I don't understand. If I call the cycles $(1,2,3)(4,5,6)(7,8)(9,10)$ $s_1s_2s_3s_4$ then we have $4!$ ways to arrange these cycles. In each arrangement we still have $2$ $2$-cycles and $2$ $3$-cycles. Each arrangement corresponds to a distinct permutation of the numbers $1$ through $10$ but the same element of $S_{10}$ up to conjugation. – QuartelQuartz Aug 13 '21 at 09:48
  • You're not dealing correctly with this. Apply your argument to the case of elements $(ab)(c)$ in $S_3$. You will want to get $3!$ divided by $2$ (because $(ab)=(ba)$ and also by $2!$ because there are two cycles: that doesn't work, does it? – ancient mathematician Aug 13 '21 at 14:27

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Conjugate permutations have the same shape, and permutations of the same shape are conjugate. This is a trivial consequence of the fact that if $\sigma$ sends letter $a$ to letter $b$ then $\tau\sigma\tau^{-1}$ sends letter $\tau(a)$ to letter $\tau(b)$.

Hence the size of a conjugacy class is the number of permutations of that shape.

For example: How many different permutations $(abc)(def)(gh)(ij)$ are there?

Well there are $10!$ ways to list $a, \dots, j$ in order; but this overcounts.

(i) We can write $(abc)=(bca)=(cab)$ and so we must divide by $3$. The same is true for $(def)$, and also similarly for $(gh)$ and $(ij)$; so in fact we need to divide by $3^2 \cdot 2^2$.

(ii) Also we always have $(abc)(def)=(def)(abc)$ (these are disjoint cycles of course), so we also need to divide by $2!$. The same is true for $(gh)(ij)$ so again we need to divide by $2!$.

Putting it all together there are $\dfrac{10!}{3^2\cdot 2!\cdot 2^2\cdot 2!}$ such permutations.

ancient mathematician
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