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How many elements of order 6 are in $S_5$?

Here is my solution: possible cycle structures are $[2,3]$. Applying class equation gets $5! \over 2!3! $ = $10$. Is my solution correct?

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How did you get that out of the class equation? IOW: No. – Jyrki Lahtonen Feb 17 '12 at 4:41
up vote 6 down vote accepted

No, your solution is incorrect: you are undercounting.

You are correct that an element of order $6$ in $S_5$ must be the product of a $2$-cycle and a disjoint $3$-cycle. $$\binom{5}{2} = \frac{5!}{2!3!}$$ counts the number of ways of selecting $2$ elements from $5$; in essence, you are counting the number of $2$-cycles; of course, the $3$-cycle will then consist of the three remaining elements... but there are two possibilities, not one.

For example, if we select $1$ and $2$ for the $2$-cycle, you get the elements $(1\;2)(3\;4\;5)$ and $(1\;2)(3\;5\;4)$. These are distinct elements.

The order of the two elements chosen for the $2$-cycle does not matter, since $(a\;b)=(b\;a)$; but the order of the elements of the $3$-cycle does matter: $$(a\;b\;c) = (b\;c\;a) = (c\;a\;b)\neq (a\;c\;b)=(b\;a\;c) = (c\;b\;a).$$

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oops, I $5! \over 6$ I think. – Mark Feb 17 '12 at 4:58
In other words, two for every $2$-cycle. Yes. – Arturo Magidin Feb 17 '12 at 4:59

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