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I have a feeling you use Sylow's Theorems but I'm not sure where to start, any hints?

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    To comfort you about your idea: Using Sylow's Theorems would be the best bet if you were in a group other than a symmetric group, and looking for elements of prime power order.2012-06-05

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If you write a permutation in disjoint cycle notation:

$(\alpha_1 \alpha_2 ... \alpha_{n_1})(\beta_1 ... \beta_{n_2})...$

then the order of the permutation is the lowest common multiple of the $n_i$.

So it is clear that elements of order $10$ in $S_7$ must have cycle type $(a b)(c d e f g)$.

How many of these are there? Well there are $7$ choices for $a$, then for each choice there are $6$ choices for $b$ etc.

We get $7!$ choices for $a,b,c,d,e,f,g$. Divide by $2$ to account for counting $(a b)$ and $(b a)$ as the same. Divide by $5$ to do the same for the $5$-cycle $(c d e f g)$.

Thus there are $7!/10 = 504$ elements of order $10$ in $S_7$.

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    None of this is group theory anymore, it is basic combinatorics. The number of $6$-cycles is $7!/6$ and the number of $3,2$-cycles is $7!/(3\times 2)$. It just was a coincidence that these numbers are the same so that it was enough to just double $7!/6$.2012-06-05
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HINT

There are no 10-cycles. Any subgroup with an element containing, say, a 6-cycle will have an order divisible by 6. But there are plenty of 2 cycles and 5 cycles.

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    Naturally there wouldn't be a 10 cycle: there are only seven symbols and it is physically impossible to produce a cycle of length ten.2012-06-05