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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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Think about possible cycle strucures for an element of order 10. – WimC Jun 5 '12 at 10:37
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. – rschwieb Jun 5 '12 at 11:50

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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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Ah ok, so for example if we're looking for the number of elements of order $12$ then these must have cycle type $3^14^1$ so we have $7! / 12 = 420$ elements? – user26069 Jun 5 '12 at 11:08
Yes, that's correct. The only problem is when the specific cycle types have more than one component of the same length, e.g. $(ab)(cd)$. In this case you have to divide by an extra $2$ to account for reordering of the cycles. – fretty Jun 5 '12 at 11:50
Ah ok, and also (sorry one more example!) for the elements of order say $15$ this has cycle type $3^15^1$ of which there are none. So am I write in saying that there are no elements of order 15? – user26069 Jun 5 '12 at 12:03
Yes, clearly you couldn't have 7 numbers appearing in 8 slots without repetition. This shows the converse of Lagrange's theorem not working. The number $15$ does divide $7!$ however, there is no such element of order $15$. – fretty Jun 5 '12 at 13:00
Also there is not always just one cycle type matching what you need. For elements of order $6$ you would have to consider both $6$-cycles and $3,2$-cycles. – fretty Jun 5 '12 at 13:02
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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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What do you mean by no 10 cycles? – user26069 Jun 5 '12 at 10:46
The cycle types of $S_7$ are: $1, 2, 2^2, 2^3, 3, 3^2, 2\, 3, 2^2\, 3, 4, 2\, 4, 3\, 4, 5, 2\, 5, 6, 7$ – user26069 Jun 5 '12 at 10:47
So do you mean there isn't any '10' cycle types in there? – user26069 Jun 5 '12 at 10:48
I refer to the cycle as in 'cycle notation.' – mixedmath♦ Jun 5 '12 at 11:09
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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. – rschwieb Jun 5 '12 at 11:47

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