Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
2  
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

2 Answers 2

up vote 3 down vote accepted

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 $S_7$.

share|improve this answer
    
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

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.

share|improve this answer
    
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
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.