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May 28th, 2017, 03:25 AM   #1
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Symmetric Groups

I want to know how do we find the number of elements in the Symmetric Group $S_5$ whose order is 2?

Please help!

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May 28th, 2017, 09:06 AM   #2
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Pretty much, this is just a counting problem. The following also computes the number of elements of order m for any possible m in $S_5$.

You need to know a couple of facts.
1. Every element of $S_n$ is a product of disjoint cycles.
2. The order of a product of disjoint cycles is the least common multiple of the lengths of the cycles.

So an element g of $S_n$ has order 2 iff g is the product of disjoint cycles of length 2. Thus in $S_5$, g=(a,b) or g=(a,b)(c,d). Now it's just a counting problem. There are ${5 \choose 2}$=10 two cycles. For the count of the product of two cycles (a,b)(c,d), it is
$${{5 \choose 2}{3 \choose 2}\over2}=15$$
The 2 in the dedominator occurs since (a,b)(c,d)=(c,d)(a,b). Thus there are 25 elements of order 2.

In $S_n$ the number of m cycles is
$${n\cdot (n-1)\cdot\cdots\cdot (n-m+1)\over m}$$
The m occurs in the denominator since for examplle (1,2,3,4)=(2,3,4,1)=(3,4,1,2)=(4,1,2,3).

So in $S_5$:
20 elements of order 3
30 elements of order 4
24 elements of order 5
20 elements of order 6
1 element of order 1

Last edited by johng40; May 28th, 2017 at 10:00 AM.
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May 28th, 2017, 09:55 PM   #3
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Quote:
Originally Posted by johng40 View Post
Pretty much, this is just a counting problem. The following also computes the number of elements of order m for any possible m in $S_5$.

You need to know a couple of facts.
1. Every element of $S_n$ is a product of disjoint cycles.
2. The order of a product of disjoint cycles is the least common multiple of the lengths of the cycles.

So an element g of $S_n$ has order 2 iff g is the product of disjoint cycles of length 2. Thus in $S_5$, g=(a,b) or g=(a,b)(c,d). Now it's just a counting problem. There are ${5 \choose 2}$=10 two cycles. For the count of the product of two cycles (a,b)(c,d), it is
$${{5 \choose 2}{3 \choose 2}\over2}=15$$
The 2 in the dedominator occurs since (a,b)(c,d)=(c,d)(a,b). Thus there are 25 elements of order 2.

In $S_n$ the number of m cycles is
$${n\cdot (n-1)\cdot\cdots\cdot (n-m+1)\over m}$$
The m occurs in the denominator since for examplle (1,2,3,4)=(2,3,4,1)=(3,4,1,2)=(4,1,2,3).

So in $S_5$:
20 elements of order 3
30 elements of order 4
24 elements of order 5
20 elements of order 6
1 element of order 1
I understood upto calculating the number of cycles but after that the calculations does it have any specific formula for it ?

Please let me know if so.
Thank you
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