May 28th, 2017, 02:25 AM  #1 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2  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! Thanks 
May 28th, 2017, 08:06 AM  #2 
Member Joined: Jan 2016 From: Athens, OH Posts: 88 Thanks: 47 
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 $${n\cdot (n1)\cdot\cdots\cdot (nm+1)\over m}$$ 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 09:00 AM. 
May 28th, 2017, 08:55 PM  #3  
Senior Member Joined: Nov 2015 From: hyderabad Posts: 232 Thanks: 2  Quote:
Please let me know if so. Thank you  

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