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September 21st, 2014, 07:13 AM  #1 
Newbie Joined: Aug 2012 Posts: 22 Thanks: 0  group of permutations
Hello! Please answer this question: $\displaystyle \text{How to prove that} \\ \forall \sigma \in S_n,\exists k \in \mathbb{N},k\neq 0 \text{ so that }\sigma^k=e, \\ \text{where } S_n \text{ is the group of permutations.}$ I need a proof that does not concern group theory and finite groups theory. 
September 21st, 2014, 12:56 PM  #2  
Global Moderator Joined: May 2007 Posts: 6,558 Thanks: 602  Quote:
$\displaystyle \sigma^i = \sigma^j, j>i$ Therefore $\displaystyle \sigma^k=e$ where k=ji.  
September 23rd, 2014, 10:36 PM  #3 
Senior Member Joined: Mar 2012 Posts: 294 Thanks: 88 
Write $\sigma$ as a product of disjoint cycles. Suppose that these cycles are of lengths: $k_1,k_2,\dots k_r$. Let $m = \text{lcm}(k_1,k_2,\dots,k_r)$. Then $\sigma^m$ fixes every element of $\{1,2,\dots,n\}$, so is the identity permutation, QED. 

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