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September 21st, 2014, 07:13 AM   #1
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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.
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September 21st, 2014, 12:56 PM   #2
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Quote:
Originally Posted by matemagia View Post
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.
Since the group is finite, after a sufficient number of steps:
$\displaystyle \sigma^i = \sigma^j, j>i$ Therefore $\displaystyle \sigma^k=e$ where k=j-i.
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September 23rd, 2014, 10:36 PM   #3
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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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