Ok, this is rather a question than a problem. So, if we have $\sigma\:\in S_n$ a permutation, why there is a natural number, say $p$, such that $\sigma^p=e$, where $e$ is the identical permutation?
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1It would improve your posts of Questions to work out an example or two, so that you can provide Readers with a better indication of where you got with understanding the underlying mechanism, e.g. in this case taking a power of a permutation of $n$ things. – hardmath Oct 05 '19 at 16:35
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Is $n$ finite? (For instance, $n$ could be an infinite cardinal. There are permutation groups for such sets.) – Eric Towers Oct 06 '19 at 03:19
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The set $$\{\sigma,\sigma^2, \sigma^3,...\}$$ is finite so at some points we have $\sigma ^m=\sigma^n$
That is $$\sigma ^{m-n}=e$$
Mohammad Riazi-Kermani
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In any finite group every element has finite order by Lagrange. Hence for every $\sigma\in S_n$ there exists a $p$ such that $\sigma ^p=id$. However, the largest possible order of an element in $S_n$ is much smaller than the group order, see here:
Dietrich Burde
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Lagrange is at the very beginning of group theory, and you don't have to know group theory for it. – Dietrich Burde Oct 05 '19 at 16:53
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One more methodical idea is the following:
Let $\sigma = (a_1a_2\dots a_k)$. Consider $$\sigma^k = (a_1a_2\dots a_k)(a_1a_2\dots a_k)\cdots (a_1a_2\dots a_k).$$ Then \begin{align} a_1 \to a_2 \to &\cdots \to a_k \to a_1,\\ a_2 \to a_3 \to &\cdots \to a_1 \to a_2,\\ &\;\; \vdots\\ a_k \to a_1 \to &\cdots \to a_1 \to a_k, \end{align} sending every $a_i$ to itself. Think of this as sort of chaining together $a_i \to a_{i+1}, a_{i+1} \to a_{i+2},\dots$.
An explicit example: If we have $(123)^3 = (123)(123)(123)$, then \begin{align} 1 \to 2 \to 3 \to 1\\ 2 \to 3 \to 1 \to 2\\ 3 \to 1 \to 2 \to 3 \end{align}
Hendrix
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