algebra permutations 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? 
 A: The set $$\{\sigma,\sigma^2, \sigma^3,...\}$$ is finite so at some points we have $\sigma ^m=\sigma^n$
That is $$\sigma ^{m-n}=e$$
A: 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:
Element of Largest Order in $S_n$
A: 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}
