Permutation and order I know that the order of a group is the number of the elements, then if we have a permutation what does the order of permutation mean? The number of distinct elements?
 A: Order of a permutation $f$ is defined as the least positive integer $r$ for which $f^r=e$ where $e$ is the identity permutation.
A: The order of a permutation $\sigma \in S_n$ is the smallest positive integer $k$ such that $\sigma^k$ is the identity permutation in $S_n$. For example, the order of the permutation $(1,3,4)(2,5)(6) \in S_6$ is the least common multiple of 3 and 2 and hence is equal to 6.
A: The term “order” is used in group theory in two distinct ways:


*

*the order of a finite group is the number of its elements;

*the order of an element $g\in G$ ($G$ a group) is usually defined as the least positive integer $k$ such that $g^k=1$, if a positive integer with this property exists.
For permutations, it is no different than from general groups: a permutation on $\{1,2,\dots,n\}$ is an element of the symmetric group $S_n$.
However, there is a different way to look at the order of an element: the order of $g\in G$ is the number of elements in the subgroup
$$
\langle g\rangle=\{g^n:n\in\mathbb{Z}\}
$$
generated by $g$ (it can be infinite). If the subgroup $\langle g\rangle$ is finite, then the number of its elements is the same as the least positive integer $k$ such that $g^k=1$.
From this point of view, the two usages of the term “order” are not so different: the order of an element is the number of elements of the subgroup it generates.

I believe it's instructive to prove the statement above.
Consider the map $\varphi_g\colon\mathbb{Z}\to G$ defined by $\varphi_g(n)=g^n$. Since $\varphi_g(m+n)=g^{m+n}=g^mg^n=\varphi_g(m)\varphi(n)$, the map is a group homomorphism and its image is exactly $\langle g\rangle$.
By the homomorphism theorem, $\langle g\rangle$ is isomorphic to $\mathbb{Z}/\ker\varphi_g$ and, by known results about the group of integers, $\ker\varphi_g=k\mathbb{Z}$, for a unique $k\ge0$.
If $k=0$, then $\varphi_g$ is injective and $\langle g\rangle$ is infinite. Thus, if $\langle g\rangle$ is finite, we have $k>0$ and so
$$
\langle g\rangle\cong\mathbb{Z}/k\mathbb{Z}
$$
has $k$ elements. Moreover $k\in k\mathbb{Z}=\ker\varphi_g$, so $g^k=\varphi_g(k)=1$ and, for $0<r<k$, $r\notin k\mathbb{Z}=\ker\varphi_g$, so $g^r\ne1$. Hence $k$ is the least positive integer such that $g^k=1$.

Note that this characterization of the order provides a very simple proof of the fact that, if $|G|=n$ and $g\in G$, then $g^n=1$. Indeed, by Lagrange's theorem, if $k=|\langle g\rangle|$ (which is obviously finite), then $k$ divides $n$, so $n=kh$ for some integer $h$. Therefore $g^n=(g^k)^h=1^h=1$.
Choosing between the two possible definitions of order of an element is a question of taste: the one with the powers doesn't use machinery such as the homomorphism theorem and the structure of subgroups of $\mathbb{Z}$, so it can be seen as more “elementary”. On the other hand, the subgroup based definition does use those tools, which is a good way for showing their usefulness.
