Determine the orders of elements in a permutation group. I was working on a problem that is about finding all possible orders of elements in $S_7$ and $A_7$. At first, I thought $S_7$ should take all elements from order $1$ to order $12$, since the maximum order of element formed by disjoint cycles is $lcm(3,4)=12$ and the least order of element it can form is the single cycle $(1)$. And since $A_7$ which takes all even permutation of $S_7$ is a subgroup of $S_7$, so $A_7$ should take all elements of odd orders, such as $1$, $3$, $5$, $\dots$,$11$. However, I am not sure if I am correct or not.
As a matter of fact, I also find out the elements formed by all the transpositions which share a common number has a higher order than the element formed by the disjoint cycle in the case when for example $|(12)(32)|>|(23)(14)|$. So I wonder how I can also include the elements formed by the joint cycles in the consideration toward the answer(Ps: not just the case I mention for the transposition, but also like in some general case such as $(134)(235)$)and conduct it properly?
And I want to know the rigorous proof towards this problem of finding orders of elements for the permutation group and also if it is possible tell me some general method that I can use for finding orders not just in the case of $S_7$ and $A_7$, but also in all the other cases. Please don't make it too advance because I am just a beginner in studying abstract algebra.
 A: You already know the crucial pieces of information: the order of a cycle is its length, and the order of two group elements that commute is the least common multiple of their orders. So all you need to do is write down the possible cycle patterns for permutations in $S_7$.
Your guess that all orders up to $12$ would occur can't be right since the order of any element must divide the order of the group, and $11$ does not divide $7!$.
More help, in response to the OP's comment.
The possible patterns when you write an element of $S_7$ as a product of disjoint cycles (which you can always do) are
(xxxxxxx)
(xxxxxx)(x)
(xxxxx)(xx)
(xxxxx)(x)(x)
...
(xx)(x)(x)(x)(x)(x)
(x)(x)(x)(x)(x)(x)(x)

The last entry is the identity permutation.
You should be able to complete this list and find the orders of the elements of each type.
$S_7$ has $7 \times 6 \times 5 \times 4\times 3 \times 2$ elements. That number is not a multiple of the prime $11$. See Lagrange's theorem, in wikipedia:

A consequence of the theorem is that the order of any element $a$ of a
finite group (i.e. the smallest positive integer number $k$ with $a^k = e$,
where $e$ is the identity element of the group) divides the order of
that group, since the order of $a$ is equal to the order of the cyclic
subgroup generated by $a$.

A: Your solution is mostly correct. However, you forgot to consider the fact that the order of any element of $S_7$ must divide $7!$.
First of all, note that $|S_7| = 7!$. Therefore, it follows from Lagrange's Theorem that $| \langle s \rangle |$ (the order of the cyclic group generated by $s$) divides $7!$ (the order of $S_7$) for all $s \in S_7$. But $| \langle s \rangle |$ is equal to the order of $s$. As such, the order of $s$ divides $7!$, for any $s \in S_7$.
