Exercise :
(A.1) Show that there exists an element of order $6$ in the group of permutations $S_5$.
(A.2) Show that there does not exist an element of order $8$ in the group of permutations $S_5$.
(A.3) Find all the possible orders of the elements of the group $S_5$.
Attempt :
We know that the group $S_5$ is the group of permutations of $5$ objects.
If I started in a reverse way and found every possible disjoint cycle product for $S_5$ I could answer all the $3$ questions. Take the permutations of $(12345)$. Then, starting off and writing them as a product of disjoint cycles to figure out every possible order / combination :
$$(1,2) \space \text{ of order } \space 2$$ $$(1,2,3) \space \text{ of order } \space 3$$ $$(1,2,3,4) \space \text{ of order } \space 4$$ $$(1,2)(3,4) \space \text{ of order } \space 2$$ $$(1,2,3,4,5) \space \text{ of order } \space 5$$ $$(1,2,3)(4,5) \space \text{of order } \space 6$$ $$\text{The identity permutation of order 1}$$
These are all the possible combinations of products of disjoint cycles for elements in $S_5$, from which we can see that there exists an element of order $6$ in $S_5$ and that there does not exist an element of order $8$. Also, this way (A.3) is instantly solved.
QUESTION : I guess though, that the exam exercise does not expect from a student to be solved that way. What would be another approach into avoiding the complete elaboration of disjoint cycles until (A.3) ? How would one show in another way the first two questions ?
If you play around with cycle decomposition you can show A2:
Say you have an element $x$ of order $8$. If you write the element as product of distinct cycles say $x=c_1c_2c_3c_4c_5$ (some being trivial cycles) you can show that $ord(x)=lcm(l_i)$ with $l_i$ being the length of each cycle. So for $x$ to have order $8$ you must have some cycles with order $8$ but this is non-sense since each cycle cannot be of order bigger than 5.
You can use the same reasoning for A3 too.
Comment: $S_n$ isn't the group of symmetries of the $n$-gon, $D_n$ is (except of course if $n=3$ where they are the same group).