Find all numbers $n$ such that $S_7$ contains an element of order $n.$
Identity is the only element of order $1.$So $n=1$ is possible.
Case 1: Elements that can be written as a unique cycle of length$≥2$:Cycles of length 2 to 7 can exists whence the corresponding values of n are 2 to 7.
Case 2: Elements that can be written as a product of two disjoint cycles of length$≥2$:Here the possible elements are $σ_1 σ_2$ where $σ_1$ and $σ_2$ are disjoint cycles of length$≥2$,such that $4≤|σ_1 |+|σ_2 |≤7$. So in this case the possible values of n are {lcm(|σ_1 |,|σ_2 | ): σ_1 & σ_2 are disjoint cycles of length≥2 with 4 ≤|σ_1 |+|σ_2 |≤7}={lcm(n_1,n_2 ): n_1,n_2∈Z^+-{1} with 4≤n_1+n_2≤7} (Since there always exist disjoint cycles of length n_1 and n_2 such that n_1,n_2∈Z^+-{1} with 4≤n_1+n_2≤7)={2,6,4,10,3,12}.
Case 3: Elements that can be written as a product of three disjoint cycles of length≥2:Here the possible elements are σ_1 σ_2 σ_3 where σ_1 and σ_2 are disjoint cycles of length≥2,such that 6≤|σ_1 |+|σ_2 |+|σ_3 |≤7. So in this case the possible values of n are {lcm(|σ_1 |,|σ_2 |,|σ_3 | ): σ_1,σ_2,σ_3 are disjoint cycles of length≥2 with 6≤|σ_1 |+|σ_2 |+|σ_3 |≤7}={lcm(n_1,n_2,n_3 ): n_1,n_2,n_3∈Z^+-{1} with 4≤n_1+n_2+n_3≤7} (Since there always exist disjoint cycles of length n_1,n_2,n_3 with n_1,n_2,n_3∈Z^+-{1} with 4≤n_1+n_2+n_3≤7)={2,6}. We further note that there^' s no element in S_5 whose representation as a product of disjoint cycles contains 4 or more cycles of length≥2.Consequently the above three cases exhaust all the posibilites for the order of the elements of S_5.Thus n=1,2,3,4,5,6,7,10,12.
Am I correct?
I, by using a computational software in Group Theory called GAP, could checked that: