Subgroups of order $9$ isomorphic to $C_3 \times C_3$ in $S_6$ In this question, people discussed that there cannot be cyclics subgroups of order 9 in $S_6$. 
How does one reason that the subgroups isomorphic to $C_3 \times C_3$ (the other candidates to be subgroups of order 9) are precisely the ones given by two disjoint permutations $\langle (123),(456) \rangle$
 A: If you already know a bit about basic group theory (in this case, roughly up the Sylow theorems), you first see that $H:=\langle(1\,2\,3),\,(4\,5\,6)\rangle\cong C_3\times C_3$ is a subgroup of order $9$ in $S_6$ and is thus a Sylow-3 subgroup. Therefore by Sylow's theorem, all subgroups of $S_6$ of order $9$ are conjugate to $H$ and are thus isomorphic to $C_3\times C_3$; moreover, since conjugation preserves cycle structure, all such subgroups are generated by two disjoint $3$-cycles.
A: I hope this answer will be helpful to you.
The group $C_3 \times C_3$ is generated by two distinct elements of order three which commute with one another.
Note that the order of a permutation is the least common multiple of the lengths of its cycles when the permutation is written in disjoint cycle notation.  
So the order of, for example, $(1 4 5)(2 6)(3 7 9)$ is 6.
Inside of $S_6$, the only permutations of order 3 that exist are the $3$-cycles and the products of two disjoint $3$-cycles.  
To find one example of a $C_3 \times C_3$ subgroup inside of $S_6$, you can take the subgroup generated by $(1 2 3)(4 5 6)$ and $(4 5 6)$.  This would make the same subgroup as the subgroup generated by $(1 2 3)$ and $(4 5 6)$.  To find another subgroup of $S_6$ which is isomorphic to $C_3 \times C_3$, you can take the subgroup generated by $(1 4 5)$ and $(2 3 6)$.  
All of these $C_3 \times C_3$ subgroups of $S_6$ are generated by two distinct permutations of order 3 that commute with one another.  Can you determine how many different $C_3 \times C_3$ subgroups there are inside of $S_6$?
