How many different subgroups of order $3$ are in alternating group $A_{4}$?
Using direct listing is doable, but primitive. I wonder how to apply combinatorial approach (intuitively it would be much better approach than direct listing)? Thanks for help in advance.
Every non-identity element of these subgroups will be of order $3$ per Lagrange's theorem. One can show that the elements of order $3$ in $A_4$ are precisely the $3$ cycles. Furthermore, each subgroup will be disjoint sans the identity. Thus, counting the total number of possible $3$-cycles will almost get you the total number of order $3$ subgroups (you'll have to do an easy correction for the over-counting).
Alternatively, if you're familiar with the Sylow theorems, they'll give you an answer pretty much immediately without hardly having to think since there're only two factors of $|A_4| = 12$ congruent to $1 \pmod{3}$ .