There are useful theorems for the existence of subgroups, such as the Sylow-theorem and the Hall-theorem. But even if the desired order divides the group order, subgroups need not exist.
Given a group and an order dividing the group order. What is the easiest way to verify whether a subgroup with the desired order exists. In particular, how can we show that there is no such subgroup ? Are there any useful sufficient conditions for the non-existence of such a subgroup ?
Well, in the case of the smallest example $A_4$, not satisfying the converse to Lagrange (since it has no subgroup of order $6$) the reasoning appears to be pretty specialized. See this $A_4$ has no subgroup of order $6$?.
A $3$-cycle is used to get a contradiction
In addition to Sylow and Hall, we know by Cauchy's theorem that there's a subgroup of order $p$ for every prime dividing the order of the group. So we can look elsewhere.
Of course, if we could come up with a criterion that works on subgroups of subgroups of $S_n$, we'd be in business, by Cayley's theorem.
My guess is that one might need to play around with specific elements of $S_n$ from case to case, as in the example above.