[Math] Example where a finite group $G$ of order $n$ has no subgroup of order $m$

Question

Using the Fundamental Theorem of Abelian Groups, one can prove that if $G$ is a finite abelian group of order $n$ such that $m$ is a positive integer that divides $n$, then $G$ contains a subgroup of order $m$. But what are some examples of non-abelian groups $G$ of order $n$ and $m$ a factor of $n$ such that $G$ has no subgroup of order $m$?

Answer 1

Possibly the smallest group where this occurs is the alternating group $A_{4},$ which has order $12$ and yet has no subgroup of order $6$. Note that this is a solvable group.