TITLE: Example where a finite group $G$ of order $n$ has no subgroup of order $m$
QUESTION [5 upvotes]: 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 $m$?

REPLY [3 votes]: Maybe good to know if you are interested: finite groups for which the property does hold (that is, for every divisor $d$ of the order $|G|$, $G$ has a subgroup of order $d$) have been studied extensively. They are called CLT (Converse Lagrange Theorem) groups. It is known that a CLT group must be solvable and that every supersolvable group is a CLT group: however solvable groups exist, which are not CLT and CLT groups which are not supersolvable. For example Geoff's example $A_4$ is solvable but not CLT.