Example where a finite group $G$ of order $n$ has no subgroup of order $m$ 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 $m$? 
 A: It is not possible for $G$ to be finite and abelian for this to be true. Many examples are given with $n=2m$ by simple groups: any subgroup of index 2 is normal, hence no simple group can have a subgroup of index 2. Furthermore, any nonabelian simple group is of even order, so any nonabelian finite simple group gives an example.
As Hagen von Eitzen points out, a concrete example is given by $A_5$, which is of order 60 and has no subgroup of order 30.
A: 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. 
A: 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.
