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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 $m$?

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  • $\begingroup$ I take it that this means the counterexample must be a non-abelian group, right? $\endgroup$ – Libertron Jan 17 '15 at 22:42
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    $\begingroup$ Yes, Libertron. $\endgroup$ – Namaste Jan 17 '15 at 22:45
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    $\begingroup$ @Libertron At the beginning you say it's easy to see this is true for abelian groups, but at the end you ask if $G$ is necessarily abelian? I don't understand it at all. $\endgroup$ – user2345215 Jan 17 '15 at 22:47
  • $\begingroup$ I just realized the logical error there. I meant to say a finite non-abelian example. $\endgroup$ – Libertron Jan 17 '15 at 22:49
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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.

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    $\begingroup$ So a concrete example is that the order $60$ group $A_5$ does not have a subgroup of order $30$. $\endgroup$ – Hagen von Eitzen Jan 17 '15 at 22:43
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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.

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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.

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