Q:

When does every group with order divisible by $n$ have a subgroup of order $n$?

According to Sylow's theorem, every finite group with order divisible by $p^k$ for some prime $p$ has a subgroup of order $p^k$. Is this the best possible result in this direction? That is, if $n$ is not a power of a prime, does there always exist a group with order divisible by $n$ that does not have a subgroup of order $n$?
EDIT: Just to clarify, I am aware that groups like this exist. The standard example seems to be $A_4$, which has order divisible by $6$ but no subgroup of order $6$. What I am looking for is a proof that a counterexample exists for any $n$ that is not a power of a prime.

A:

Here is a proof that the answer is yes. Suppose first that $n = p^aq^b$ with $p,q$ prime, $a,b>0$, and suppose that $p^a > q^b$. Let $c$ be minimal such that $p^a$ divides $q^c-1$ - so clearly $c > b$. Then a faithful irreducible module for the cyclic group of order $p^a$ over the field of order $q$ has dimension $c$. (You can define the action explicitly as multiplication by an element $x$  in the field of order $q^c$, where $x$ has multiplicative order $p^a$.)
Now let $G = Q \rtimes P$ be the semidirect product of an elementary abelian group $Q$ of order $q^c$ by a cyclic group $P$ of order $p^a$, using this module action. So $Q$ is the unique minimal normal subgroup of $G$. A subgroup of $G$ of order $p^aq^b$ would have a normal subgroup of order $q^b$ which would also be normal in $Q$ and hence normal in $G$, contradiction, so there is no such subgroup.
For the general case, let $n = p^aq^br$ where $r$ is coprime to $p$ and $q$. Then a direct product of $G$, as constructed above, with a cyclic group of order $r$ has no subgroup of order $n$.

A:

Good question! And, you might also look at research on the following: a CLT (Converse Lagrange Theorem) group is a finite group with the property that for every divisor of the order of the group, there is a subgroup of that order. 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. 

A:

I don't know if you are familiar with Hall's theorem which gives a further partial answer to your question.
A Hall-subgroup  $H$ in $G$ with regard to a set of primes $\Pi$ has the property that the index of $|G:H|$ is coprime to every element in $\Pi$.
Hall's theorem states that for solvable groups Hall-subgroups exist for every set of primes. Furthermore for a given set of primes two Hall-subgroups are conjugate.

