TITLE: When does every group with order divisible by $n$ have a subgroup of order $n$?
QUESTION [28 upvotes]: 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.

REPLY [4 votes]: 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.