Does a finite group has subgroups of order of all divisors of its order?

Question

If G is a group of order d, then does there exist a subgroup of order c for all divisors c of d? I cannot find a proper countetexample. Could anyone provide me with one?

Answer 1

The alternating group $A_4$ has no subgroup of order $6.$

More examples are the alternating groups $A_n$ for $n \ge 5$ which are known to be simple. So they can have no subgroup of half their order, since a subgroup of index $2$ is normal.

Answer 2

Such group is called CLT (Converse of Lagrange Theorem) group. One can prove that any nilpotent group is a CLT group. Also, for any $n\geq 5$, $S_{n}$ is not a CLT group. (For the proof, see here.)

Answer 3

In fact, there is a result due to Bray that all supersolvable finite groups are CLT, and all CLT groups are solvable. See https://msp.org/pjm/1968/27-2/pjm-v27-n2-p02-p.pdf .

The inclusions are strict since $A_4$ is solvable but not CLT and $S_4$ is CLT but not supersolvable.

(Note that a "supersolvable" group is a group with a normal series with cyclic factors such that every group in the series is normal in the whole group. For a finite group, this means its chief series has cyclic factor groups of prime order, the chief series being a normal series of normal subgroups that cannot be refined. In general, solvable finite groups have chief series with elementary abelian factors, but they need not be cyclic.)