If a group contains a subgroup with the order of each of its divisors, is it abelian? Let $G$ be a group  that has a subgroup of  size $d$ for every $d$ that divides $|G|$. Must $G$ be abelian?
It can be shown using complete induction that the converse of the above statement is true, but my alg-fu is too underdeveloped to tackle the above problem.
 A: Consider the quaternion group.
A: No, take the quaternion group $Q_8 = \{ \pm 1, \pm i, \pm j, \pm k \}$ for example. It has a subgroup of order 2, namely $\{ \pm 1 \}$, it has a subgroup of order 4 e.g. $\{ \pm i \}$ and of course has a subgroup of order 1 and of order 8.
A: The question of Banarama is an interesting one and refers to the class of finite groups known as CLT groups, where CLT stands for Converse Lagrange Theorem:
$G$ is a CLT group if for each positive integer $d$ dividing $|G|$, $G$ has at least one subgroup of order $d$.
These groups have been studied extensively. It turns out for example that all supersolvable groups are CLT, and all CLT groups are solvable. See also one of the early papers of Henry G. Bray, Pac. J. Math 27 (1968).
A: No. Choose any nonabelian group of order $pq$ for $p < q$ distinct primes - such a group exists (and is unique) if $p | q - 1$ (e.g. $21$). The divisors of the group order are than $1, p, q$ and $pq$; by Cauchy's theorem, there are subgroups of order $p$ and $q$, since there are elements of these orders.
Alternatively, to be explicit, $S_3$ is a group satisfying this.
A: Dihedral groups, or semidirect product groups of order $p^3$
