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.

-
but if it has uniqe such subgroup for each $d$ , you can say it is cyclic. – mesel Aug 1 '14 at 21:16
@mesel: one can also strengthen this to: if it has a unique normal subgroup for each $d$ - see here – zcn Aug 2 '14 at 1:22
The various answers, taken together, actually show that every non-abelian group of order $<12$ has subgroups of all orders that divide the group order. So if you had started looking for a counterexample, you would very probably have found one. The smallest non-abelian group that doesn't have subgroups of all orders that divide the group order is the alternating group on $4$ objects. It has order $12$ and has no subgroup of order $6$. – Andreas Blass Aug 2 '14 at 1:25

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.

-
Oh, thanks for the short answer, I am having touble with why there is a unique nonabelian group of order $pq$ – Carry on Smiling Aug 1 '14 at 20:54
It's constructed using some sort of semidirect product, IIRC. It's given in Dummit and Foote, I know. – user61527 Aug 1 '14 at 20:55
well, I guess the fact that it exists is enough, why is it unique? – Carry on Smiling Aug 1 '14 at 20:58
I don't recall off-hand, but it's in D&F. The construction is also given in Hungerford. – user61527 Aug 1 '14 at 21:00

Consider the quaternion group.

-

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.

-

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).

-

Dihedral groups, or semidirect product groups of order $p^3$

-