If two subgroups have a complete set of left coset representatives in common, then  Let $H,K$ be proper subgroups of a group $G$ having a complete set $S$ of representatives of left cosets in common, that is,
$$
G = \bigsqcup_{s \in S} sH = \bigsqcup_{s \in S} sK
$$
It seems in general one cannot expect any serious relation on $H,K.$ But I am afraid of overlooking some general result here. Any information on the subject will be warmly accepted. Regards, Olod

|S| = [G:H] = [G:K], so at least their indexes are equal. I don't think much else is true, because of the following example:
If G = 2 × 2 is the Klein four group, then H = 2 × 1 and K = 1 × 2 are two subgroups with a common set of coset representatives: S = { (0,0), (1,1) }.  H and K are not conjugate.
In general if G is a semi-direct product H⋉N, then H has S=N as a set of coset representatives.  It is very possible for N to have more than one "complement" H, that is, another subgroup K such that G=K⋉N.  In nice situations like G nonabelian of order 6, all complements are conjugate, but in general they need not be as the dihedral 2-groups (including the Klein four group) show.
It would be nice to have an example where H and K are not even isomorphic, but semi-direct products won't do that.  I don't believe H, K need to be complemented to have a common transversal, but I guess that is another reasonable guess to rule out.
Edit: Well, the dihedral group G of order 8 has a cyclic normal subgroup H and Klein four normal subgroup K, the union of which is not all of G.  Since a set of coset representatives S has only 2 elements, we just need to take the identity and an element neither in H nor K.  In particular, H need not be isomorphic to K.
Also the dihedral group of order 16 has a similar pair (H cyclic order 4, K a four-group), and so neither H nor K need be complemented.
One can also use Abelian examples as Arturo points out, and one can extend the D8 example to S4 as Steve points out.  A fun problem.
