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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

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@Olod: What did you try. $s_{1}H=s_{H}$ implies $s_{1}s_{2}^{-1} \in H$ –  anonymous Aug 12 '10 at 16:30
    
What do you mean by "set of representatives"? The usual meaning I know for that phrase is in the context of taking a quotient, and you don't have a quotient here. –  Noah Snyder Aug 12 '10 at 16:31
    
What do you mean by complete set of representatives S? If you mean coset representatives, then sH=tH for s,t in S implies s=t, so your "moreover" condition is redundant. –  Jack Schmidt Aug 12 '10 at 16:33
    
Sorry for mistakes and uncertaities, editing. –  Olod Aug 12 '10 at 16:38
    
Noah, coset reps don't need normal subgroups. In any transitive group action, thought of as a left multiplication action on a coset space, one may pick coset representatives for a computation. I think you've seen that before. (Explicit example: transfer homomorphism in group theory.) –  KCd Aug 12 '10 at 19:02
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|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.

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If H is the Klein 4-group and K is <(1234)> in $S_4$, then they both have the coset reps. { (1), (123), (132), (12), (13), (23)}. –  user641 Aug 12 '10 at 17:11
    
Thank you. I also thought about two subgroups H,K of index two in G; indeed, in this situation any s∈G∖in (H∪K) gives S={1,s}, which is a complete set of representatives for both H and K. But in what way two subgroups of index could be related in general? Still, what if some more (natural, or none too natural) conditions are to be added to SH=SK=G?.. –  Olod Aug 12 '10 at 17:15
    
I think two subgroups of the same index don't need to be related at all to each other. I don't know of any extra conditions to put on H and K that would use the common set of coset representatives. It seems that some groups G have lots of H,K, and some groups have none. It might be interesting to investigate why a group falls into either category. –  Jack Schmidt Aug 12 '10 at 17:24
    
I sincerely hope that the group I am working with is in the second category. :) "My" H is kind of "stabilizer" of some "small" set. –  Olod Aug 12 '10 at 17:48
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If $S$ is a complete set of (left) coset representatives for $H$, then for every $x\in G$ there exists $s\in S$ such that $xH=sH$, and moreover, if $s_1,s_2\in S$ are such that $s_1H=s_2H$, then $s_1=s_2$. That is: there is one and exactly one representative from each coset of $H$ in $G$ in the set $S$. As such, your condition is trivially satisfied under the assumption that $S$ is a complete set of coset representatives for both $H$ and $K$, since for $s_1,s_2\in S$, you have $$s_1H=s_2H \Longleftrightarrow s_1=s_2\Longleftrightarrow s_1K=s_2K.$$

So either you meant something else, or you are just asking for condition under which two subgroups $H$ and $K$ can have the same complete set of coset representatives.

Jack Schmidt already gave an example with $H$ and $K$ not conjugate, and asked if there is an example in which $H$ and $K$ are not isomorphic. I think this does it: take $G=\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_4$, the product of two cyclic groups of order two and one of order four. Take $H={0}\times{0}\times\mathbb{Z}_4$, and $K={0}\times\mathbb{Z}_2\times\langle 2\rangle$ (so $H$ is cyclic of order $4$, and $K$ is the Klein $4$-group). Let $S={(0,0,1), (1,0,1), (1,1,0), (0,1,0)}$. If I did not make some silly mistake, then this is a complete set of coset representatives for both $H$ and $K$.

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Hmm... I was writing this as you were editing. But your condition new edit still misses the requirement that no two distinct elements of $S$ represent the same coset. –  Arturo Magidin Aug 12 '10 at 16:42
    
No, since in a complete set of l.c. representatives any two distict elements are, by definition, in distinct left cosets. Jack Schmidt pointed that out above as well. –  Olod Aug 12 '10 at 16:49
    
Arturo means that in your "that is" you only said union, not disjoint union. I think I get what you are asking, and I think you are right to suspect there is not too strong of a relationship between H and K. –  Jack Schmidt Aug 12 '10 at 16:52
    
Your example looks very good to me. It's nice to see that abelian examples suffice to show there is not much relation. –  Jack Schmidt Aug 12 '10 at 17:15
    
@Arturo Magidin: thanks. –  Olod Aug 12 '10 at 17:33
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