The number of double cosets divides the order of the group

group-theory

I am working on the following question, but am not 100% convinced my answer to the third part is correct.
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I have found counter examples to (b)(i) and (b)(ii)

Then for (b)(iii)

Let X denote the set of double cosets of H,K.

Then we define a group action on X by $f(HgK) \rightarrow HfgK$

By the orbit-stabilizer theorem, we then get:

$|orb(HeK)||stab(HK)| = |G|$, where e is the identity element

As |orb(HeK)| = X, we get |x||stab(HK)| which concludes the proof.

Secondly, would the above proof (if it is correct), work for if H and K were not subgroups? It seems to me it would still work if H,K were just subsets of G, which is what makes me question whether the proof is valid.

Best Answer

Your action is not well-defined in general.

Let $H$ be a subgroup of order 2 in the dihedral group of order 10. Then there are three double cosets $HgH$, and 3 does not divide 10.

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