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Let $G_1, G_2 $ and $G_3$ be subgroups of $G$ with finite indexes. Suppose that there are $x_1,x_2,x_3\in G$ such that

  • $G_1x_1\cap G_2x_2=G_2x_2\cap G_3x_3=G_1x_1\cap G_3x_3=\emptyset$,
  • $G_1G_2x_2\cup G_1G_3x_3=G_2x_2\cup G_3x_3$,
  • $G_2G_1x_1\cup G_2G_3x_3=G_1x_1\cup G_3x_3$, and
  • $G_3G_1x_1\cup G_3G_2x_2=G_1x_1\cup G_2x_2$.

Must it be that at least two of $|G:G_1|$, $|G:G_2|$, $|G:G_3|$ are equal?

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Where $x_1$, $x_2$, and $x_3$ are what? Arbitrary elements of $G$? –  Zev Chonoles Dec 21 '10 at 2:03
Also, it would help to have some of the motivation for this question. Where did it come up? –  Zev Chonoles Dec 21 '10 at 2:04
yes, $x_i$ is arbitrary. –  Rulin Dec 21 '10 at 2:14
Rulin, please let me know if I edited incorrectly. –  Andres Caicedo Dec 21 '10 at 2:44
very good, correctly. –  Rulin Dec 21 '10 at 3:16

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