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?

  • 1
    $\begingroup$ Where $x_1$, $x_2$, and $x_3$ are what? Arbitrary elements of $G$? $\endgroup$ Dec 21 '10 at 2:03
  • 4
    $\begingroup$ Also, it would help to have some of the motivation for this question. Where did it come up? $\endgroup$ Dec 21 '10 at 2:04
  • $\begingroup$ yes, $x_i$ is arbitrary. $\endgroup$
    – Rulin
    Dec 21 '10 at 2:14
  • $\begingroup$ Rulin, please let me know if I edited incorrectly. $\endgroup$ Dec 21 '10 at 2:44
  • $\begingroup$ very good, correctly. $\endgroup$
    – Rulin
    Dec 21 '10 at 3:16

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