cosets question

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?

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