In 'Buildings and Finite $BN$-Pairs', Jacques Tits gives the following statement which is left as an easy exercise.

Let $G_1,G_2,G_3$ be three subgroups of a group $G$. Then the following conditions are equivalent.

1. $G_2G_1 \cap G_3G_1 = (G_1 \cap G_3) G_1$

2. $(G_1 \cap G_2) \cap (G_1 \cap G_3) = (G_2G_3) \cap G_1$

3. If the three cosets $xG_1$, $yG_2$ and $zG_3$ have pairwise nonempty intersection, then $xG_1 \cap yG_2 \cap z G_3 \neq \emptyset$.

I know it is not usual to ask for the solution of an exercise from a book on MO. However after hours of trying hard with friends, I dedided to post it anyway. Indeed, the problem seems not easy at all to me! Therefore I think it is worth to ask it here.

Does anyone know a reference for a proof or know a solution?

As usual, thanks in advance.

In 'Buildings and Finite $BN$-Pairs', Jacques Tits gives the following statement which is left as an easy exercise.

Let $G_1,G_2,G_3$ be three subgroups of a group $G$. Then the following conditions are equivalent.

1. $G_2G_1 \cap G_3G_1 = (G_2 \cap G_3) G_1$

2. $(G_1 \cap G_2) \cdot (G_1 \cap G_3) = (G_2G_3) \cap G_1$

3. If the three cosets $xG_1$, $yG_2$ and $zG_3$ have pairwise nonempty intersection, then $xG_1 \cap yG_2 \cap z G_3 \neq \emptyset$.

I know it is not usual to ask for the solution of an exercise from a book on MO. However after hours of trying hard with friends, I dedided to post it anyway. Indeed, the problem seems not easy at all to me! Therefore I think it is worth to ask it here.

Does anyone know a reference for a proof or know a solution?

As usual, thanks in advance.