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.



Proof of an 'easy' exercise in a book of Tits
(1) implies (2):
Indeed, the left side of (2) is obviously contained in the right side. So pick an arbitrary $h=g\_2g\_3=g\_1 \in (G\_2G\_3)\cap G\_1$. Now, $g\_3=g\_2^{-1}g\_1$ is contained in $G\_2G\_1$ but also in $G\_3G\_1$ and hence by (1) in $(G\_2\cap G\_3)G\_1$. So there is $g\_{23}\in G\_2\cap G\_3$ and $\tilde{g}\_1\in G\_1$ such that $g\_3=g\_2^{-1}g\_1=g\_{23}\tilde{g}\_1$. Thus $g\_2g\_{23}=g\_1\tilde{g}\_1^{-1}\in G\_1\cap G\_2$ and $g\_{23}^{-1}g\_3=\tilde{g}\_1\in G\_1\cap G\_3$, therefore
$h=g\_2g\_3 = (g\_2g\_{23})(g\_{23}^{-1}g\_3)\in (G\_1\cap G\_2)\cdot(G\_1\cap G\_3)$.


(2) implies (3):
Wlog assume x=1. From $G\_1\cap yG\_2\neq\emptyset$ follows $y\in G\_1G\_2$; similarly we find $z\in G\_1G\_3$.  

Thus $y=g\_1g\_2$, $z=\tilde{g\_1}g\_3$ for suitable elements $\tilde{g\_1},g\_1\in G\_1$, $g\_2\in G\_2$, $g\_3\in G\_3$.


From $yG\_2\cap zG\_3\neq\emptyset$ we get $y^{-1}z\in G\_2G\_3$, hence
$$
g\_2^{-1} g\_1^{-1} \tilde{g\_1}g\_3 \in G\_2 G\_3 
\implies
g\_1^{-1} \tilde{g\_1} \in (G\_2 G\_3) \cap G\_1 \overset{(2)}= (G\_1\cap G\_2)(G\_1\cap G\_3).
$$
Hence there are $u\in G\_1\cap G\_2$ and $v\in G\_1\cap G\_3$ with $g\_1^{-1} \tilde{g\_1} = uv$. Then one readily verifies $g\_1u \in G\_1\cap g\_1 G\_2 \cap \tilde{g\_1} G\_3 = xG\_1\cap yG\_2 \cap zG\_3 \neq \emptyset$.


(3) implies (1):
The right side of (1) is clearly contained in the left side, so we just need to prove the reverse inclusion. So assume $h\in G\_2 G\_1 \cap G\_3 G\_1$. Then $h \in G\_2 y \cap G\_3 z$ for suitable $y,z\in G\_1$. Thus $y\in G\_1\cap yG\_2$ and $z\in G\_1\cap zG\_3$. Thus by (3) there exists $g\_1 \in G\_1 \cap G\_2 y \cap G\_3 z$, and we have $1 \in G\_2 y g\_1^{-1} \cap G\_3 z g\_1^{-1}$. This implies $yg\_1^{-1} \in G\_2$ and $zg\_1^{-1}\in G\_3$. Therefore $hg\_1^{-1} \in G\_2 yg\_1^{-1} \cap G\_3 zg\_1^{-1} = G\_2 \cap G\_3$, and thus $h\in (G\_2 \cap G\_3) G\_1$ as claimed.


UPDATE 2023-02-13: replaced the "second wlog" in "(2) implies (3)" (sorry for the confusion, I never noticed the comments here before)


