Proof of an 'easy' exercise in a book of Tits 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.


*

*$G_2G_1 \cap G_3G_1 = (G_2 \cap G_3) G_1$

*$(G_1 \cap G_2) \cdot (G_1 \cap G_3) = (G_2G_3) \cap G_1$

*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.
 A: (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)
