# 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.

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?

-
This occurs on page 5 of the book Buildings of Spherical Type and Finite BN-Pairs, Lect. Notes in Math. 386 (1974), but with simpler notation $X, Y, Z$ for the subgroups which makes for easier typesetting. The three equivalent conditions are called (1), (2), (3). I suspect what Max has written down works. Even though mistakes in mathematics are always possible, the book by Tits has been around for a long time and this "easy" exercise has been looked at by many people. – Jim Humphreys Apr 19 '11 at 11:25
Whether my proof is right or not, I am also absolutely sure that the equivalence holds. This is strongly related to the concept of residual connectedness in incidence / diagram geometries, and many people used this over the years (me included). – Max Horn Apr 19 '11 at 12:54

(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 $$g_2^{-1} g_1^{-1} \tilde{g_1}g_3 \in G_2 G_3 \implies g_1^{-1} \tilde{g_1} \in (G_2 G_3) \cap G_1.$$ But using (2), wlog we may assume $g_1\in G_1\cap G_2$ and $\tilde{g}_1\in G_1\cap G_3$. Thus $y\in G_2$ and $z\in G_3$, and so $xG_1\cap yG_2 \cap zG_3 = G_1\cap G_2 \cap G_3 \neq \emptyset$.

(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.

-
Looks nice, but I'm not comfortable with two things: (a) In the proof of (2) => (3), why can we "wlog assume"? (b) In the proof of (3) => (1), why does there exist $g_1\in G_1\cap G_2y\cap G_3z$ ? I only see that there exists some element of $G_1\cap yG_2\cap G_3z$. – darij grinberg Apr 19 '11 at 14:17
@darij: wlog $x=1$ is just the usual symmetry: If $xG_1,yG_2,zG_3$ are pairwise non-disjoint than $G_1,x^{-1}yG_2,x^{-1}zG_3$ are as well. By the above proof $G_1\cap x^{-1}yG_2\cap x^{-1}zG_3$ is non-empty and hence $xG_1\cap yG_2\cap zG_3$ is non-empty. I don't see what's the other problem. If there is an element of $G_1\cap G_2 y \cap G_3 z$ (you have $yG_2$ there, that's just a typo, isn't it? But that doesn't matter, you can always change sides by inverting) and you can name it $g_1$. What's unclear about it? – Johannes Hahn Apr 19 '11 at 15:28
@Johannes: sorry, there are two "wlogs" in that proof and I meant the second one: "wlog we may assume $g_1\in G_1\cap G_2$ and $\tilde{g}_1\in G_1\cap G_3$." – darij grinberg Apr 19 '11 at 16:57
Sorry for my typo, but it should be $G_1\cap yG_2\cap zG_3$, and that's quite a difference to $G_1\cap G_2y\cap G_3z$. – darij grinberg Apr 19 '11 at 16:58
@Johannes: I'm stuck at the same points as Darij. The second "wlog" in the answer is not clear to me, and inverting an element from $G_1\cap yG_2\cap z G_3$ (whose existence follows from (3)) gives something in $G_1\cap G_2y^{-1}\cap G_3 z^{-1}$, not the claimed $g_1\in G_1\cap G_2y\cap G_3 z$. – Thomas Kalinowski Apr 19 '11 at 22:26