Proof. Let $\rho _A$, and $\rho _B$ be the corresponding retractions for $G_A$, and $G_B$ respectively. Consider the compositions $\rho _A \circ \rho _B$ and $\rho _B \circ \rho _A$. When applying them to $v \in V$, we notice that $(\rho _A \circ \rho _B)(v) = \rho _{A \cap B}(v) = (\rho _B \circ \rho _A)(v)$. Extending to morphisms on the group $G_{\Gamma }$, we obtain a commutative diagram of retractions, in the form: $\rho _A\circ \rho _B = \rho _B\circ \rho _A = \rho _{A \cap B}.$

As $G_{A \cap B} \subseteq G_A$ and $G_{A \cap B} \subseteq G_B$ one has $G_{A \cap B} \subseteq G_A \cap G_B$.

To show the other inclusion $G_A \cap G_B \subseteq G_{A \cap B}$, pick an element $x \in G_A \cap G_B$. One has $x \in G_A$ and $x \in G_B$, so $\rho _A(x) = \rho _B(x) = x$. Now using that retractions commute, we obtain:

\begin{align*} \rho_{A \cap B}(x) = (\rho_A\circ \rho_B) (x) = \rho_A(\rho_B (x)) = \rho_A(x) = x. \end{align*}

As $\rho _{A \cap B}$ is a retraction, we have $x \in G_{A \cap B}$, as required.

Proof. Conjugating by $h^{-1}$, we can write the proper inclusion $gG_Ag^{-1} \subsetneq hG_B h^{-1}$ in the equivalent form $fG_Af^{-1} \subsetneq G_B$, for $f = h^{-1}g$. Applying $\rho _B$ we obtain:

\begin{align*} fG_Af^{{-}1} = \rho_B(fG_Af^{{-}1}) = \rho_B(f)G_{A\cap B}\rho_B(f)^{{-}1} \subsetneq G_B. \end{align*}

So, the proper inclusion $fG_Af^{-1} \subsetneq G_B$ is equivalent to the proper inclusion

\begin{align*} \rho_B(f)G_{A\cap B}\rho_B(f)^{{-}1} \subsetneq G_B, \end{align*}

which after conjugating by $\rho _B(f)^{-1}$ becomes equivalent to $G_{A\cap B} \subsetneq G_B$, and this implies that $A \cap B \subsetneq B$.

Instead, applying $\rho _A$ to $fG_Af^{-1} \subsetneq G_B$, we obtain

\begin{align*} G_A = \rho_A(f)G_A\rho_A(f)^{{-}1} = \rho_A(fG_Af^{{-}1}) \subseteq \rho_A(G_B) = G_{A\cap B}. \end{align*}

The inclusion $G_A \subseteq G_{A\cap B}$ implies $A\subseteq A\cap B$. Ultimately $A\subseteq A\cap B \subsetneq B$, which means that $A \subsetneq B$, as required.

Proof. One has the equality

\begin{align*} fG_Af^{{-}1} \cap gG_B g^{{-}1} = f[G_A \cap (f^{{-}1}g) G_B (f^{{-}1}g)^{{-}1}]f^{{-}1}. \end{align*}

Set $h = f^{-1}g$ and consider $P = G_A \cap hG_B h^{-1}$. Using $P\subseteq G_A$, and $G_A \cap G_B = G_{A \cap B}$ (see Lemma 3.1), we obtain:

\begin{align*} P = \rho_A(P) = \rho_A(G_A \cap hG_B h^{{-}1}) & \subseteq \rho_A(G_A)\cap \rho_A(hG_B h^{{-}1}) \\ & = G_A \cap \rho_A(h)\rho_A(G_B)\rho_A(h^{{-}1}) \\ & = \rho_A(h)G_{A\cap B}\rho_A(h)^{{-}1}. \end{align*}

Setting $a= \rho _A(h)\in G_A$ and $A\cap B = C$, we can write the inclusion above as $P\subseteq aG_{C}a^{-1}$, and we notice that $aG_{C}a^{-1} \subseteq G_A$. Also, $P = G_A \cap hG_B h^{-1}$, so we have

\begin{align*} P = (G_A \cap hG_B h^{{-}1}) \cap aG_{C}a^{{-}1} & = h G_B h^{{-}1} \cap (G_A \cap aG_{C}a^{{-}1}) \\ & = h G_B h^{{-}1} \cap aG_{C}a^{{-}1}. \end{align*}

Multiplying the last equation by $h^{-1}$ and denoting $P' = h^{-1}Ph$, $k = h^{-1}a$, we obtain: $P' = G_{B} \cap kG_{C} k^{-1}.$

Applying the same procedure as for $P$ above, we obtain:

\begin{align*} P' = \rho_B(P') & = \rho_B(G_B \cap kG_{C} k^{{-}1}) \\ & \subseteq \rho_B(G_B)\cap \rho_B(kG_{C} k^{{-}1}) \\ & = \rho_B(k)G_{B\cap C}\rho_B(k)^{{-}1}\\ & = \rho_B(k)G_{C}\rho_B(k)^{{-}1}. \end{align*}

Setting $b = \rho _B(k)\in G_B$ we express the inclusion above as $P' \subseteq bG_{C} b^{-1} \subseteq G_B$. Putting together $P' = G_{B} \cap kG_{C} k^{-1}$ and $P' \subseteq bG_{C} b^{-1}$, we have:

\begin{align*} P' = (G_{B} \cap kG_{C} k^{{-}1}) \cap bG_{C} b^{{-}1} = h^{{-}1}aG_{c} a^{{-}1}h \cap (G_{B} \cap bG_{C} b^{{-}1}). \end{align*}

Using $G_{B} \cap bG_{C} b^{-1} = bG_{C} b^{-1}$, and $P' = h^{-1}Ph$, we ultimately have:

\begin{align*} P = aG_{C} a^{{-}1} \cap hbG_{C} b^{{-}1}h^{{-}1}. \end{align*}

Turning back, we have $fG_Af^{-1}\cap g G_Bg^{-1} = fPf^{-1}$, and $h=f^{-1}g$, so we obtain:

\begin{align*} fG_Af^{{-}1}\cap g G_Bg^{{-}1} = faG_Ca^{{-}1}f^{{-}1}\cap gb G_Cb^{{-}1}g^{{-}1}, \end{align*}

where $C= A\cap B$, as desired.

Proof. We have that $gG_Ag^{-1}\leqslant hG_Ah^{-1}$ if and only if $h^{-1}gG_Ag^{-1}h\leqslant G_A$. In particular, $h^{-1}gG_Ag^{-1}h= \rho _A(h^{-1}gG_Ag^{-1}h) =G_A$. The lemma follows.

Proof. Let $A\subseteq V$, $g\in G_\Gamma$, and $x\in V {\setminus} A$ with the property $\operatorname {Link}(x)\not \supseteq A$ be as in the hypothesis. Consider $P = G_A \cap gG_Ag^{-1}$.

If $G_{V {\setminus} \{x\}} = gG_{V {\setminus} \{x\}}$, then $g \in G_{V {\setminus} \{x\}}$. This means that both $G_A$ and $gG_Ag^{-1}$ are parabolic subgroups in $G_{V {\setminus} \{x\}}$, and hence $G_A \cup gG_Ag^{-1}$ is contained in the proper parabolic subgroup $G_{V {\setminus} \{x\}}$ of $G$. This contradicts the assumptions of the proposition, so suppose that $G_{V {\setminus} \{x\}} \neq gG_{V {\setminus} \{x\}}$. From the presentation, one has a splitting of $G_\Gamma$ as an amalgamated free product:

\begin{align*} G_\Gamma = G_{\operatorname{Star}(x)}*_{G_{\operatorname{Link}(x)}} G_{V {\setminus} \{x\}}. \end{align*}

Consider the Bass–Serre tree $T$ corresponding to this splitting (see for example [Reference Dicks and Dunwoody6]). There are two types of vertices in $T$: left cosets of $G_{\operatorname {Star}(x)}$, and left cosets of $G_{V {\setminus} \{x\}}$ in $G$. Only vertices of different type can be adjacent in $T$. The group $G$ acts naturally on $T$, without edge inversions. Moreover, the vertex stabilizers correspond to conjugates of $G_{\operatorname {Star}(x)}$ and conjugates of $G_{V {\setminus} \{x\}}$ for the respective type of vertices, while the edge stabilizers are conjugates of ${G_{\operatorname {Link}(x)}}$.

In the tree $T$, both $G_{V {\setminus} \{x\}}$ and $gG_{V {\setminus} \{x\}}$, are distinct vertices of the same type. Their stabilizers are $G_{V {\setminus} \{x\}}$, and $gG_{V {\setminus} \{x\}}g^{-1}$ respectively. As we are on a tree, there is a unique geodesic $p$ in $T$ connecting $G_{V {\setminus} \{x\}}$ and $gG_{V {\setminus} \{x\}}$.

Since $x\not \in A$, we have: $G_A \subseteq G_{V {\setminus} \{x\}}$ and $gG_Ag^{-1} \subseteq gG_{V {\setminus} \{x\}}g^{-1}$, which means that our parabolic subgroups $G_A$ and $gG_Ag^{-1}$ stabilize the vertices corresponding to the cosets $G_{V {\setminus} \{x\}}$ and $gG_{V {\setminus} \{x\}}$, respectively. The intersection $G_A \cap gG_Ag^{-1}$ stabilizes the geodesic $p$ connecting those vertices, and hence it stabilizes any edge belonging to $p$. Since stabilizers of edges in $T$ are conjugates of ${G_{\operatorname {Link}(x)}}$, we have:

\begin{align*} P = G_A \cap gG_Ag^{{-}1} \subseteq hG_{\operatorname{Link}(x)}h^{{-}1} \end{align*}

for some $h\in G$. Now one can write $P$ as:

\begin{align*} P= G_A \cap gG_Ag^{{-}1} \cap hG_{\operatorname{Link}(x)}h^{{-}1} = (G_A \cap hG_{\operatorname{Link}(x)}h^{{-}1}) \cap (gG_Ag^{{-}1} \cap hG_{\operatorname{Link}(x)}h^{{-}1}) \end{align*}

By Lemma 3.3, one can express $G_A \cap hG_{\operatorname {Link}(x)}h^{-1}$ as an intersection of two parabolic subgroups over $\operatorname {Link}(x)\cap A \subsetneq A$ (because $\operatorname {Link}(x)\not \supseteq A$), hence $P$ is contained in a parabolic subgroup over a proper subset of $A$. This completes the proof.

Proof. Suppose that the inclusion $G_A\cup gG_Ag^{-1} \subseteq tG_\Delta t^{-1}$ holds. Multiplying by $t^{-1}$, we obtain $t^{-1}G_A t \cup t^{-1} gG_A g^{-1}t\subseteq G_\Delta$. Applying $\rho _\Delta$, and recalling that $A\subseteq V_\Delta$, we get:

\begin{align*} t^{{-}1}G_A t = \rho_\Delta(t^{{-}1}) G_{A} \rho_\Delta(t)\text{ and }t^{{-}1} gG_A g^{{-}1}t = \rho_{\Delta}( t^{{-}1} g)G_A \rho_{\Delta}(g^{{-}1}t). \end{align*}

Let $f_1=\rho _\Delta (t^{-1})$ and $f_2=\rho _\Delta (t^{-1} g)\in G_\Delta$. We have that

\begin{align*} t^{{-}1}G_A t \cap t^{{-}1} gG_A g^{{-}1}t = f_1G_Af_1^{{-}1} \cap f_2G_A f_2^{{-}1}. \end{align*}

Observe that $t^{-1}G_A t \cap t^{-1} gG_A g^{-1}t$ is $\Delta$-parabolic (respectively contained in a parabolic subgroup over a subset of $A$) if and only if $G_A\cap gG_Ag^{-1}$ is $\Delta$-parabolic (respectively contained in a parabolic subgroup over a subset of $A$).

We will take $h=f_1^{-1}f_2\in G_\Delta$. Observe that $f_1G_Af_1^{-1} \cap f_2G_A f_2^{-1}$ is $\Delta$-parabolic (respectively contained in a parabolic subgroup over a subset of $A$) if and only if $G_A\cap hG_Ah^{-1}$ is $\Delta$-parabolic (respectively contained in a parabolic subgroup over a subset of $A$).

We prove (i). Note that $G_A=gG_Ag^{-1}\Leftrightarrow t^{-1}G_A t = t^{-1}gG_Ag^{-1}t \stackrel {(*)}{\Leftrightarrow } \rho _{\Delta } (t^{-1}G_A t) = \rho _\Delta ( t^{-1}gG_Ag^{-1}t ) \Leftrightarrow f_1 G_A f_1^{-1} = f_2 G_A f_2^{-1} \Leftrightarrow G_A = hG_Ah^{-1}$. The equivalence $(*)$ uses that $t^{-1}G_At$ and $t^{-1}gG_A g^{-1}t$ are contained in $G_\Delta$.

To show (ii), by the previous discussion, it is enough to show that $t^{-1}G_A t \cap t^{-1} gG_A g^{-1}t$ is $\Gamma$-parabolic if and only if $f_1G_Af_1^{-1} \cap f_2G_A f_2^{-1}$ is $\Delta$-parabolic.

As $\Delta$ is a subgraph of $\Gamma$, any $\Delta$-parabolic subgroup of $G_\Delta$ is a $\Gamma$-parabolic subgroup of $G_\Gamma$. Thus, if $f_1 G_Af_1^{-1}\cap f_2 G_A f_2^{-1}$ is $\Delta$-parabolic, then it is also $\Gamma$ parabolic. Conversely, assume there is some $B\subseteq V\Gamma$ and some $d\in G_\Gamma$ such that, $f_1G_Af_1^{-1}\cap f_2G_Af_2^{-1}= d G_Bd^{-1}$. As $f_1G_Af_1^{-1} \cap f_2G_A f_2^{-1}\subseteq G_\Delta$, applying $\rho _\Delta$, and noting that $B\subseteq A \subseteq \Delta$ we get that $f_1G_Af_1^{-1}\cap f_2G_Af_2^{-1}=f_3G_{B}f_3^{-1}$ where $f_3=\rho _\Delta (d)$.

A similar argument as above shows (iii).

Proof. Let $\Gamma \in {\mathcal C}$ and let $P,\,Q$ be two parabolic subgroups of $G_\Gamma$. By Lemma 3.3, we can suppose that there is $A\subseteq V$ and $h_1,\,h_2\in G_\Gamma$ such that $P\cap Q = h_1 G_Ah_1^{-1} \cap h_2 G_A h_2^{-1}$. Therefore, $P\cap Q$ is parabolic if and only if $G_A\cap gG_Ag^{-1}$ is parabolic, where $g=h_1^{-1}h_2$.

If $G_A\cup gG_Ag^{-1}$ is contained in a proper parabolic subgroup of $G_\Gamma$, then by Lemma 3.7, we can replace $\Gamma$ by a proper subgraph $\Delta$ and replace $g$ by some $h\in G_\Delta$. Note that $\Delta$ is still in the class ${\mathcal C}$.

Therefore, we can return to the initial notation and further assume that $G_A\cup gG_Ag^{-1}$ is not contained in a proper parabolic of $G_\Gamma$.

We will show that for every $A\subseteq V$ finite and any $g\in G_\Gamma$, $G_A\cap g G_Ag^{-1}$ is parabolic. Our proof is by induction on $|A|$. If $|A|=0$, then $G_A$ is trivial and the result follows. We now assume that $|A|>0$ and that for parabolic subgroups over smaller sets the result holds. We remark that the induction hypothesis is equivalent to saying that for any $B\subseteq V$, $|B|<|A|$ and any $g_1,\,g_2\in G_\Gamma$, $g_1G_Bg_1^{-1} \cap g_2G_Bg_2^{-1}$ is parabolic.

If there is $x\in V {\setminus} A$ such that $A$ is not contained in $\operatorname {Link}(x)$, then Lemma 3.6 implies that $G_A\cap gG_Ag^{-1}\leqslant dG_Bd^{-1}$ for some $B\subsetneq A$ and some $d\in G_\Gamma$. Therefore, by Lemma 3.3, there are $a,\,a'\in G_A$ and $b,\,b'\in G_B$ such that

\begin{align*} G_A\cap gG_Ag^{{-}1} & = G_A\cap gG_Ag^{{-}1} \cap d G_B d^{{-}1}\\ & = (G_A \cap d G_B d^{{-}1}) \cap (gG_Ag^{{-}1} \cap d G_B d^{{-}1})\\ & = (aG_{A\cap B}a^{{-}1} \cap d bG_{A\cap B} b^{{-}1} d^{{-}1})\cap (ga'G_{A\cap B} {a'}^{{-}1}g^{{-}1}\\& \quad \cap db'G_{A\cap B}(b')^{{-}1} d^{{-}1} )\\ & = (aG_{B}a^{{-}1} \cap d G_{B} d^{{-}1})\cap (ga'G_{B} {a'}^{{-}1}g^{{-}1} \cap dG_{ B} d^{{-}1} ). \end{align*}

As $|B|<|A|$, by induction, $aG_{B}a^{-1} \cap d G_{B} d^{-1}$ and $aG_{B}a^{-1} \cap d G_{B} d^{-1}$ are parabolic subgroups of $G_\Gamma$ over subsets of $B$. Say that $aG_{B}a^{-1} \cap d G_{B} d^{-1} = g_1 G_{B_1} g_1^{-1}$ and $aG_{B}a^{-1} \cap d G_{B} d^{-1} = g_2 G_{B_2} g_2^{-1}$. Thus, using again Lemma 3.3, we get that

\begin{align*} G_A\cap gG_Ag^{{-}1} = g_1 G_{B_1} g_1^{{-}1} \cap g_2 G_{B_2} g_2^{{-}1} = g_1xG_C x^{{-}1}g_1^{{-}1}\cap g_2yG_Cy^{{-}1}g_2^{{-}1} \end{align*}

where $x\in G_{B_1}$, $y\in G_{B_2}$ and $C=B_1\cap B_2$. As $|C|<|A|$, using again induction, we get that $G_A\cap gG_Ag^{-1}$ is parabolic.

So, we assume that for all $x\in V {\setminus} A$, $A\subseteq \operatorname {Link}(x)$. We now argue by induction on $N=\sharp \{x\in V {\setminus} A : \operatorname {Star}(x)\neq V\}$. In the case $N=0$, as we are in the class ${\mathcal C}$ that satisfies (3.1), we have that either $G_A=gG_Ag^{-1}$ and hence $G_A\cap gG_Ag^{-1}$ is parabolic, or $G_A\cap gG_Ag^{-1}\leqslant d G_B d^{-1}$ for some $B\subsetneq A$. As before, the latter implies that $G_A\cap gG_Ag^{-1}$ is an intersection of four parabolics over $B$, with $|B|<|A|$ and arguing as above, we get that $G_A\cap gG_Ag^{-1}$ is parabolic. So assume that $N>0$ and the result is known for smaller values of $N$ and $A$.

As $N > 0$ and $A\subseteq \operatorname {Link}(x)$ for all $x\in V {\setminus} A$, there are $x,\,y\in V {\setminus} A$ not linked by an edge. Setting $X=\operatorname {Star}(x)$, $Y=V {\setminus} \{x\}$, and $Z=\operatorname {Link}(x)$, we obtain an amalgamated free product:

\begin{align*} G = G_X*_{G_Z} G_{Y}, \end{align*}

and there is an associated Bass–Serre tree $T$ corresponding to this splitting.

Consider the edges $G_Z$ and $gG_Z$ on $T$. Let $G_Z,\, g_1G_Z,\, \ldots,\, g_nG_Z,\, gG_Z$ be a sequence of edges in the unique geodesic in $T$ connecting $G_Z$ and $gG_Z$. If $G_Z = gG_Z$ (i.e $n=0$) and taking into account that $A\subseteq Z$, we have that $G_A\cup g G_Ag^{-1}$ is contained in $G_Z$, which is a proper parabolic of $G_\Gamma$ and this contradicts our hypothesis. So we assume that $n\geq 1$. By the construction of $T$, one has either $g_i^{-1}g_{i+1} \in G_{X}$ or $g_i^{-1}g_{i+1} \in G_{Y}$, for any $i = 0,\, \ldots,\, n$ where $g_0 = 1$ and $g_{n+1} = g$. The intersection $G_A\cap gG_A g^{-1}$ stabilizes the endpoints of the geodesic path, hence it stabilizes the whole path. As the stabilizer of a geodesic in a tree is the intersection of stabilizers of its edges, we have the equality

\begin{align*} G_A\cap gG_A g^{{-}1} = G_A\cap g_1G_Z g_1^{{-}1} \cap \ldots \cap g_nG_Zg_n^{{-}1}\cap gG_A g^{{-}1}. \end{align*}

By Lemma 3.3, (applied to $G_A\cap g_i G_Zg_i^{-1})$, and the fact that $A\subseteq Z$ we have that there are $z_i\in G_Z$ such that $G_A\cap g_iG_Zg_i^{-1}$ is equal to $G_A\cap g_iz_i G_A z_i^{-1}g_i^{-1}$.

Note that $(g_iz_i)^{-1}(g_{i+1}z_{i+1}) = z_i^{-1}(g_i^{-1}g_{i+1})z_{i+1}$, so replacing $g_iz_i$ by $g_i$ we still have that $g_i^{-1}g_{i+1} \in G_{X}$ or $g_i^{-1}g_{i+1} \in G_{Y}$, for any $i = 0,\, \ldots,\, n$ where $g_0 = 1$ and $g_{n+1} = g$. Hence:

\begin{align*} G_A\cap gG_A g^{{-}1} = G_A\cap g_1G_Ag_1^{{-}1} \cap \ldots \cap g_nG_Ag_n^{{-}1}\cap gG_A g^{{-}1}. \end{align*}

The intersections $g_iG_Ag_i^{-1} \cap g_{i+1}G_Ag_{i+1}^{-1}$ can be expressed as:

\begin{align*} g_iG_Ag_i^{{-}1} \cap g_{i+1}G_Ag_{i+1}^{{-}1} = g_i[G_A \cap g_i' G_Ag_i'^{{-}1}]g_i^{{-}1}, \end{align*}

where $g_i' = g_i^{-1}g_{i+1}$ is either in $G_{X}$, or in $G_{Y}$. As the number of vertices in $X {\setminus} A$ (respectively $Y {\setminus} A$) whose star is not $X$ (respectively $Y$) is less than $N$, we can apply induction and the intersections $G_A \cap g_i' G_Ag_i'^{-1}$ are either equal to $G_A$ or are contained in a parabolic subgroup over a proper subset $B$ of $A$. If we have equality for $i=1,\,\dots,\, n$, then $G_A\cap g G_Ag^{-1}=G_A$. Otherwise, $G_A\cap g G_Ag^{-1}$ is contained in a parabolic subgroup over a proper subset $B$ of $A$ and we at the desired conclusion by induction on $|A|$.

Proof. Let ${\mathcal C}$ be the family of finite right-angled Artin graphs. Clearly ${\mathcal C}$ is closed under subgraphs. Now take $A\subseteq V,\, g\in G_\Gamma$ such that for all $x\in V {\setminus} A,\, \, \operatorname {Star}(x)=V$. Then $G_\Gamma$ is a direct product of $G_A$ and $G_{V {\setminus} A}$ and thus, $G_A=gG_Ag^{-1}$. Then ${\mathcal C}$ satisfies (3.1) and the corollary follows from Proposition 3.8.

Proof. Let $\rho _x\colon G_\Gamma \to \langle x \rangle$. Both $\langle a \rangle$ and $g\langle a \rangle g^{-1}$ lie on $\ker \rho _x$. From § 4.1 we know that $\ker \rho _x$ is free with basis $a_0,\, a_1,\,\dots,\, a_{k-1}$ where $a_i=x^{i}ax^{-i}$. Write $g = h x^{s}$ where $s=\rho _x(g)$ and $h\in \ker \rho _x$. Following Equation (4.1), we have that $x^{s}ax^{-s} = \sigma _a^{l(s)}a_r \sigma _a^{-l(s)}$ for some $l(s)\in \mathbb {Z}$, $0\leq r < k$ and $\sigma _a= a_0a_1a_2\cdots a_{k-1}$. In particular,

\begin{align*} \langle a \rangle \cap g\langle a \rangle g^{{-}1}= \langle a_0 \rangle \cap h \sigma_a^{l(s)}\langle a_r \rangle \sigma_a^{{-}l(s)}h^{{-}1}. \end{align*}

Now, the intersection is trivial if $r\neq 0$. If $r=0$, as $\langle a_0\rangle$ is a malnormal subgroup (even more a free factor) of $\ker \rho _x$, the intersection is trivial unless $h\sigma _a^{l(s)}\in \langle a_0\rangle$, and in that case, the intersection is the whole $\langle a_0\rangle = \langle a \rangle$.

Proof. If $A$ is empty, then $G_A=\{1\}$ and $G_A=gG_Ag^{-1}$. So we assume that $A$ is non-empty.

Let $N$ be the number of edges from $A$ to $V {\setminus} A$ with label greater than $2$. We will argue by induction on $N$.

If $N=0$, then for all $x\in V {\setminus} A$ and all $a\in A$, the label of $\{x,\,a\}$ is $2$, and hence $G = G_{V {\setminus} A}\times G_A$ and $gG_Ag^{-1}=G_A$ for all $g\in G$.

So assume that $N>0$ and the theorem holds for smaller values of $N$.

Consider first the case when there is $x\in V {\setminus} A$ and $a\in A$ such that the label of $\{x,\,a\}$ is $2k_a$ with $k_a >1$ and $A\subseteq \operatorname {Star}(a)$. In this case, $\operatorname {Star}(x) = \operatorname {Star}(a) = V$, and for all $z\in Z := V {\setminus} \{a,\,x\}$ we have that $m_{x,z}=m_{a,z}=2$, which yields $G_\Gamma = G_{\{x,a\}}\times G_Z$. Write $g=(g_1,\,g_2)$ with $g_1\in G_{\{x,a\}}$ and $g_2\in G_Z$. Then $G_A\cap gG_Ag^{-1}$ is equal to the direct product of the subgroup $\langle a \rangle \cap g_1\langle a\rangle g_1^{-1}$ of the direct factor $G_{\{x,a\}}$ and the subgroup $G_{A {\setminus} \{a\}}\cap g_2 G_{A {\setminus} \{a\}}g_2^{-1}$ of the direct factor $G_Z$. By Lemma 5.1 we have that $\langle a \rangle \cap g_1\langle a\rangle g_1^{-1}$ is either trivial or equal to $\langle a \rangle$. Let $A'= A {\setminus} \{a\}$. Note that for all $z\in Z {\setminus} A'$, $\operatorname {Star}(z)=Z$ and the number of vertices $z\in Z {\setminus} A'$ with an edge with label $>2$ is less than $N$ (as $Z$ spans a subgraph of $\Gamma$ that consists of deleting the vertices $x,\,a$ and the edge $\{x,\,a\}$). Therefore, by induction, either $G_{A'}=g_2 G_{A'}g_2^{-1}$ or $G_{A'}\cap g_2 G_{A'}g_2^{-1}$ is contained in a parabolic subgroup over a proper subset of $A'$. The theorem follows in this case.

So let us consider the case when there is $x\in V {\setminus} A$ and $a\in A$ such that the label of $\{x,\, a\}$ is $2k_a$ with $k_a >1$ (in particular $N\geq 1$), $A\not \subseteq \operatorname {Star}(a)$ and that the theorem holds for smaller values of $N$. We remark that the condition $A\not \subseteq \operatorname {Star}(a)$ will not be used until Case 3.2 below.

Recall from the previous section, that there exists a finite Artin graph $\Delta$, such that $\ker \rho _{x}$ is isomorphic to $G_\Delta$. By the notation of § 4.1, $V_\Delta = \{z_0,\,\dots,\, z_{k_z-1} : z\in V {\setminus} \{x\}\}$, $E_\Delta = \{\{u_i,\, v_j\}\subseteq V_\Delta : u_i\neq v_j \textrm { and } \{u,\,v\}\in E\},$ and $m^{\Delta }_{u_i,v_j}=m_{u,v}$. Recall that the vertices of $V_\Delta$ are indexed. We will write $A_0$ to denote the vertices of type $v\in A$ and index $0$, i.e. $A_0=\{b_0: b\in A\}$. Observe that the vertices $y_i$ of $V_\Delta {\setminus} A_0$ such that $\operatorname {Link}(y)$ does not contain $A_0$ are exactly the vertices $b_1,\,\dots,\, b_{k_b-1}$ with $b\in A$ and $k_b>1$. Indeed, if $k_b>1$, $b_0,\,\dots,\, b_{k_b-1}$ span a subgraph with no edges of $\Delta$ and thus $b_0\notin \operatorname {Link}(b_i)$ for $i>0$. On the other hand, if $y\in V {\setminus} A$, as $\operatorname {Star}(y)=V$, we have that $k_y=1$ (since $y,\,x,\,a$ form a triangle) and then in $\Delta$ we only have a vertex $y_0$ of type $y$ and by definition of $\Delta$, we have that $\operatorname {Star}(y_0)=V_\Delta$.

We note that $G_{A_0}$ is an index-parabolic subgroup of $G_\Delta$ and it is equal to the subgroup $G_A$ of $G_\Gamma$.

Write $g= hx^{s}$ where $h\in \ker \rho _{x}$ and $s=\rho _{x}(g)$. Let $Q=x^{s} G_A x^{-s}\leqslant \ker \rho _x$. We note that $Q$ might not be a parabolic subgroup of $\ker \rho _x$ although we can give a very precise description using Equation (4.1): $Q$ is generated by $\{ x^{s} b x^{-s} :b\in A \}$. If $k_b=1$, then $x^{s} b x^{-s} = b_0$. If $k_b>1$ then $x^{s} b x^{-s}$ is equal to $\sigma _b^{l(s,b)} b_i \sigma _b^{-l(s,b)}$ where $i\in \{0,\,\dots,\, k_b-1\}$, $i\equiv s \mod k_b$, $l(s,\,b)\in \mathbb {Z}$ and $\sigma _b=b_0b_1 \dots b_{k_b-1}$.

Now $G_A\cap gG_Ag^{-1} = G_{A_0} \cap hQh^{-1}$. Note that even if $Q$ is not parabolic, we are reduced to show that either $G_{A_0} = hQh^{-1}$ or that $G_{A_0} \cap hQh^{-1}$ is contained in a $\Delta$-parabolic subgroup over a proper subset of $A_0$. Indeed, in the latter case, as $\Delta$-parabolics over subsets of $A_0$ are $\Gamma$-parabolics, we also get that $G_A\cap gG_Ag^{-1}$ is contained in a $\Gamma$-parabolic subgroup over a proper subset of $A$.

We consider three cases:

Case 1: $s=0$. Then $Q=G_{A_0}$ is a parabolic subgroup of $G_\Delta$. By Lemma 3.7 we can reduce the problem to a subgraph $\Delta '$ of $\Delta$ and $h'\in G_{\Delta '}$ such that $G_{A_0}\cup h'G_{A_0}(h')^{-1}$ is not contained in a proper parabolic subgroup of $G_{\Delta '}$. We need to show that either $G_{A_0}=h'G_{A_0}(h')^{-1}$ or $G_{A_0}\cap h'G_{A_0}(h')^{-1}$ is contained in a $\Delta '$-parabolic subgroup over a proper subset of $A_0$.

If $b_{i}\in V_{\Delta '}$ for some $b\in A$, $k_b>1$ and $i>0$, then Lemma 3.6 implies that $G_{A_0}\cap h'G_{A_0}(h')^{-1}$ is contained in a $\Delta '$-parabolic subgroup over a proper subset of $A_0$ and we are done. So, we can assume that $V_{\Delta '}\subseteq A_0 \cup \{y_0: y\in V {\setminus} A\}$. Note that in this case, $\Delta '$ is a finite Artin graph of even FC-type, for all $y_0\in V_{\Delta '} {\setminus} A_0$ we have that $\operatorname {Star}(y_0)=V_{\Delta '}$ and the number of edges from $V\Delta ' {\setminus} A_0$ to $A_0$ with label $>2$ is less than $N$ (in fact, it is less than $N$ minus the number of edges $\{x,\,b\}$ with label $> 2$). By induction, we get that either $G_{A_0} = {h'}Q(h')^{-1}$ or $G_{A_0} \cap {h'}Q(h')^{-1}$ is contained in a $\Delta '$-parabolic subgroup over a proper subset of $A_0$ and we are done.

Case 2: $s\notin k_a\mathbb {Z}$. Then $\rho _{A_0}(hQh^{-1})\leqslant G_{A_0 {\setminus} \{a_0\}}$ and therefore, $G_{A_0}\cap hG_Qh^{-1} \leqslant G_{A_0 {\setminus} \{a_0\}}$ and the lemma holds.

Case 3: $s\in k_a\mathbb {Z}$, $s\neq 0$. Recall that $Q$ is generated by $\{ x^{s} b x^{-s} :b\in A \}$, and the element $x^{s} bx^{-s}$ is equal (in $G_\Delta$) to $\sigma _b^{l(s,b)} b_{i(s,b)} \sigma _b^{-l(s,b)}$ where $i(s,\,b) \equiv s \mod k_b$, $l(s,\,b)\in \mathbb {Z}$ and $\sigma _b = b_0b_1\dots b_{k_b-1}$(see Equation(4.1)). If some $i(s,\,b)\neq 0$, we lie in Case 2. So we assume that $i(s,\,b)=0$ for all $b\in A$.

For simplifying our notation and argumentsFootnote ^† , consider the automorphism

\begin{align*} \phi\colon G_{\Delta}\to G_{\Delta}\qquad \phi(v)=\begin{cases}v & v\neq a_1\\ \sigma_a & v=a_1.\end{cases} \end{align*}

We need to show that this is well defined. By construction, the only edges of $\Delta$ adjacent to $a_1$ are of the form $\{z_0,\,a_1\}$ with $z\in \operatorname {Link}_\Gamma (a) {\setminus} \{x\}$. Moreover, as $\operatorname {Star}(x)=V$ and $k_a>1$, necessarily $m_{z_0,a_1}=2$ for $z\in \operatorname {Link}_\Gamma (a) {\setminus} \{x\}$. Thus, we need to check that $\sigma _a$ commutes with $z_0$, $z\in \operatorname {Link}_\Gamma (a) {\setminus} \{x\}$. But this holds, as by construction $m_{z_0,a_i}=2$ for all $i=0,\,1,\,\dots,\, k_a-1$ (recall that $\sigma _a=a_0a_1\cdot a_{k_a-1}$). Thus, $\phi$ is well defined. It is easy to check that $\phi$ is bijective.

We apply $\phi ^{-1}$ to $G_{A_0}$, $Q$ and $h$, and we get $G_{A_0}$,

(5.1)\begin{equation} P=\langle \{a_1^{l(s,a)}a_0 a_1^{{-}l(s,a)}\} \cup \{\sigma_b^{l(s,b)} b_0 \sigma_b^{{-}l(s,b)}: b_0\in A_0 {\setminus} \{a_0\}\}\rangle \end{equation}

and $f=\phi ^{-1}(h)$ respectively. Note that as $G_{A_0}$ is fixed by $\phi$, we have that $G_{A_0}=hQh^{-1}$ if and only if $G_{A_0}=fP f^{-1}$ and that $G_{A_0}\cap hQh^{-1}$ is contained in a $\Delta$-parabolic subgroup over a proper subset of $A_0$ if and only if the same holds for $G_{A_0}\cap fPf^{-1}$. For simplicity, we set $l=l(s,\,a)$. Note that as $s\neq 0$, $l\neq 0$.

Let $D= V_\Delta {\setminus} \{a_0\}$ and $\rho _D$ the corresponding retraction. Then $G_\Delta = \ker \rho _D \rtimes G_D$. Let $\rho _{a_1}\colon G_\Delta \to \langle a_1\rangle$ be the canonical retraction. We have now two subcases.

Case 3.1: $\rho _{a_1}(fa_1^{l}) \neq 0$.

We are going to show that

\begin{align*} (G_{A_0}\cap \ker \rho_D)\cap (fPf^{{-}1}\cap \ker \rho_D)=\{1\}. \end{align*}

This implies that $G_{A_0}\cap fPf^{-1}\leqslant G_D$ and therefore, $G_{A_0}\cap fPf^{-1}\leqslant G_{A_0\cap D}$ and we are done, as $A_0\cap D = A_0 {\setminus} \{a_0\}$. Note that

\begin{align*} (G_{A_0}\cap \ker \rho_D) = \langle a_0^{G_{A_0}}\rangle \quad\text{ and }\quad (fPf^{{-}1}\cap \ker \rho_D)= \langle (fa_1^{l} a_0 a_1^{{-}l}f^{{-}1})^{fPf^{{-}1}}\rangle. \end{align*}

Let $L=\operatorname {Link}_\Delta (a_0)$. As $k_a>1$ every vertex in $L$ commutes with $a_0$. Let $\rho _L$ be the canonical retraction $\rho _L \colon G_{D}\to G_L$, and $T=\ker (\rho _L)$. Recall from § 4.2 that $\ker \rho _D$ is free with free basis $\{ta_0 t^{-1} : t\in T\}$. Note that if $g\in G_D$, then there are unique $g_L=\rho _L(g)$ and $g'\in \ker \rho _L$ such that $g=g'g_L$ and $ga_0g^{-1}= g'a_0 g'^{-1}$.

Claim 1: $\langle a_0^{G_{A_0}}\rangle$ is free with basis $T_{A_0}=\{ta_0t^{-1} : t\in \ker \rho _L \cap G_{A_0}\}$.

Notice that $\langle T_{A_0}\rangle \leqslant \langle a_0^{G_{A_0}}\rangle$. So it is enough to show that $\langle a_0^{G_{A_0}}\rangle \leqslant \langle T_{A_0}\rangle$ to prove Claim 1. In order to show it, pick $g\in G_{A_0}$. We need to show that $ga_0g^{-1}\in \langle T_{A_0}\rangle.$ Write $g$ as $g = g_1a_0^{m_1}g_2a_0^{m_2}\dots g_n a_0^{m_n}$ where $n\geq 0$, $g_i\in (G_D\cap G_{A_0}) {\setminus} \{1\}$ for $i=1,\,2,\,\dots,\, n$, $m_i\in \mathbb {Z} {\setminus} \{0\}$ for $i=1,\,2,\,\dots,\, n-1$ and $m_n\in \mathbb {Z}$. We can further write $g_i$ as $c_ih_i$ where $h_i\in \ker \rho _D$ and $c_i\in G_L$, that is

\begin{align*} g = c_1h_1a_0^{m_1}c_2h_2a_0^{m_2}\dots c_n h_n a_0^{m_n}, \end{align*}

which rewriting $c_1 \cdots c_i h_i c_i^{-1}\cdots c_1$ as $h_i'$ and using that the $c_i$'s commute with $a_0$, we get that

\begin{align*} g = h_1'a_0^{m_1}h'_2a_0^{m_2}\dots h'_n a_0^{m_n}c_1c_2\cdots c_n, \end{align*}

notice that $ga_0g^{-1}$ is equal to $g'a_0(g')^{-1}$ where

\begin{align*} g' = h_1'a_0^{m_1}h'_2a_0^{m_2}\dots h'_n. \end{align*}

We can write $g'a_0(g')^{-1}$ as a product of elements of $T_{A_0}=\{ta_0t^{-1} : t\in \ker \rho _L \cap G_{A_0}\}$. Indeed:

\begin{align*} g'a_0(g')^{{-}1} & = (h_1'a_0 (h_1')^{{-}1})^{m_1} \cdot (h_1' h_2' a_0 (h_1'h_2')^{{-}1})^{m_2}\\ & \quad \cdots (h_1'\cdots h_{n-1}' a_0 (h'_{1}\cdots h'_{n-1})^{{-}1})^{m_{n-1}} \cdot \\ & \quad \cdot h_1'\cdots h_n' a_0 (h_1'\cdots h_n')^{{-}1} \cdot \\ & \quad \cdot (h_1'\cdots h_{n-1}' a_0 (h'_{1}\cdots h'_{n-1})^{{-}1})^{{-}m_{n-1}}\cdots (h_1' h_2' a_0 (h_1'h_2')^{{-}1})^{{-}m_2}\\& \quad \cdot(h_1'a_0 (h_1')^{{-}1})^{{-}m_1} . \end{align*}

This completes the proof of Claim 1.

Claim 2: $\langle (fa_1^{l} a_0 a_1^{-l}f^{-1})^{fPf^{-1}}\rangle$ is free with basis $T_P= \{ta_0t^{-1}: t \in f'Pa_1^{l}\cap \ker \rho _L\}$ where $f'$ is the unique element of $\ker \rho _L$ such that $f=f'\rho _L(f)$.

The proof of the claim is very similar to the previous one. One has that $\langle T_P\rangle \leqslant \langle (fa_1^{l} a_0 a_1^{-l}f^{-1})^{fPf^{-1}}\rangle$ so it is enough to show that for any $g\in P$ the element

\begin{align*} (fgf^{{-}1})(fa_1^{l}a_0a_1^{{-}l}f^{{-}1})(fgf^{{-}1}) =fga_1^{l}a_0a_1^{{-}l}g^{{-}1}f^{{-}1} \end{align*}

lies in $\langle T_P\rangle$. Recall that from Equation (5.1) that a generating set of $P$ is

\begin{align*} \{a_1^{l}a_0a_1^{{-}l}\}\cup \{\sigma_b^{l(s,b)}b_0\sigma_b^{{-}l(s,b)}: b\in A_0 {\setminus} \{a_0\}\}. \end{align*}

In a similar way as before, we can write $g$ as

\begin{align*} g= c_1h_1(a_1^{l}a_0a_1^{{-}l})^{m_1}c_2h_2(a_1^{l}a_0a_1^{{-}l})^{m_2}\dots c_n h_n (a_1^{l}a_0a_1^{{-}1})^{m_n} \end{align*}

where $n\geq 0$, $c_i\in P\cap G_L$ and $h_i\in P\cap \ker \rho _L$ for $i=1,\,\dots n$. Let $f_L=\rho _L(f)$. Rewriting $f_Lc_1 \cdots c_i h_i c_i^{-1}\cdots c_1^{-1} f_L^{-1}$ as $h_i'$ and using that the $c_i$'s and $f_L$ commute with $a_0,\, a_1$, we get that

\begin{align*} fg =f' h_1'(a_1^{l}a_0a_1^{{-}l})^{m_1}h'_2(a_1^{l}a_0a_1^{{-}l})^{m_2}\dots h'_n (a_1^{l}a_0a_1^{{-}l})^{m_n}f_Lc_1c_2\cdots c_n. \end{align*}

Now notice that

\begin{align*} fga_1^{l}a_0 a_1^{{-}l}g^{{-}1}f^{{-}1} = f'g'a_1^{l}a_0 (f'g'a_1^{l})^{{-}1} \end{align*}

where

\begin{align*} g'= h_1'(a_1^{l}a_0a_1^{{-}l})^{m_1}h'_2(a_1^{l}a_0a_1^{{-}l})^{m_2}\dots h'_n. \end{align*}

Now we can write $f'g'a_1^{l}a_0 (f'g'a_1^{l})^{-1}$ as a product of elements of $T_{P}=\{ta_0t^{-1} : t\in \ker \rho _L \cap (f'Pa_1l)\}$.

\begin{align*} f'g'a_1^{l}a_0 (f'g'a_1^{l})^{{-}1} & = (f'h_1'a_1^{l}a_0 (f'h_1'a_1^{l})^{{-}1})^{m_1} \cdot ((f'h_1' h_2'a_1^{l}) a_0 (f'h_1'h_2'a_1^{l})^{{-}1})^{m_2} \cdot \\ & \quad \cdots ((f'h_1'\cdots h_{n-1}'a_1^{l}) a_0 (f'h'_{1}\cdots h'_{n-1}a_1')^{{-}1})^{m_{n-1}} \cdot \\ & \quad \cdot (f'h_1'\cdots h_n'a_1^{l}) a_0 (f'h_1'\cdots h_n'a_1^{l})^{{-}1} \cdot \\ & \quad \cdot ((f'h_1'\cdots h_{n-1}'a_1^{l}) a_0 (f'h'_{1}\cdots h'_{n-1}a_1^{l})^{{-}1})^{{-}m_{n-1}}\cdot \\ & \quad \cdots ((f'h_1' h_2'a_1^{l}) a_0 (f'h_1'h_2'a_1^{l})^{{-}1})^{{-}m_2}\cdot((f'h_1'a_1^{l})a_0 (f'h_1'a_1^{l})^{{-}1})^{{-}m_1} . \end{align*}

This completes the proof of Claim 2.

Now, if $\rho _{a_1}(fa_1^{l})\neq 0$, then $T_{A_0}\cap T_P=\emptyset$ and both are subsets of a free basis of $\ker \rho _D$. Therefore, $\langle T_{A_0}\rangle \cap \langle T_P\rangle = \{1\}$.

Case 3.2: $\rho _{a_1}(fa_1^{l}) = 0$. Note that $G_{A_0}\leqslant \ker \rho _{a_1}$ and $fPf^{-1}\leqslant \ker \rho _{a_1}$. As every $z\in \operatorname {Link}(a_1)$ commutes with $a_1$, we are in case (b) of § 4.1 and $\ker \rho _{a_1}$ is isomorphic to $G_\Lambda$ where $\Lambda$ is an even, FC-type, Artin graph (possibly infinite). Recall that

\begin{align*} V_\Lambda = \{b_{i,0}: b_i\in \operatorname{Link}_\Delta(a_1)\} \cup \{z_{i,j}: z_i\in V_\Delta {\setminus} \operatorname{Star}_\Delta(a_1), j\in \mathbb{Z}\} \end{align*}

and there is an edge $\{v_{i,j},\, u_{s,t}\}$ in $\Lambda$ if and only if there is and edge $\{v_i,\, u_s\}$ in $\Delta$ and the label of both edges is the same.

Let $A_{0,0}=\{b_{0,0}: b_0\in A_0\}$ be the vertices of $\Lambda$ of level $0$ and type $A_0$. Note that $G_{A_{0,0}}\leqslant G_\Lambda$ is the subgroup $G_{A_0}$ of $G_\Delta$ and the subgroup $G_{A}$ of $G_\Gamma$.

As $\rho _{a_1}(fa_1^{l}) = 0$, we have that $\rho _{a_1}(f)\neq 0$ (recall that $l\neq 0$). Write $f$ as $f'a_1^{\alpha }$, with $\alpha =\rho _{a_1}(f)\in \mathbb {Z}$. Consider the canonical retraction $\rho _{A_{0,0}}\colon G_\Lambda \to G_{A_{0,0}}$. Now, we have that $fPf^{-1}$ is equal to $f' P' (f')^{-1}$ where $P'$ is generated by

\begin{align*} \{a_{0,0} \}\cup \{\tau_{c}^{l(s,c)} a_1^{\alpha} c_{0} a_1^{-\alpha}\tau_{c}^{{-}l(s,c)} : c\in A {\setminus} \{a\} \} \end{align*}

where $\tau _{c}= a_1^{\alpha }\sigma _{c}a_1^{-\alpha }$ is some element of $\langle \{ v_{i,j} : v_{i,j}\textrm { of type }c_i\in V_\Delta \} \rangle$. Moreover, using Equation (4.1) in the setting of $\rho _{a_1}$ we have that

\begin{align*} a_1^{\alpha} c_{0} a_1^{-\alpha}= \beta_c^{l(\alpha,c)} c_{0,i(\alpha,c)} \beta_c^{{-}l(\alpha,c)} \end{align*}

for some word $\beta _c\in \langle \{ v\in V_{\Lambda }: \,$v$\textrm { of type } c_{0}\}\rangle$ and some $i(\alpha,\,c)\in \mathbb {Z}$. Observe that

\begin{align*} \rho_{A_{0,0}} (\tau_c^{l(s,b)}a_1^{\alpha} c_{0}a_1^{-\alpha}\tau_c^{{-}l(s,b)}) = \begin{cases} c_{0,0} & \text{if }i_{\alpha,c}=0\\ 1 & \text{otherwise}. \end{cases} \end{align*}

Recall that we are assuming $A\not \subseteq \operatorname {Link}(a)$ and therefore, there exists some $b\in A$ such that $b$ is not linked to $a$. Thus, $b_0$ is not linked to $a_i$, $i=0,\,1,\,\dots,\, k_a-1$ in $\Delta$. As $b_0$ is not linked to $a_1$, we have that $a_1^{\alpha } b_{0} a_1^{-\alpha }=b_{0,\alpha }$. And we get that $\rho _{A_{0,0}}(P') \leqslant G_{A_{0,0} {\setminus} \{b_{0,0}\}}.$ In particular $G_{A_{0,0}}\cap f'P'(f')^{-1}\leqslant \rho _{A_{0,0}}(f') G_{A_{0,0} {\setminus} \{b_{0,0}\}}\rho _{A_{0,0}}(f')^{-1}$. Note that $G_{A_{0,0} {\setminus} \{b_{0,0}\}}=G_{A_0 {\setminus} \{b_0\}}$. So, there is $d\in G_{\Delta }$ such that $\rho _{A_{0,0}}(f') G_{A_{0,0} {\setminus} \{b_{0,0}\}}\rho _{A_{0,0}}(f')^{-1}=d G_{A_0 {\setminus} \{b_0\}}d^{-1}$, and thus, $G_{A_0}\cap fPf^{-1}$ is contained in $dG_{A_0 {\setminus} \{b_0\}}d^{-1}$, a parabolic over a proper subset of $A_0$. This completes the proof in this case.

Proof of Theorem 1.1. Let ${\mathcal C}$ be the class of finite, even, FC-type Artin graphs. Then ${\mathcal C}$ is closed under taking subgraphs and satisfies (3.1) by Theorem 5.2. The theorem now follows from Proposition 3.8.

Proof. Let ${\mathcal P}$ be the set of parabolic subgroups in $G$. Note that as $\Gamma$ is finite, ${\mathcal P}$ is countable. For an arbitrary indexing set $I$, we want to show that:

\begin{align*} Q = \bigcap_{i\in I, P_i \in {\mathcal P}} P_i \end{align*}

is a parabolic subgroup. If $I$ is finite, the claim follows from Theorem 1.1 and induction. So, we can assume that the indexing set $I$ is countable, and we can index its elements by natural numbers. Write:

\begin{align*} \bigcap_{i\in I} P_i = \bigcap_{n \in \mathbb{N}} \left( \bigcap_{i \leq n} P_i \right), \end{align*}

and set $Q_n = \bigcap _{i \leq n} P_i$. We know that $Q_n$ is a parabolic subgroup for any $n$. Moreover, we have a chain of parabolic subgroups:

\begin{align*} Q_1 \supseteq Q_2 \supseteq Q_3 \supseteq \cdots \end{align*}

where the intersection of all members $Q_i$ of the chain above is equal to $Q$. We cannot have an infinite chain of nested distinct parabolic subgroups. Indeed, using Lemma 3.2, we have that $gG_Ag^{-1} \subsetneq hG_Ah^{-1}$ implies $A \subsetneq B$. Hence there are at most $|V| + 1$ distinct parabolic subgroups in the chain above.

Ultimately, $Q$ is an intersection of at most $|V| + 1$ parabolic subgroups and hence it is a parabolic subgroup.