Question

Let $\Gamma (G;(G_i)_{i \in I})$ be a coset geometry (in the sense of Buekenhout) firm, residually connected and flag transitive with Borel subgroup $B$. Consider $H$ any subgroup of $B$. I want to find all the elements $\rho_i$ in the minimal parabolics subgroups $G_{I-{i}}$ such that $[ \langle H ,\rho_i \rangle : H] = 2$.

I have been given an idea: consider the quotient $G_{I-{i}}/B$ together with the canonical homomorphism $\phi_i : G_{I-{i}} \to G_{I-{i}}/B$. Then keep all the elements $\rho_i \in G_{I-{i}}-B$ such that $\phi_i(\rho_i)$ is an involution. In order to find them, take all involutions in $G_{I-{i}}/B$, then take their preimage by $\phi_i$ and it then suffices to multiply the preimage by the elements in the kernel of $\phi_i$ and keep those which normalize $H$ to get the elements satisfying the condition given above.

However, I have two problems. First, I don't really see why this works. Second is that I have no guarantee that $B$ is normal in each minimal parabolic subgroup. Worse, most of the time it is not. Hence, the set of right cosets of $B$ in $G_{I-{i}}$ does not have a group structure. Therefore, I have some problem with this solution.

Does anyone have a suggestion?

Thanks in advance.

Answer 1

First compute the normalizer $N_{P_i}(H)$ of $H$ in each minimal parabolic subgroup $P_i$ of $\Gamma$. Then consider the quotient $N_{P_i}(H)/H$ and the natural homomorphism $\varphi_i:N_{P_i}(H) \to N_{P_i}(H)/H$. Now define the set $$S_i := { \rho \in N_{P_i}(H) \mid (\varphi_i(\rho))^2 = 1 } $$ of elements of $N_{P_i}(H)$ each of which has an image by $\varphi_i$ in $N_{P_i}(H)/H$ that is an involution. We then have $[\langle H,\rho \rangle : H] = 2$ for any $\rho \in S_i$.

Indeed, since $(\varphi_i(\rho))^2=1$, one has $H\rho^2=H$ and since $\rho \in N_{P_i}(H)$, one has $\rho H = H \rho$. Now take $x \in \langle H,\rho \rangle$. One can write $x=h_1 \rho^{p_1} \dots h_k \rho^{p_k}$ for some $h_i \in H$ and some powers $p_i \in \mathbb{Z}$, where $i \in {1,\ldots, k}$. Let us consider $Hx= Hh_1 \rho^{p_1} \dots h_k \rho^{p_k}$. Using the fact that $\rho$ normalizes $H$, one gets $Hh_1 \rho^{p_1} \dots h_k \rho^{p_k} = H \rho^{p_1} \dots \rho^{p_k}$. One concludes quickly that $Hx=H$ or $Hx=H\rho$, i.e. there are only two cosets of $H$ in $\langle H,\rho \rangle$.