Consider a quasi-split reductive group $G$ over a field $k$. Let $B$ be a Borel subgroup of $G$, and let $P, Q$ be two parabolic subgroups of $G$ that contain $B$. Is the product set $PQ = \{xy| x \in P(k), y \in Q(k)\}$ a group?
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1$\begingroup$ I think that the unique possibilty is $P\subset Q$ or $Q\subset P$ and this should be a general property of groups with a $BN$ pair (or Tits system). Attached to your context is a Weyl group $W$ and and a distinguished generating subset $S\subset W$ of involutions. Any parabolic subgroup $M$ containing $B$ writes $M=B<T_M >B$ for a well determined subset $T_M$ of $S$. The game is to use the fact that $PQ$ is a group to prove that $S_P \subset S_Q$, or $S_Q\subset S_P$. For this use the properties of Tits systems for the product of cosets $(Bw_1 B).(Bw_2 B)$. $\endgroup$– Paul BroussousCommented Oct 19, 2015 at 11:42
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1$\begingroup$ @PaulBroussous: I suppose you mean that, if $PQ$ is a group, then either $P \subset Q$ or $Q \subset P$? (I misunderstood your first sentence on a first reading.) $\endgroup$– Tom De MedtsCommented Oct 19, 2015 at 13:25
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3$\begingroup$ @Paul Broussous: The OP does not specify that $G$ is simple, so I think there are other examples. For instance, for the split semisimple group $G=\textbf{SL}_2\times \textbf{SL}_2$ with Borel subgroup $B\times B$, then $P=\textbf{SL}_2\times B$ and $Q=B\times \textbf{SL}_2$ gives one example. $\endgroup$– Jason StarrCommented Oct 19, 2015 at 18:07
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1$\begingroup$ You're perfectly right. One needs to assume that the root system of $G$ is irreducible. $\endgroup$– Paul BroussousCommented Oct 20, 2015 at 8:17
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$\begingroup$ Thank you for the clarifications! I have a follow up question: In the above setting, could it instead be that the product $N_P N_Q$ of the unipotent parts of $P, Q$ is a group? $\endgroup$– azxCommented Oct 20, 2015 at 14:49
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