Let $G$ be a group with subgroup $B$.
Suppose we have $\Delta=\{Pg : B\subseteq P \leq G , g \in G \}$ and we partially order this set by reverse inclusion meaning that $Pg \preceq P’g’$ if and only if $P’g’ \subseteq Pg $.
I am trying to show that for any two elements $Pg, Pg’$ of $\Delta$ that there is a greatest lower bound and it is the subgroup $P’’=\langle Pg, P’g’ \rangle $. I understand that here a greatest lower bound is a minimal coset in $\Delta$ that could rains $Pg$ and $P’g’$.
I can show that if $P_1g_1 $ is any lower bound of $Pg$ and $P’g’$ then we must have $\langle P, P’ \rangle \subseteq P_1 $.
Can’t really progress much from here. Is there some reason why the greatest lower bound must be a subgroup and not just some coset that isn’t a subgroup?
EDIT
The problem I’m actually trying to solve is showing that if $Pg, P’$ are elements of $\Delta$ then the greatest lower bound is $P’’=\langle Pg, P’ \rangle $ and then this apparently implies that any pair of elements of $\Delta$ have a greatest lower bound. Not sure how though.
If a coset in a group contains a subgroup, then it contains the identity element, and so it must be a subgroup. So a lower bound of $P'$ and $Pg$ is a subgroup containing both of them, and hence the greatest lower bound must be $\langle P',Pg \rangle$ (which contains $P$, so it is actually equal to $\langle P,P',g \rangle$).
It follows that for any $g' \in G$ the greatest lower bound of $P'g'$ and $Pgg'$ is $\langle P',Pg \rangle g'$, so any two cosets have a greatest lower bound: the greatest lower bound of $Pg$ and $P'g'$ is $\langle P,P',g'g^{-1} \rangle g = \langle P,P',g'g^{-1} \rangle g'$.