Lower bounds of poset consisting of cosets

Question

Let $G$ be a group and let $B$ be a subgroup of $G$. Let $\Delta $ be the set of all right cosets where $\Delta = \{Pg: P \text{ is a subgroup of } G \text{ containing } B, g \in G \}$. We order $\Delta$ by reverse inclusion so we write $Pg \preceq P'g'$ if and only if $P'g' \subseteq Pg $.

Now given $Pg \in \Delta$ and subgroup $P' \in \Delta$ I have shown the the greatest lower bound of $Pg$ and $P'$ is $P''= \langle P', Pg \rangle$. The proof I am reading then goes on to say that this implies that any two elements of $\Delta$ has a greatest lower bound. How is this the case.

I thought maybe somehow using the greatest lower bound of $Pg$ and $P'$ and then taking the greatest lower bound of $P'g'$ and $P$ and somehow using this to give a greatest lower bound of $Pg$ and $P'g'$ but I don't think that works.

Any ideas here on what to do?

Answer 1

Let $Pg,P'g'\in \Delta$. We must find a greatest lower bound.

To do this, observe the following statements are equivalent:

1. $P''g''$ is a lower bound for $Pg$ and $P'g'$.
2. $P''g''g^{-1}$ is a lower bound for $P=Pgg^{-1}$ and $P'g'g^{-1}$.
3. $P''g''g^{-1}\supseteq H:=\langle P,P'g'g^{-1}\rangle$.
4. $P''g''\supseteq Hg$.

To see this, note that (1) and (2) are equivalent by the fact that right multiplication preserves inclusions, and (3) and (4) are equivalent for the same reason. (2) and (3) are equivalent by the fact that $H$ is the greatest lower bound for for $P$ and $P'g'g^{-1}$ under reverse inclusion, as you have already proved. Thus all four statements are equivalent.

But now the fact that (1) and (4) are equivalent says precisely that $Hg$ is the greatest lower bound for $Pg$ and $P'g'$.

(Keep in mind that we know $Hg$ is itself a lower bound since if we let $P''g''=Hg$, then (4) is trivially satisfied, hence so is (1).)

Answer 2

First we will prove the following claim: Let $X$ and $Y$ be cosets in $\Delta$, and suppose they have GLB $Z\in\Delta$. Then given $g\in G$, the GLB of $Xg$ and $Yg$ is $Zg$.

First of all, since $Z\supseteq X$ and $Z\supseteq Y$, we clearly have $Zg\supseteq Xg,Yg$, so that $Zg$ is indeed a lower bound for $Xg$ and $Yg$. Thus, it remains to show that $Zg$ is the greatest lower bound of $Xg$ and $Yg$, i.e., if $W\in\Delta$ contains $Xg$ and $Yg$, then $W$ contains $Zg$.

To see this, first of all, note that since $W\supseteq Xg$ and $W\supseteq Yg$, it follows that $Wg^{-1}$ contains $X$ and $Y$. Then since $Z$ is the GLB of $X$ and $Y$, we have that $Wg^{-1}\supseteq Z$. Thus, $W=Wg^{-1}g$ contains $Zg$, as desired.

Now that we have shown our claim, we will prove the desired statement. Let $Pg,P'g'\in\Delta$. Then first of all, by what you have shown, we know that the GLB of $P$ and $P'g'g^{-1}$ is $\langle P,P'g'g^{-1}\rangle$. Then it follows by the above claim that the GLB of $Pg$ and $P'g'=P'g'g^{-1}g$ is $\langle P,P'g'g^{-1}\rangle g$. We are done.