Lattice: Examples

In fact, this lattice corresponds to boolean logic, as can be seen when we replace b with true and a with false in the following "truth table".

Let $H,K\in N\text{-}SubG$. Then $H\wedge K=H\cap K$. We first note that $H\cap K$ is a subgroup of $G$. For let $a,b\in H\cap K$. Since $H$ is a group, $a\in H$, and $b\in H$, we have $ab\in H$. Likewise, $ab\in K$. Combining these, we have $ab\in H\cap K$, and so $H\cap K$ is satisfies the closure requirement for subgroups. Since $H$ and $K$ are groups, $a\in H$, and $a\in K$, we have $a^{-1}\in H$ and $a^{-1}\in K$. Hence, $a^{-1}\in H\cap K$, and so $H\cap K$ satisfies the inverses requirement for subgroups. Since $H$ and $K$ are subgroups of $G$, we have $e\in H$ and $e\in K$. Hence, we have $e\in H\cap K$, and so $H\cap K$ satisfies the identity requirement for subgroups.

Furthermore, $H\cap K$ is a normal subgroup, because for all $a\in G$, $a^{-1}(H\cap K)a=a^{-1}Ha\cap a^{-1}Ka=H\cap K$. It's clear from the definition of intersection that $H$ and $K$ do not share a common subset larger than $H\cap K$.

For $H,K\in N\text{-}SubG$, we have $H\vee K=HK=\{hk\mid h\in H,k\in K\}$.

First we will show that $HK$ is a group. For $hk,h^{\prime }{k}^{\prime }\in HK$, since $kH=Hk$, there is some $h^{\prime \prime }\in H$ such that $k{h}^{\prime }=h^{\prime \prime }k$. Hence, $hk{h}^{\prime }{k}^{\prime }=h{h}^{\prime \prime }k{k}^{\prime }\in HK$, and so $HK$ is closed under $G$'s group action. For $hk\in HK$, we have $(hk{)}^{-1}=k^{-1}{h}^{-1}\in k^{-1}H=H{k}^{-1}\subseteq HK$, and so $HK$ is closed under inversion. Since $e\in H$ and $e\in K$, we have $e=ee\in HK$. Finally, $HK$ inherits its associativity from $G$.

To see that $HK$ is a normal subgroup of $G$, let $a\in G$. Then $a^{-1}HKa=H{a}^{-1}Ka=HK{a}^{-1}a=HK$.

There is no subgroup $F$ of $G$ smaller than $HK$ which contains both $H$ and $K$. If there were such a subgroup, there would exist some $h\in H$ and some $k\in K$ such that $hk\notin F$. But $h\in F$ and $k\in F$, and so from $F$'s group closure we conclude $hk\in F$, a contradiction.