# Lattice: Examples

https://arbital.com/p/poset_lattice_examples

by Kevin Clancy Jul 7 2016 updated Jul 16 2016

Here are some additional examples of lattices. $\newcommand{\nsubg}{\mathcal N \mbox{-} Sub~G}$

## A familiar example

Consider the following lattice. Does this lattice look at all familiar to you? From some other area of mathematics, perhaps?

%%hidden(Reveal the truth):

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". %%comment:

Latex source:

\begin{tabular} {| c | c | c | c |}
\hline
$x$ & $y$ & $x \vee y$ & $x \wedge y$ \\ \hline
$a$ & $a$ & $a$ & $a$ \\ \hline
$a$ & $b$ & $b$ & $a$ \\ \hline
$b$ & $a$ & $b$ & $a$ \\ \hline
$b$ & $b$ & $b$ & $b$ \\ \hline
\end{tabular}

%%

%%

## Normal subgroups

Let $G$ be a group, and let $\nsubg$ be the set of all normal subgroups of $G$. Then $\langle \nsubg, \subseteq \rangle$ is a lattice where for $H, K \in \nsubg$, $H \wedge K = H \cap K$, and $H \vee K = HK = \{ hk \mid h \in H, k \in K \}$.

%%hidden(Proof):

Let $H,K \in \nsubg$. 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 $H, K \in \nsubg$, we have $H \vee K = HK = \{ hk \mid h \in H, k \in K \}$.

First we will show that $HK$ is a group. For $hk, h'k' \in HK$, since $kH = Hk$, there is some $h'' \in H$ such that $kh' = h''k$. Hence, $hkh'k' = hh''kk' \in HK$, and so $HK$ is closed under $G$'s group action. For $hk \in HK$, we have $(hk)^{-1} = k^{-1}h^{-1} \in k^{-1}H = Hk^{-1} \subseteq HK$, 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 $a^{-1}HKa = Ha^{-1}Ka = HKa^{-1}a = HK$.

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 $hk \not\in F$. But $h \in F$ and $k \in F$, and so from $F$'s group closure we conclude $hk \in F$, a contradiction.

%%