# Lattice: Examples

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?

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”. <div><div>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}

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## 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 $$H \vee K = HK = \{ hk \mid h \in H, k \in K \}$$.

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.

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