Lattice: Examples

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

A fa­mil­iar example

Con­sider the fol­low­ing lat­tice.

Suspicious Lattice Hasse Diagram

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”.

lattice 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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Parents:

  • Lattice (Order Theory)

    A poset that is closed un­der bi­nary joins and meets.

    • Order theory

      The study of bi­nary re­la­tions that are re­flex­ive, tran­si­tive, and an­ti­sym­metic.