Lattice: Examples

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Written by Kevin Clancy last updated 17th Jul 2016

Requires: Lattice (Order Theory)

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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?

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

& & & \ \hline & & & \ \hline & & & \ \hline & & & \ \hline & & & \ \hline \end{tabular}

Normal subgroups

Let

G

be a group, and let be the set of all normal subgroups of . Then is a lattice where for , , and .

Proof

Let

H,K \in \nsubg

. Then . We first note that is a subgroup of . For let . Since is a group, , and , we have . Likewise, . Combining these, we have , and so is satisfies the closure requirement for subgroups. Since and are groups, , and , we have and . Hence, , and so satisfies the inverses requirement for subgroups. Since and are subgroups of , we have and . Hence, we have , and so satisfies the identity requirement for subgroups.

Furthermore,

H \cap K

is a normal subgroup, because for all , . It's clear from the definition of intersection that and do not share a common subset larger than .

For

H, K \in \nsubg

, we have .

First we will show that

HK

is a group. For , since , there is some such that . Hence, , and so is closed under 's group action. For , we have , and so is closed under inversion. Since and , we have . Finally, inherits its associativity from .

To see that

HK

is a normal subgroup of , let . Then .

There is no subgroup

F

of smaller than which contains both and . If there were such a subgroup, there would exist some and some such that . But and , and so from 's group closure we conclude , a contradiction.

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Here are some additional examples of lattices.

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?

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".

lattice truth table

Normal subgroups

Let $G$ be a group, and let $N - S u b G$ be the set of all normal subgroups of $G$ . Then $⟨ N - S u b G, \subseteq ⟩$ is a lattice where for $H, K \in N - S u b G$ , $H \land K = H \cap K$ , and $H \lor K = H K = {h k ∣ h \in H, k \in K}$ .

Proof

Let $H, K \in N - S u b G$ . Then $H \land K = H \cap K$ . 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 $a b \in H$ . Likewise, $a b \in K$ . Combining these, we have $a b \in H \cap K$ , 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 $a \in G$ , $a^{- 1} (H \cap K) a = a^{- 1} H a \cap a^{- 1} K a = H \cap K$ . It's clear from the definition of intersection that $H$ and $K$ do not share a common subset larger than $H \cap K$ .

For $H, K \in N - S u b G$ , we have $H \lor K = H K = {h k ∣ h \in H, k \in K}$ .

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

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

There is no subgroup $F$ of $G$ smaller than $H K$ 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 $h k \notin F$ . But $h \in F$ and $k \in F$ , and so from $F$ 's group closure we conclude $h k \in F$ , a contradiction.

Parents:

Lattice (Order Theory)