Definition 12. If s is a sublattice of L, L is an extension of s In some case, we have the following special sublattice Definition13. The subset s of the lattice l is called convex纩fa,b∈S,c∈L,anda≤c≤b imply that c∈S Given two lattices, they sometimes have the same topology structure by means of Hasse diagram. Consider the following example Example 6. Given a poset ((1, 2, 3, 6],), it is isomorphic to(P(a,b1),9) They have the Hasse diagrams as shown in Figure 4. It is easy to construct a bijection between b} (a)<{1,2,3,6},|> (b)<P({a,b}),g> Figure 4: Finite Boolean algebra example 1, 2, 3, 6) and P(a, b). And the mapping is not unique Lattice is a specail type of a partial order. Generally, we have the assertion Theorem 14. Given two lattice L and L, a bijection f: L,L froml to L is an isomorphism f and only if a≤ b in L implies f(a)≤f(b)in. Proof. If f is isomorphic, f is order preserving. Suppose a b, we have a Ub=b. Then we have ∫(a∪b)=f(a)∪f(b)=f(b), which means f(a)≤f(b). Conversely,f(a)≤f(b) also implies a≤b because f is also a bijection Suppose f is an order preserving bijection, f is isomorphic. Given a, bE L, let d= aUb, we have f(a)Uf(b)≤∫(d) because a,b≤ d leads f(a),f(b)≤f(d). For any f(e)=e∈ L such that f(a),f(b)≤e', we have a,b≤e. So aub=d≤e. Then we have f(d)≤f(e)=e, which means f(aUb)=f(a)∫(b). Similarly, we can prove f(an∩b)=f(a)∩f(b) Exercies 1. Let A be the set of finitely generated subgroups of a group G, ordered by set inclusion. Prove that(A, S) is a join-semilattice, but not necessarily a lattice 2. Show that every chain is a lattice 3. Prove that the absorption identities imply the idempotency of U and nDefinition 12. If S is a sublattice of L, L is an extension of S. In some case, we have the following special sublattice. Definition 13. The subset S of the lattice L is called convex if a, b ∈ S, c ∈ L, and a ≤ c ≤ b imply that c ∈ S. Given two lattices, they sometimes have the same topology structure by means of Hasse diagram. Consider the following example. Example 6. Given a poset h{1, 2, 3, 6}, |i, it is isomorphic to hP({a, b}), ⊆i. They have the Hasse diagrams as shown in Figure 4. It is easy to construct a bijection between 1 2 3 6 (a) < {1, 2, 3, 6}, | > ∅ {a} {b} {a, b} (b) < P({a, b}), ⊆> Figure 4: Finite Boolean algebra example {1, 2, 3, 6} and P({a, b}). And the mapping is not unique. Lattice is a specail type of a partial order. Generally, we have the assertion. Theorem 14. Given two lattice L and L 0 , a bijection f : L → L 0 from L to L 0 is an isomorphism if and only if a ≤ b in L implies f(a) ≤ f(b) in L 0 . Proof. If f is isomorphic, f is order preserving. Suppose a ≤ b, we have a ∪ b = b. Then we have f(a∪b) = f(a)∪f(b) = f(b), which means f(a) ≤ f(b). Conversely, f(a) ≤ f(b) also implies a ≤ b because f −1 is also a bijection. Suppose f is an order preserving bijection, f is isomorphic. Given a, b ∈ L, let d = a ∪ b, we have f(a) ∪ f(b) ≤ f(d) because a, b ≤ d leads f(a), f(b) ≤ f(d). For any f(e) = e 0 ∈ L 0 such that f(a), f(b) ≤ e 0 , we have a, b ≤ e. So a ∪ b = d ≤ e. Then we have f(d) ≤ f(e) = e 0 , which means f(a ∪ b) = f(a) ∪ f(b). Similarly, we can prove f(a ∩ b) = f(a) ∩ f(b). Exercies 1. Let A be the set of finitely generated subgroups of a group G, ordered by set inclusion. Prove that hA, ⊆i is a join-semilattice, but not necessarily a lattice. 2. Show that every chain is a lattice. 3. Prove that the absorption identities imply the idempotency of ∪ and ∩. 6