Algebraic Structures: Semigroups, Monoids, Groups & Subgroups
“Group axioms, abelian groups, permutation groups, subgroup criteria, and Lagrange's coset theorem.”
1.The Hierarchy of Algebraic Structures
§ 3.1.An Algebraic Structure consists of a non-empty set G paired with one or more binary operations *:
1. Groupoid (Magma): Closed under (forall a, b in G, a b in G).
2. Semigroup: Groupoid that satisfies Associativity: (a b) c = a (b c).
3. Monoid: Semigroup that has an Identity element e in G such that a e = e a = a.
4. Group: Monoid where every element a in G has a unique Inverse a^{-1} in G such that a a^{-1} = a^{-1} a = e.
5. Abelian (Commutative) Group: Group that also satisfies a b = b a for all a, b in G.
2.Subgroups, Cosets & Lagrange's Theorem
§ 3.2.A subset H subseteq G is a Subgroup (H leq G) if H itself forms a group under the operation of G.
One-Step Subgroup Test: H is a subgroup iff forall a, b in H, a * b^{-1} in H.
Cosets: For a subgroup H leq G and element g in G:
- Left Coset: gH = { g * h : h in H }.
- Right Coset: Hg = { h * g : h in H }.
The left cosets of H form an exact partition of G into disjoint subsets, each of cardinality |H|.
Lagrange's Theorem:
If G is a finite group and H is a subgroup of G, then |H| divides |G|:
|G| = [G : H] * |H|
where [G : H] is the index of H in G (the number of distinct cosets).
3. Worked Problem: Subgroup Order Analysis using Lagrange Theorem
Possible subgroup orders must be divisors of |G|. Non-divisors can never be subgroup orders.
Problem: A group G has order |G| = 30. What are all possible orders of non-trivial proper subgroups of G? Can G have a subgroup of order 7?
Step 1: State Lagrange's theorem condition:
If H is a subgroup of G, then |H| must divide |G| = 30.
Step 2: Find all positive divisors of 30:
Divisors of 30 = {1, 2, 3, 5, 6, 10, 15, 30}.
Step 3: Identify non-trivial proper subgroups:
- Trivial subgroup order = 1 ({e}).
- Improper subgroup order = 30 (G itself).
- Non-trivial proper subgroup orders are divisors strictly between 1 and 30:
Possible orders = {2, 3, 5, 6, 10, 15}.
Step 4: Check subgroup of order 7:
Since 7 does NOT divide 30 (30 % 7 = 2 != 0), by Lagrange's theorem, G CANNOT have any subgroup of order 7.
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