Groups and Monoids
Short Notes
Algebraic structures consisting of a non-empty set combined with one or more binary operations.
- Closure: \(a * b \in S\).
- Associativity: \((a * b) * c = a * (b * c)\).
- Identity: \(\exists e\) such that \(a * e = e * a = a\).
- Inverse: \(\exists a^{-1}\) such that \(a * a^{-1} = e\).
Hierarchy
- Semigroup: Closure + Associativity.
- Monoid: Semigroup + Identity.
- Group: Monoid + Inverse.
- Abelian Group: Group + Commutative (\(a * b = b * a\)).
Key Theories & Formulas
1. Group Properties
- In a group, the identity element is unique.
- Every element has a unique inverse.
- Idempotent Law: The only idempotent element (\(x*x=x\)) in a group is the identity \(e\).
- Cancellation Law: \(a * b = a * c \implies b = c\) (holds in groups).
2. Subgroups & Order
- Order of Group \(|G|\): Number of elements in \(G\).
- Order of Element \(|a|\): Smallest \(n\) s.t. \(a^n = e\).
- Lagrange's Theorem: Order of a subgroup \(H\) divides the order of group \(G\). (\(|H|\) divides \(|G|\)).
- Cyclic Group: A group generated by a single element (\(G = \langle a \rangle\)). Every cyclic group is Abelian.
- Number of generators of a cyclic group of order \(n\) is \(\phi(n)\) (Euler's Totient Function).
Example Problems
Problem: Find the order of element 3 in the group \((\mathbb{Z}_7^*, \times_7)\).
- \(\mathbb{Z}_7^* = \{1, 2, 3, 4, 5, 6\}\). Identity is 1.
- \(3^1 \equiv 3 \pmod 7\)
- \(3^2 \equiv 9 \equiv 2 \pmod 7\)
- \(3^3 \equiv 2 \times 3 \equiv 6 \pmod 7\)
- \(3^4 \equiv 6 \times 3 \equiv 18 \equiv 4 \pmod 7\)
- \(3^5 \equiv 4 \times 3 \equiv 12 \equiv 5 \pmod 7\)
- \(3^6 \equiv 5 \times 3 \equiv 15 \equiv 1 \pmod 7\)
- Order is 6. This means 3 is a generator for \(\mathbb{Z}_7^*\).
Hardest GATE Questions
Topic: Number of Subgroups
Tricky Question: How many subgroups does a group \(G\) of order 6 have, if \(G\) is non-abelian?
- Analysis:
- A non-abelian group of order 6 must be isomorphic to \(S_3\) (Symmetric group of degree 3).
- Elements: \(\{e, (12), (13), (23), (123), (132)\}\).
- Orders:
- Order 1: \(\{e\}\) (1 subgroup)
- Order 2: \(\{(12)\}, \{(13)\}, \{(23)\}\). Each generates a subgroup of size 2. (3 subgroups)
- Order 3: \(\{(123), (132)\}\). Together with \(e\), they make a subgroup of size 3. (1 subgroup)
- Lagrange's Theorem says subgroup sizes can be 1, 2, 3, 6.
- Subgroup of order 6 is \(G\) itself. (1 subgroup)
- Total: \(1 + 3 + 1 + 1 = 6\) subgroups.
- Why it's hard: Requires knowing the structure of \(S_3\) and exhaustively counting based on element orders.