Binary operations and stable parts
A binary operation (also called an internal composition law) on a set M is a rule ∗ that takes any two elements a, b of M and gives exactly one element a ∗ b also in M.
- + and × on the integers Z are binary operations.
- Subtraction on the natural numbers N is not: 2 − 5 = −3 is not in N.
- a ∗ b = a + b − ab on R is a binary operation (you can always compute it).
A subset H of M is a stable part (closed) if a, b ∈ H always gives a ∗ b ∈ H. Example: the even numbers are stable under +; the odd numbers are not (3 + 5 = 8).
Useful properties of ∗: commutative (a ∗ b = b ∗ a), associative ((a ∗ b) ∗ c = a ∗ (b ∗ c)), identity e (a ∗ e = e ∗ a = a), inverse a′ (a ∗ a′ = a′ ∗ a = e). For a small set we show ∗ in an operation table (Cayley table): row a, column b, cell a ∗ b.
What is a group? Examples
A group (G, ∗) is a set G with a binary operation that is closed, associative, has an identity, and every element has an inverse. If also commutative, the group is abelian.
Number groups
- (Z, +), (Q, +), (R, +), (C, +): identity 0, inverse −a.
- (Q*, ×), (R*, ×): identity 1, inverse 1/a. (Z, ×) is not a group: 2 has no integer inverse.
Zn, the clock groups
Zn = {0, 1, …, n − 1} with addition mod n is an abelian group of order n. The non-zero classes of Zp (p prime) form a group under multiplication.
Matrix groups
Invertible 2×2 matrices (det ≠ 0) form a group under matrix multiplication. It is not abelian: AB ≠ BA in general.
Permutation groups
All rearrangements of {1, 2, 3} form S3, with 3! = 6 elements and the operation "do one, then the other" (composition). S3 is the smallest non-abelian group. The symmetries of an equilateral triangle (3 rotations, 3 flips) behave exactly like S3.
Subgroups, order of an element and Lagrange's theorem
A subgroup H of G is a non-empty subset that is itself a group with the same operation. Quick test: H ≠ ∅ and a, b ∈ H ⇒ a ∗ b′ ∈ H.
The order of an element a is the smallest k ≥ 1 with a ∗ a ∗ … ∗ a (k times) = e. In Zn with +, the order of a is n / gcd(a, n). The powers of a make the cyclic subgroup ⟨a⟩.
Lagrange's theorem: in a finite group, the order of every subgroup divides the order of the group. So the order of every element divides |G|, and a|G| = e. A group of prime order p is cyclic.
Example: in Z6, ⟨2⟩ = {0, 2, 4} (order 3), ⟨3⟩ = {0, 3} (order 2), and 1 and 5 generate all of Z6.
Morphisms and isomorphisms
A group morphism (homomorphism) f : (G, ∗) → (H, ∘) keeps the operation: f(a ∗ b) = f(a) ∘ f(b). Then f(e) = e′ and f(a′) = f(a)′.
An isomorphism is a bijective morphism. Isomorphic groups have the same table, just with renamed elements. Example: f(x) = ln x is an isomorphism from (0, ∞) with × to (R, +), because ln(ab) = ln a + ln b. Another: the rotations of a square ≅ Z4.
To show two finite groups are not isomorphic, find a property one has and the other does not: different orders, one abelian and one not, or different numbers of elements of each order.
Rings and fields
A ring (A, +, ·) has two operations: (A, +) is an abelian group, · is associative, and · distributes over +. Usually it also has a unit 1. Examples: Z, Q, R, C, Zn, the ring of n×n matrices, the ring of real functions.
A field is a commutative ring where every non-zero element has a multiplicative inverse. Examples: Q, R, C and Zp for p prime. Z is not a field (2 has no inverse). Z6 is not a field: 2 · 3 = 0, so it has zero divisors.
A ring morphism keeps both operations: f(a + b) = f(a) + f(b), f(ab) = f(a)f(b), f(1) = 1. Example: complex conjugation on C.
Key formulas and definitions
- Group axioms: closure, associativity, identity e, inverse a′ (CAIN)
- Abelian: a ∗ b = b ∗ a for all a, b
- Zn: a + b mod n; order of a = n / gcd(a, n)
- Subgroup test: H ≠ ∅, a, b ∈ H ⇒ a ∗ b′ ∈ H
- Lagrange: |H| divides |G|; a^|G| = e
- Morphism: f(a ∗ b) = f(a) ∘ f(b); isomorphism = bijective morphism
- |Sn| = n!
- Field: commutative ring, every a ≠ 0 has a⁻¹; Zn is a field ⇔ n is prime
Worked examples
1. Is (Z, ∗) a group, where a ∗ b = a + b − 3?
Closed: yes, integers. Associative: (a∗b)∗c = a + b + c − 6 = a∗(b∗c). Identity: a + e − 3 = a ⇒ e = 3. Inverse: a + a′ − 3 = 3 ⇒ a′ = 6 − a, an integer. So yes, it is an abelian group with identity 3.
2. Find the order of every element of Z8 (addition).
Order of a = 8 / gcd(a, 8). 0 → 1; 1, 3, 5, 7 → 8; 2, 6 → 4; 4 → 2. All divide 8, as Lagrange says. The generators are 1, 3, 5, 7.
3. Can a group of order 10 have a subgroup of order 4?
No. By Lagrange the order of a subgroup must divide 10. 4 does not divide 10.
4. Solve 3x = 4 in Z7 (multiplication mod 7).
Find 3⁻¹: 3 · 5 = 15 = 1 (mod 7), so 3⁻¹ = 5. x = 5 · 4 = 20 = 6 (mod 7). Check: 3 · 6 = 18 = 4. So x = 6.
5. Show f : (R, +) → (0, ∞), f(x) = 2^x is an isomorphism with ×.
f(x + y) = 2^(x+y) = 2^x · 2^y = f(x) f(y), so it is a morphism. It is injective (2^x is strictly increasing) and surjective (every positive y is 2^(log₂ y)). So it is an isomorphism.
6. In S3, let σ swap 1 and 2, and τ swap 2 and 3. Is στ = τσ?
στ (do τ first): 1→1→2, 2→3→3, 3→2→1, so στ = (1 2 3). τσ: 1→2→3, 2→1→1, 3→3→2, so τσ = (1 3 2). They are different, so S3 is not abelian.
Common mistakes
- Forgetting to check closure first: subtraction on N fails before you even look for an identity.
- Using the usual identity (0 or 1) for a new operation. Always solve a ∗ e = a for the given rule.
- Thinking every group is commutative. Matrix groups and S3 are not.
- Treating Zn as a field for every n. In Z6, 2 · 3 = 0, so 2 has no inverse; only prime n gives a field.