Group in Maths: Group Theory (original) (raw)

Last Updated : 4 Jun, 2026

A group is a basic concept from abstract algebra. It describes a set of elements together with an operation that combines any two elements to form another element in the same set, following certain rules.

A group is a set(G) equipped with a single binary operation( * ) that satisfies the following four properties:

**1. Closure: For every pair of elements a and b in G, the result of the operation a * b is also in G.

a ∗ b ∈ G ,∀ a, b ∈ G

**2. Associativity: For every three elements a, b, and c in G, the equation (a * b) * c = a * (b * c) holds.

(a ∗ b) ∗ c = a ∗ (b ∗ c) ,∀ a, b, c ∈ G

**3. Identity Element: There exists an element e in G (called the identity element) such that the given equation holds.

e ∗ a =a ∗ e = a , ∃ e ∈ G such that ∀ a ∈ G

**4. Inverses: For every in G, there exists an element a−1 in G (called the inverse of a ) such that

a ∗ a−1 = a−1 ∗ a = e , ∀ a ∈ G, ∃ a-1 ∈ G

Examples of Group

Some of the important examples of groups are discussed below:

**Integers under Addition (Z, +)

• Set: Z, the set of all integers {…, −3, −2, −1, 0, 1, 2, 3,…}.
• Operation: Addition (+).
• Identity Element: 0 (since n + 0 = n for any integer n).
• Inverse Element: For any integer n, the inverse is −n (since n + (−n) = 0).

**Non-Zero Real Numbers under Multiplication (R, ⋅)

• Set: R, the set of all non-zero real numbers.
• Operation: Multiplication (⋅).
• Identity Element: 1 (since n ⋅ 1 = n for any non-zero real number n).
• Inverse Element: For any non-zero real number n, the inverse is 1/n​ (since n * 1/n = n).

**Cyclic Group Z n

• Set: The set of integers (modulo n), {0, 1, 2, …, n−1}.
• Operation: Addition modulo n.
• Identity Element: 0 (since (a+0) mod n = a for any integer a).
• Inverse Element: For any integer a, the inverse is n−a (since (a+ (n−a)) mod n = 0 , only when a ≠ 0)

Algebraic Structure

An algebraic structure is a set of elements equipped with one or more operations that combine elements of the set in a specific way.

A non-empty set S is called an algebraic structure with a binary operation (∗) if it follows the closure axiom.

**Closure Axiom: For a, b in S, if (a*b) belongs to S, then S is closed for operation *.

Let's consider an example for better understanding.

• S = {1,- 1} is an algebraic structure under multiplication.
• Set of Integers; Z = {...,−3,−2,−1,0,1,2,3,...} with addition as well as multiplication.
• Set of non-zero real numbers (R, ∗) with multiplication.

Other algebraic structures related to groups are:

• Abelian Group
• Semi Group
• Monoid

Abelian Group or Commutative group

For a non-empty set S, (S, *) is called a Abelian group if it follows the following axiom:

• Closure: (a*b) belongs to S for all a, b ∈ S.
• Associativity: a*(b*c) = (a*b)*c where a ,b ,c belongs to S.
• Identity Element: There exists e ∈ S such that a*e = e*a = a where a ∈ S
• Inverses: For a ∈ S there exists a -1 ∈ S such that a*a -1 = a -1 *a = e
• Commutative: a*b = b*a for all a, b ∈ S

For finding a set that lies in which category one must always check axioms one by one starting from closure property and so on.

Every Abelian group is a group, monoid, semigroup, and algebraic structure.

Semi Group

A non-empty set S, (S,*) is called a semigroup if it follows the following axiom:

• Closure: (a*b) belongs to S for all a, b ∈ S.
• Associativity: a*(b*c) = (a*b)*c where a, b ,c∈ S.

A semi-group is always an algebraic structure.

**Example: (Set of integers, +), and (Matrix ,*) are examples of semigroup.

Monoid

A non-empty set S, (S,*) is called a monoid if it follows the following axiom:

• Closure: (a*b) belongs to S for all a, b ∈ S.
• Associativity: a*(b*c) = (a*b)*c ? a, b, c ∈ S.
• Identity Element: There exists e ∈ S such that a*e = e*a = a , a ∈ S

A monoid is always a semi-group and algebraic structure.

**Examples of Various Algebraic Structures

(Set of integers, *) is Monoid as 1 is an integer which is also an identity element. (Set of natural numbers, +) is not Monoid as there doesn’t exist any identity element. But this is Semigroup. But (Set of whole numbers, +) is Monoid with 0 as identity element.

Here is a Table with different nonempty set and operation:

Where,

• N: Set of Natural Number
• Z: Set of Integer
• R: Set of Real Number
• E: Set of Even Number
• O: Set of Odd Number
• M: Set of Matrix
• +, -, ×, ÷ are the operations

Solved Examples

**Problem 1: Prove that in a group G, if (ab)² = a²b² for all a, b ∈ G, then G is abelian.

Given: (ab)² = a²b² for all a, b ∈ G

Expand (ab)²: ab.ab = a²b²

Multiply both sides by a⁻¹ on the left: a⁻¹(abab) = a⁻¹(a²b²)

Simplify: bab = ab²

Multiply both sides by b⁻¹ on the right: (bab)b⁻¹ = (ab²)b⁻¹

Simplify: ba = ab

Therefore, G is abelian.

**Problem 2: Let G be a group of order 15. Prove that G is cyclic.

By Lagrange's theorem, the possible orders of elements in G are 1, 3, 5, and 15.

If there exists an element of order 15, then G is cyclic.

If not, then G has elements of order 3 and 5 (since it can't be all identity elements).

Let a be an element of order 3 and b be an element of order 5.

Consider the subgroup H = <a, b>. Its order must divide 15.

|H| ≠ 3 or 5 because it contains elements of both orders.

|H| ≠ 1 because it's not just the identity.

Therefore, |H| = 15, which means H = G.

By the fundamental theorem of finite abelian groups, G ≅ Z₃ ⊕ Z₅ ≅ Z₁₅.

Thus, G is cyclic.

**Problem 3: Let G be a group and H be a subgroup of G. Prove that if [G:H] = 2, then H is normal in G.

[G:H] = 2 means there are only two cosets of H in G.

These cosets are H itself and some aH where a ∉ H.

For any g ∈ G, gH must equal either H or aH.

If gH = H, then g ∈ H, so gHg⁻¹ = H.

If gH = aH, then g = ah for some h ∈ H.

In this case, gHg⁻¹ = ahH(ah)⁻¹ = aHa⁻¹

But aHa⁻¹ must be either H or aH.

If aHa⁻¹ = aH, then Ha⁻¹ = H, which means a ∈ H, contradicting our choice of a.

Therefore, aHa⁻¹ = H.

Thus, for all g ∈ G, gHg⁻¹ = H, so H is normal in G.

**Problem 4: Let G be a group and let a, b ∈ G. Prove that if ab = ba, then (ab)ⁿ = aⁿbⁿ for all n ∈ Z.

We'll use induction on n.

Base case: For n = 0, (ab)⁰ = e = a⁰b⁰

For n = 1, (ab)¹ = ab = a¹b¹

Inductive hypothesis: Assume (ab)ᵏ = aᵏbᵏ for some k ≥ 1

Inductive step: Consider (ab)ᵏ⁺¹

(ab)ᵏ⁺¹ = (ab)ᵏ(ab) = (aᵏbᵏ)(ab) = aᵏ(bᵏa)b = aᵏ(abᵏ)b = aᵏ⁺¹bᵏ⁺¹

By induction, (ab)ⁿ = aⁿbⁿ for all n ≥ 0

For negative integers, note that (ab)⁻ⁿ = ((ab)ⁿ)⁻¹ = (aⁿbⁿ)⁻¹ = b⁻ⁿa⁻ⁿ = a⁻ⁿb⁻ⁿ

Therefore, (ab)ⁿ = aⁿbⁿ for all n ∈ Z.

**Problem 5: Let G be a group of order pq, where p and q are distinct primes. Prove that G is cyclic if and only if gcd(p-1, q-1) = 1.

First, prove that if G is cyclic, then gcd(p-1, q-1) = 1:

If G is cyclic, it has an element of order pq.

By the structure theorem of cyclic groups, G ≅ Zₚq

The number of elements of order pq in Zₚq is φ(pq) = (p-1)(q-1)

This number must equal the number of generators of G, which is φ(pq)

For this to be true, we must have gcd(p-1, q-1) = 1

Now, prove that if gcd(p-1, q-1) = 1, then G is cyclic:

By Sylow's theorems, G has a unique subgroup P of order p and a unique subgroup Q of order q

Let a generate P and b generate Q

Consider the order of ab:

(ab)ᵖq = aᵖqbᵖq = (aᵖ)q(bq)ᵖ = e

So the order of ab divides pq

If ord(ab) = p or q, then ab ∈ P or ab ∈ Q, which is impossible

Therefore, ord(ab) = pq, and G is cyclic

Thus, G is cyclic if and only if gcd(p-1, q-1) = 1.

Practice Problems

1. Let G be a group of order pq, where p and q are distinct primes. Prove that G is abelian.

2. Prove that if G is a group of order p², where p is prime, then G is abelian if and only if it has p + 1 subgroups of order p.

3. Let G be a finite group and H be a proper subgroup of G. Prove that the union of all conjugates of H cannot be equal to G.

4. Let G be a group and N be a normal subgroup of G. If G/N is cyclic and N is cyclic, prove that G is abelian.

5. Prove that in any group G, the set of elements of finite order form a subgroup of G.

**Answer Key

1. Abelian
2. True
3. False
4. True
5. True

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