Group theory is one of the most important branches of abstract algebra which is concerned with the concept of the group. A group consists of a set equipped with a binary operation that satisfies four key properties: specifically, it includes property of closure, associativity, the existence of an identity element, and the existence of inverse elements. This framework leads to an understanding of mathematical structures and their invariance and such a concept finds its application in the fields of physics, chemistry, computer sciences, and cryptography.
What is Group Theory?
Group theory is a branch of abstract algebra that studies the algebraic structures known as groups. A group is a set equipped with a single binary operation that satisfies certain axioms. Group theory has profound applications in various fields, including physics, chemistry, computer science, and cryptography. Understanding the basic properties and concepts of groups provides a foundation for exploring more complex algebraic structures and their applications.
Group Definition
A set G together with a binary operation * (often called multiplication or addition) is called a group if it satisfies the following four properties:
Closure: For every pair of elements aaa and b in G, the result of the operation a * b is also in G.
∀ a, b ∈ G, a ∗ b ∈ G
Associativity: For every three elements a, b, and c in G, the equation (a * b) * c = a * (b * c) holds.
∀ a, b, c ∈ G, (a ∗ b) ∗ c = a ∗ (b ∗ c)
Identity Element: There exists an element e in G (called the identity element) such that for every element a in G, the equations e * a = a and a * e = a hold.
∃ e ∈ G such that ∀ a ∈ G, e ∗ a = a and a ∗ e = a
Inverses: For every element a in G, there exists an element b in G (called the inverse of a, often denoted a −1 ) such that a * b = e and b * a = e.
∀ a ∈ G, ∃ b ∈ G such that a ∗ b = e and b ∗ a = e
Note: A group is always a monoid, semigroup, and algebraic structure.
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).
Cyclic Group Z (Integers Modulo 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).
Quaternion Group
- Set: The set of quaternions {±1, ±i, ±j, ±k}.
- Operation: Quaternion multiplication.
- Identity Element: 1.
- Inverse Element: The inverse of each element is its negative or the corresponding quaternion such that their product is 1.
Dihedral Group D n
- Set: The set of symmetries of a regular n-sided polygon, including rotations and reflections.
- Operation: Composition of symmetries.
- Identity Element: The identity symmetry (doing nothing).
- Inverse Element: Each symmetry has an inverse symmetry that undoes it.
What is 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
Let's discuss these in detail as follows:
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.
Note: 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 belongs to S.
Note: 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 belongs to S.
- Identity Element: There exists e ? S such that a*e = e*a = a ? a ? S
Note: 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:
Set, Operation | Algebraic Structure | Semi Group | Monoid | Group | Abelian Group |
|---|
N, + | Y | Y | - | - | - |
N, - | - | - | - | - | - |
N, × | Y | Y | Y | - | - |
N, ÷ | - | - | - | - | - |
Z, + | Y | Y | Y | Y | Y |
Z, - | Y | - | - | - | - |
Z, × | Y | Y | Y | - | - |
Z, ÷ | - | - | - | - | - |
R, + | Y | Y | Y | Y | Y |
R, - | Y | - | - | - | - |
R, × | Y | Y | Y | - | - |
R, ÷ | - | - | - | - | - |
E, + | Y | Y | Y | Y | Y |
E, × | Y | Y | - | - | - |
O, + | - | - | - | - | - |
O, × | Y | Y | Y | - | - |
M, + | Y | Y | Y | Y | Y |
M, × | Y | Y | Y | - | - |
Where,
Summary: Group Theory
| Structure | Must Satisfy Properties |
|---|
| Algebraic Structure | Closure |
| Semi Group | Closure, Associative |
| Monoid | Closure, Associative, Identity |
| Group | Closure, Associative, Identity, Inverse |
| Abelian Group | Closure, Associative, Identity, Inverse, Commutative |
Solved Examples
Problem 1: Prove that in a group G, if (ab)² = a²b² for all a, b ∈ G, then G is abelian.
Solution:
Given: (ab)² = a²b² for all a, b ∈ G
Expand (ab)²: abab = 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.
Solution:
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.
Solution:
[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.
Solution:
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.
Solution:
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 on Group Theory
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.
6. Let G be a finite group and p be the smallest prime dividing |G|. Prove that any subgroup of index p in G is normal.
7. Let G be a group and a, b ∈ G. Prove that if a⁴ = b² and ab = ba, then (ab)⁶ = e.
8. Let G be a group and H be a subgroup of G. Prove that if [G : H] = n, then for any x ∈ G, xⁿ ∈ H.
9. Let G be a finite group and p be a prime number. If G has exactly one subgroup of order p^k for each k ≤ n, where pn divides |G|, prove that G has a normal sylow p-subgroup.
10. Let G be a finite group and H be a subgroup of G. Prove that if |G| = p where p is prime and p does not divide m, and |H| = pn, then H is normal in G.
Answer -
1. Abelian
2. True
3. False
4. True
5. True
6. True
7. True
8. True
9. True
10 .True
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Conclusion
Group theory is a powerful mathematical tool with wide-ranging applications in cryptography, signal processing, robotics, chemistry, and quantum mechanics. Understanding the basic properties and structures of groups allows for deeper insights into complex systems and enhances the ability to solve diverse problems in engineering and science.
Suggested Quiz
1 Questions
In a group G, if every element is its own inverse, then G is:
Explanation:
If every element a in G satisfies a2 = e, then for any two elements a and b, (ab)2 = e ⇒ abab = e
⇒ ab = b⁻¹a⁻¹ = ba, showing that the group is Abelian.
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