Group Theory and Abstract Algebra Notes

Group

• (G,*) is a group if it satisfies:
  • Closure: If a, b ∈ G, then a * b ∈ G.
  • Associativity: The operation * is associative.
  • Identity Element: There exists an identity element $e$ such that a * e = a for all a ∈ G.
  • Inverse Element: For every a ∈ G, there exists an inverse element $a^{-1}$ such that $a * a^{-1} = e$, where $e$ is the identity element.
• Abelian Group: A group that also satisfies the commutative property (a * b = b * a for all a, b ∈ G).

Order of a Group

• The number of elements in a group.
• Denoted by O(G) or |G|.

Order of an Element

• The smallest positive integer 'n' such that $a^n = e$, where 'e' is the identity element and the operation is multiplication.

*Example:
Let $G = {1, -1, i, -i}$ be a multiplicative group with identity element 1.

• O(G) = 4 since it consists of 4 elements.

• Order of element:

  • Need to take each element from the group.

  • Check how many times we need to multiply or raise the element to get the identity element, $e$.

  • In this case, the identity element $e = 1$.

  • Take the first element from the group: 1.

    • To get the identity element $e = 1$, how many times do we need to raise 1?
    • $a^n = e$
    • $1^n = 1$
    • $n = 1$
    • $O(1) = 1$

  • $a = -1$

  • $(-1)^n = e = 1$

  • $(-1)^2 = 1$

  • $n = 2$

  • $O(-1) = 2$

  • $a = i$

  • $i^n = 1$

  • $i^4 = 1$

  • $n = 4$

  • $O(i) = 4$

  • $a = -i$

  • $(-i)^n = 1$

  • $(-i)^2 = -1$

  • $(-i)^4 = 1$

  • $n = 4$

  • $O(-i) = 4$

Subgroup

• Let H be a non-empty subset of G. Then H is a subgroup of G if H itself is a group under binary operations.

Cyclic Group

• A group (G, *) is called a cyclic group if every element of G can be expressed as some power of a particular element a ∈ G, where a is called the generator of the group.

  *Example:
  In group $G = {1, -1, i, -i}$
  Taking element "i," different values can be formed: $i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1$, so i is a generator.

Addition Modulo m ($+_m$)

• Denoted by $a +_m b = r$, where $0 ≤ r < m$. This means $a + b$ divided by m, and the remainder will be r.

Multiplication Modulo m ($x_m$)

• $a x_m b = r$, where $0 ≤ r < m$

Example Problem 1

• Prove that the set consisting of numbers ${0, 1, 2, 3, 4, 5}$ is a finite abelian group of order 6 under ordinary addition reduced modulo 6 as composition.

*Solution:

• To prove it's an abelian group, we need to prove the following:

  • Closure property
  • Associative property
  • Identity
  • Inverse
  • Commutative

*Table is created where columns contains 0 to n-1 no's. Here addition modulo 6 so, 0 to n-1 = 0 to 6-1. Hence 0 to 5 columns exist.
*Take now no in column no here addition modulo 6

*row + column/6

*Now check for all properties if elements belong to set {0,1,2,3,4,5} for closure
*Associative, (a+m b) +m c = a+m (b+m c)
*Identity, a+e = a
*Inverses a+a' = e
*Commutative a+b = b+a

Example Problem 2

• Let G be the set of all integers defined by the binary composition $a * b = a + b + 1$. Find whether G is a group or not.

*Solution:

*   Closure:
    *   $a, b ∈ G  => a * b = a + b + 1$
    *   Since both a and b are integers, and if we add 1 to it, then it will be an integer.
    *   $a + b + 1 ∈ G$
    *   Therefore, it satisfies the closure property.
*   Associative:
    *   $a * (b * c) = (a * b) * c$
    *   $a * (b + c + 1) = (a + b + 1) * c$
    *   $a + b + c + 1 + 1 = a + b + 1 + c + 1$
    *   $a + b + c + 2 = a + b + c + 2$
    *   Hence associative property holds.
*   Identity:
    *   $a * e = a$
    *   $a + e + 1 = a$
    *   $e + 1 = 0 => e = -1$
    *   $e = -1$ is the identity element.
*   Inverse:
    *   $a * a' = e$
    *   $a + a' + 1 = e$
    *   Substitute $e = -1$ that we got.
    *   $a + a' + 1 = -1$
    *   $a' = -a - 2$
*   Therefore, G is a group.

Example Problem 3

• Let
  Given the matrices, show that G is a group under multiplication.

*Solution:

• When matrices come, we need to draw a Cayley table.

• Identity matrix

  • To prove that it is a group, all the properties of closure, associativity, identity, and inverse should be satisfied. (Refer to table)

Theorem 1

*For group G:
* Identity of G is unique.
* Inverse of each element is unique.
* If a, b, c ∈ G and ab = ac, then b = c (and if ba = ca, then b = c).

Theorem 2

• Prove that the inverse of the products of two elements of a group G is the product of the inverses in reverse order.
• $(ab)^{-1} = b^{-1}a^{-1}$

Subgroup Theorems

*If G is a group off:
* a∈H, b∈H => ab∈H
* a∈H => a^-1∈H

Theorem 3

A non empty subset his a subgroup of G, If and only if $ab^{-1}$

Example Problem 4

• If every element is of order 2, then prove $G$ is abelian. ($a^2=e$)

*Solution:

• Given: every element is of order 2.

  • $n = 2$ ( $a^n = e$ )
  • $a^2 = e$
  • $b^2 = e$
  • Then $ab$ ∈ G.
  • Squaring, $(ab)^2 = e$
  • $(ab)(ab) = e$
  • $a(ba)b = e$
  • Multiplying both sides by $(ab)$
  • $[a(ba)b]ab = e(ab)$
  • $(aa)(ba)(bb) = ab$
  • $ebaeb = ab$
  • $ba = ab$

• Therefore: G is abelian.

Example Problem 5

*Prove that a cyclic subgroup G is cyclic.

*Solution:
*Cyclic group generated by a.
*Every element represented has power is n

Semigroup

• Algebraic system (S, *) is a semigroup if:
  • S is a non-empty set
  • * is a binary operation
  • * is associative
  • If * is commutative, then it is an abelian semigroup.

Monoid

• (M, *) is a semigroup with an identity element 'e'. Hence, can be written as (M, *, e).
  • * is associative
  • Element 'e' exists such that $a * e = e * a = a$
  • If * is commutative, then it's an abelian monoid. *Example:
    • To show it is an abelian semigroup, we need to prove associative, commutative and have identity element.