# Abstract Algebra – Group, Subgroup, Abelian group, Cyclic group

Abstract Algebra / Thursday, September 19th, 2019

# Abstract Algebra – Group, Subgroup, Abelian group, Cyclic group

Here in this post we will discuss about group, subgroup, abelian group, cyclic group and their properties.

## Group

A non-empty set G is said to form a group with respect to an operation o, if
G is closed under the operation i.e. a o b ∈ G for all a, b ∈ G [groupoid]
o is associative i.e. a o (b o c) = (a o b) o c for all a, b, c ∈ G [semigroup]
There exists an identity element e in G such that a o e = a for all a ∈ G [monoid]
For each element a in G, there exists an inverse element a-1 in G such that a o a-1 = e Example

The set of all n x n matrices under the operation of matrix multiplication is not a group since not every n x n matrix has its multiplicative inverse but if G is the set of all n x n non-singular matrices, then G forms a group under the operation of matrix multiplication.

## Some elementary properties of groups

1. The identity element e is unique in a group (G, o).
2. The inverse a-1 of the element a ∈ G is unique.
3. For every a ∈ G, (a-1)-1 = a.
4. For all a, b ∈ G, (a o b)-1 = b-1 o a-1.
5. The cancellation laws
a o b = a o c  ⇒ b = c [left cancellation law]
b o a = c o a  ⇒ b = c [right cancellation law] hold in G; for all a, b, c ∈ G
6. For a, b ∈ G, the linear equation a o x = b and y o a = b have unique solution in G

## Abelian group

A group (G, o) is said to be abelian or commutative if o is commutative i.e. a o b = b o a for all a, b ∈ G.

 Example

Show that the set of all rational numbers is an abelian group with respect to addition.

Solution:

Let us consider the set of all rational numbers Q with the binary operation addition. Then for any a, b, c ∈ Q we have,
i) a + b ∈ Q, as the sum of rational numbers is rational.
ii) a + (b + c) = (a + b) +c
iii) a + 0 = 0 + a = a for all a ∈ Q, so 0 ∈ Q is the identity element.
iv) a + (-a) = (-a) + a = 0 for all ∈ Q so the inverse element (-a) exists for each element a ∈ Q.
Hence the set of all rational numbers form a group with respect to addition, i.e. (Q, +) is a group.
Again a + b = b + a for all a, b ∈ Q so (Q, +) is an abelian group.

## Order of group

The order of an element a in a group G is the smallest positive integer n such that an = e, the identity element in G. Here, by an we understand a o a oo a (n factors). If there is no finite integer n such that an = e, we say that the element a has order 0. If an element a ∈ G has order n, we write O(a) = n.

The order of a group is the number of elements in the group. Thus, if a group G has n elements, then G is said to be finite group of order n, and we write O(G) = n.

 Example

If G is a group of even order, prove that it has an element a ≠ e satisfying a2 = e.

Solution:

Suppose that G is a group of order 2n, n being a positive integer. Let a1, a2, …, a2n – 1, e be the elements of G. Since in a group every element has its unique inverse and since there are odd number of elements a1, a2, …, a2n – 1 none of which is the identity element of G, it follows that there is one element a (say) among a1, a2, …, a2n – 1 whose inverse is a itself. Then it follows that a o a =e i.e. a2 = e, a ∈ G and a ≠ e.

 Example

Show that if every element of the group G is its own inverse, then G is abelian.

Solution:

Let a, b ∈ G, then a o b ∈ G. From the given condition a-1 = a, b-1 = b, (a o b)-1 = (a o b)
But (a o b)-1 = b-1 o a-1, so that a o b = b-1 o a-1
Or, a o b = b o a.
Hence G is abelian.

## Subgroup

Let (G, o) be a group and H be a non-empty subset of G. If (H, o) is a group where o is the induced composition, the (H, o) is said to be a subgroup of (G, o).

 Example

(R, +) is a group. Z is a non-empty subset of R and (Z, +) is a group. Therefore, (Z, +) is a subgroup of (R, +).

 Theorem

Let (G, o) be a group. A non-empty subset H of G forms a subgroup of (G, o) if and only if a ∈ H, b ∈ H ⇒ a o b-1H.

Proof:

Let (H, o) be a subgroup of (G, o).
Since (G, o) is a group, b ∈ H ⇒ b-1H and therefore a ∈ H, b ∈ H ⇒ a o b-1H.
Conversely,
Let H be a non-empty subset of G such that a ∈ H, b ∈ H ⇒ a o b-1H.
Let a ∈ H then a ∈ H, a ∈ H ⇒ a o a-1 = eH. Therefore H contains identity element.
Now eH, a ∈ He o a-1 = a-1H. Hence a ∈ H ⇒ a-1H. Therefore the inverse of each element in H exists in H.
Let a ∈ H, b ∈ H then a ∈ H, b-1H and by the given condition a o (b-1)-1 = a o b ∈ H. Hence a ∈ H, b ∈ H ⇒ a o b ∈ H, therefore H is closed under o.
Since H is a non-empty subset of G and o is associative on G, o is associative on H.
Therefore, (H, o) is a group and hence (H, o) is a subgroup of (G, o).

## Finite group

A group (G, o) is said to be a finite group if G contains a finite number of elements.

## Cyclic group

A group (G, o) is said to be a cyclic group if there exists an element a in G such that G = {an : n ∈ G} i.e. G = <a>, a is said to be a generator of the cyclic group.

 Example

(Z, +) is cyclic group generated by 1, -1.
Let S = {1, i, -1, –i}. Then (S, .) is a cyclic group generated by i, –i.

 Theorem

Every cyclic group is abelian.

Proof:

Let (G, o) is a cyclic group, generated by a.
Let p, q ∈ G then p = ar, q = as for some integer r and s.
p o q = ar o as = ar + s
q o p = as o ar = as + r
Since r + s = s + r,  p o q = q o p for all p, q ∈ G.
Therefore the group is abelian.

## Left Coset

Let G be a group and H be subgroup of G. Let a be an element of G for all h ∈ H, ah ∈ G. The subset { ah : h ∈ H } is called a left coset of H in G and is denoted as aH.

## Right Coset

Let G be a group and H be subgroup of G. Let a be an element of G for all h ∈ H, ha ∈ G. The subset { ha : h ∈ H } is called a right coset of H in G and is denoted as Ha.

It may easily be shown that any two right (or left) cosets of H in G are either identical or have no element in common.

 Theorem

(Lagrange) If G is a finite group and H is subgroup of G, then O(H) is a divisor of O(G).

 Theorem

Order of a cyclic group is equal to the order of its generator.

 Theorem

A subgroup of a cyclic group is cyclic.