Introduction to group theory/Part 2 Subgroups and cyclic groups

Introduction

We will learn briefly about subgroups and cyclic groups. As per the name suggest "sub" means a part of i.e; heading-subheading ,set- subset etc etc.

So subgroup is a part of a group which forms under the same binary operation as per the group. [ Note: all elements of subgroup must be elements of groups]

Definition

A non-empty subset H of a group G is said to be subgroup of G if , under the product in G, H itself forms a group.

i.e; $H\subset G$

It should follow all the properties of groups.

• Closure
• Associative
• Indentity
• Inverse
• Commutative (optional)

The following remark is clear: if H is a subgroup of G and K is a subgroup of H , then K is a subgroup of G.

LEMMA

A non empty subset H of the group G is a subgroup of G if and only if

• $a,b\in H⟹ab\in H.$
• $a\in H⟹a^{-1}\in H.$

Proof

If H is a subgroup of G , then it is obvious that (1) and (2) must hold.

In order to establish that H is subgroup, all that is needed is to verify that $e\in H$ and that the associative law holds for elements of H .

Since associative law does hold for G, it holds all more so for H, which is a subset of G.

if $a\in H$ , by part 2, $a^{-1}\in H$ and so by part 1 , $e=a{a}^{-1}\in H$

LEMMA

A non-empty finite subset H of the group G is a subgroup of G if and only if

1.$a\in H⟹a^{-1}\in H.$

Proof

The Proof is similar to the one for the previous lemma and left as an exercise for the reader.

Cyclic Group

A group is said to be a cyclic group , if there exist an element $a\in G$ such that every element of G can be expressed as some power of a.

If G is a group generated by 'a' . we can say that a is a generator of G and all the elements of G can form by some power of a.

G = (a) { Here a is the generator}

Notes

• A cyclic group may have more that one generator.
• Every group has two trivial subgroups.
  • The group containing all elements.
  • The group containing identity element only.

Congruent Modulo of a Subgroup

Let H is a subgroup of G.( $a,b\in H$)

$a\equiv b(modH)$ if $a{b}^{-1}\in H$

LEMMA

$a\equiv b(modH)$ is an equivalence relation.

proof

An equivalence relation must follows 3 properties.

1. Reflexive
2. Symmetric
3. Transitive

Reflexive

let $a\in G$

as $e\in G$

$e=a{a}^{-1}\in H$

$⟹a\equiv a(modH)$ $\mathrm{\forall }a\in G$

Symmetric

let $a,b\in G$ such that $a\equiv b(modH)$

we have to show that $b\equiv a(modH)$

$a{b}^{-1}\in H$

$(a{b}^{-1}{)}^{-1}\in H$ [ Subgroup follows closure law]

$(b^{-1}{)}^{-1}{a}^{-1}\in H$

$b{a}^{-1}\in H$

$b\equiv a(modH)$

Transitive

let $a,b,c\in G$ such that

$a\equiv b(modH)$ and $b\equiv c(modH)$

$a{b}^{-1}\in H$ and $b{c}^{-1}\in H$

$⟹a{b}^{-1}b{c}^{-1}\in H$ [Closure]

$⟹a(b^{-1}b)c^{-1}\in H$ [Associative]

$⟹ae{c}^{-1}\in H$ [Identity]

$⟹a{c}^{-1}\in H$

$⟹a\equiv c(modH)$

$\therefore Congruernce$ modulo relation is an equivalence relation.