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


 * $$a,b \in H \implies ab \in H.$$
 * $$a \in H \implies a^{-1} \in H.$$

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 , $$e = aa^{-1} \in H$$

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

1.$$a \in H \implies a^{-1} \in H.$$

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}

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

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

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 $$e \in G$$

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

$$\implies a \equiv a(modH) $$     $$\forall a \in G$$

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

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

$$ab^{-1} \in H$$

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

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

$$ba^{-1}\in H$$

$$b \equiv a(modH)$$

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

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

$$ab^{-1} \in H  $$   and $$bc^{-1} \in H$$

$$\implies ab^{-1}bc^{-1} \in H$$              [Closure]

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

$$\implies aec^{-1} \in H$$                     [Identity]

$$\implies ac^{-1} \in H$$

$$\implies a \equiv c(modH)$$

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