Subgroups and its Properties :

Lecture no 4.

Let (G, ⊕) be a group and H ⊆ G is a subgroup if

1. H is closed.

2. ∀ a ∈ H , a ^-1 ∈ H.

Proof: To show H is a group ∃

1. Closure [Follows from 1^st condition]

2. Associativity [Follows from associativity of G]

3. Inverse a ∈ H , a ^-1 ∈ H [2 ^nd condition]

                 a ⊕ a ^-1 ∈ H [1^st condition]

                ⇒ e ∈ H

Theorem 1 : A non empty closed subset of a finite group is always a subgroup.

Proof : Let (G, ⊕ ) be a finite group & H be a non empty closed subset of G. Pick an element a ∈ H & generate the sequence a , a ² , a ³ , ... where a ² = a ⊕ a , a ³ = a ² ⊕ a and so on.

This is an infinite sequence all whose members belong to finite subset H and hence all elements in the sequence cannot be distinct. Thus there must be at least 2 elements that are identical.

a ^r = a ^s ( r ≠s )

⇒  a ^{r - s} = e

⇒  a ^-1 = a ^{r - s -1} ∈ H

Thus H is a subgroup from our definition.

Definition 1: Let (G, ⊕ ) be a group and H is a subgroup of G. Between two elements a , b ∈ G we define a congruence relation as follows:

a ≡ b mod H if a ⊕ b ^-1 ∈ H

Lemma 1: Congruence relation is an equivalence relation.

Proof:

Reflexive:

We have to show a ≡ a mod H for all a ∈ G

From the definition of congruence relation a ⊕ a ^-1 = e ∈ H ⇒ a ≡ a mod H.