Let be a subgroup of . A relation R on G is defined as: xRy iff x o y-1 ∈H Prove that R is an equivalence relation.

Question:

1. Let <H,0 > be a subgroup of <0 >. A relation R on G is defined as:
xRy iff x o y-1 ∈H
Prove that R is an equivalence relation. (10%)

Summary:

Given that H is a subgroup of G, where G is a Group.

A relation ‘R’ is such that xRy iff x o y^-1 ∈ H.

The relation ‘R’ is equivalence because:

symmetric.

reflexive.

transitive.

Hence, the relation ‘R’ is equivalence.

Explanation:

Given that:

G is a group and H is a subgroup, of G.

Relation R is defined on G such that xRy iff x o y^-1∈H.

We have to prove that the relation ‘R’ is an equivalence relation.

 

A relation is said to be equivalence if it is reflexive, symmetric, and transitive.

 

A relation  R is said to be:

1. Reflexive: iff xRx.
2. Symmetric: if xRy, then yRx.
3. Transitive: if xRy and yRz, then xRz.

 

Proof:

1) To check if ‘R’ is symmetric:

Since x^-1y∈H, it’s inverse also ∈ G.

(x^-1y^-1) =xy^-1 ∈ G

Therefore, R is symmetric.

If x, y ∈ G and xRy. We have to show yRx, which is

yx^-1∈H.

But, xy^-1 ∈ H.

As per the properties of subgroup,( xy^-1)^-1∈ H.

But, (( xy^-1)^-1)=x^-1y.

Therefore, symmetric.

 

2) To check if ‘R’ is reflexive:

Let x∈ G, we need to show that xRx,

x.x^-1∈H

x.x^-1=1

1∈ H

x.x^-1∈H

Therefore, Reflexive.

 

3) To check if ‘R’ is transitive:

Let us suppose x,y,z∈ G, and if xRy and yRz.

We have to show that xRz.

Which means x.z^-1∈H.

since xRy and yRz,

x.y^-1∈H and y.z^-1∈H

Therefore y.z^-1 * x.y^-1∈H .

But, we know y.z^-1 * x.y^-1= x.z-1

x.z^-1=xRz

As xRy and yRz =xRz

Therefore it is transitive.

Therefore, the relation ‘R‘ is equivalence.

 

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