# Theorems of the Subgroup

Let (G,*) be a group and H be a non-empty subset of G.if H itself is a group with the same binary operation (*).Than H Is called a subgroup of G. In this article we will discuss the Theorems of the Subgroup

- Statement: Let (G,.) be a group, then a non
- -empty subset H of G is a subgroup if and only if
- a,b ∈H imply ab
^{-1}∈H. - Statement: The intersection of any collection of subgroups of a group G.is a subgroup of G.
- Statement: let G be a group and H a subgroup of G. Then the set aHa-1 ={aha-1 ,h∈H } is a subgroup of G.
- Statement: the union HUK be two subgroups H and K of a group G is a subgroup of G if and only if either
- H is a subset of K or k is a subset of H
- Summary of a theorem of subgroup
- Frequently Asked Question-FAQs

**Statement:** Let (G,.) be a group, then a non

## -empty subset H of G is a subgroup if and only if

## a,b ∈H imply ab^{-1} ∈H.

Proof:

Suppose H is a subgroup of G

H itself is a group

So ∀a,b∈H –>a,b^{-1} ∈H

ab** ^{-1 }**∈H

∴ H is group

**Conversely **

Suppose

a,b^{-1} ∈H —> ab^{-1} ∈H _{—> (1)}

Here we prove H is the group

**Let a∈H**

a,a ∈H –>aa^{-1 }∈H –>_{ }_{eH}

by (1)

Identity elements exist in H

e,a∈H –> ea^{ -1}∈H –>a^{-1}∈H

by (1)

The inverse of each element exists in H

Let a,b∈H –>** _{ }**a,b

^{-1}∈H

a(b^{-1})^{-1}∈H

by (1)

** _{ }**ab∈H

H is closed

Since G is a group

(ab)c=a(bc) ∀ a,b,c∈G

H is a subset of G

Associative law held in H.

By 1,2,3,4

H itself group

Hence proved;

**H is a subgroup of G**

**Statement: **The intersection of any collection of subgroups of a group G.is a subgroup of G.

**Proof:**

Let {H_{i} , i∈I } be family of subgroup of a group G.

Let ∩_{i∈I} H_{i}=H

Now we prove H is subgroup

Let a,b∈H

a,b∈H_{i }for each i∈ I

Since each H_{i} is a subgroup of G

ab^{-1} ∈H_{i} for each i∈ I

ab^{-1} ∈∩_{i∈I} H_{i}

ab^{-1} ∈H

so H is a subgroup

Thus

**∩ _{i∈I} H_{i} is a subgroup**

## Statement: let G be a group and H a subgroup of G. Then the set aHa-1 ={aha-1 ,h∈H } is a subgroup of G.

**Proof:**

Let x,y∈ aHa^{-1}

x=ah_{1}a^{-1} ,y=ah_{2}a^{-1}

h_{1},h_{2}∈H

Now

x,y^{-1}=( ah_{1}a^{-1} )( ah_{2}a^{-1})^{-1}

x,y^{-1}= ah_{1}a^{-1} .(a^{-1})^{-1} h_{2}^{-1} a^{-1}

x,y^{-1 }= ah_{1}a^{-1}ah_{2}^{-1} a^{-1}

x,y^{-1}=ah_{1}eh_{2}^{-1} a^{-1} aa^{-1}=e

x,y^{-1}= ah_{1}h_{2}^{-1} a^{-1}

H is a subgroup

and h_{1},h_{2}∈H –>h_{1}h_{2}^{-1}∈H

so

ah_{1}h_{2}^{-1} a^{-1 }= x,y^{-1 }∈H

**aHa ^{-1} is a subgroup of G.**

**Statement: **the union HUK be two subgroups H and K of a group G is a subgroup of G if and only if either

## H is a subset of K or k is a subset of H

Proof:

Let G be a group.

Let H and K be a subgroup of G

Suppose H is a subset of K or k is a subset of H

HUK=K or HUK=H

Since H and k are subgroups

HUK is a subgroup of G

**Conversely,**

Suppose HUK is a subgroup of G

To prove H is a subset of K or k is a subset of H

We suppose

H is not a subset of k

a∈H but **a** not belong to K

We also suppose

K is not a subset of H

b∈K but b does not belong to H

Since, a∈H, b∈K

a,b∈ HUK

ab∈ HUK

HUK is a subgroup

ab∈H or ab∈K

When ab∈H

a∈H,a^{-1}∈H H is a subgroup

b=a^{-1}(ab) ∈H H is a subgroup

a^{-1} a(b) ∈H

eb∈H ,b∈H

when ab∈K

b∈K,b^{-1}∈K K is a subgroup

a=(ab)b^{-1}∈K

a=a(bb^{-1}) ∈K

a=ae∈K

a ∈K

Hence

b∈H or a ∈K

Which is contradiction

Hence

H is not a subset of K.

K is not a subset of H is wrong

Thus

H is a subset of K

K is a subset of H

## Summary of a theorem of subgroup

- Two subgroups H and K of a group G, their union HUK need not be a subgroup of G
- The identity of the subgroup is the same as that of the group.
- The inverse of any element of a subgroup is the same as the inverse of that element in the group.

## Frequently Asked Question-FAQs

### what is a subgroup?

In group theory, a subset H of a group G under a binary operation ∗ is called a subgroup of G if H also forms a group.

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