# 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 âˆˆH imply ab^{-1} âˆˆH.

Proof:

Suppose H is a subgroup of G

H itself is a group

So âˆ€a,bâˆˆH –>a,b^{-1} âˆˆH

ab** ^{-1 }**âˆˆH

âˆ´ H is group

**Conversely **

Suppose

a,b^{-1} âˆˆH —> ab^{-1} âˆˆH _{—> (1)}

Here we prove H is the group

**Let aâˆˆH**

a,a âˆˆH –>aa^{-1 }âˆˆH –>_{ }_{eH}

by (1)

Identity elements exist in H

e,aâˆˆH –> ea^{ -1}âˆˆH –>a^{-1}âˆˆH

by (1)

The inverse of each element exists in H

Let a,bâˆˆH –>** _{ }**a,b

^{-1}âˆˆH

a(b^{-1})^{-1}âˆˆH

by (1)

** _{ }**abâˆˆH

H is closed

Since G is a group

(ab)c=a(bc) âˆ€ a,b,câˆˆG

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} , iâˆˆI } be family of subgroup of a group G.

Let âˆ©_{iâˆˆI} H_{i}=H

Now we prove H is subgroup

Let a,bâˆˆH

a,bâˆˆH_{i }for each iâˆˆ I

Since each H_{i} is a subgroup of G

ab^{-1} âˆˆH_{i} for each iâˆˆ I

ab^{-1} âˆˆâˆ©_{iâˆˆI} H_{i}

ab^{-1} âˆˆH

so H is a subgroup

Thus

**âˆ© _{iâˆˆI} H_{i} is a subgroup**

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

**Proof:**

Let x,yâˆˆ aHa^{-1}

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

h_{1},h_{2}âˆˆH

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}âˆˆH –>h_{1}h_{2}^{-1}âˆˆH

so

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

**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

aâˆˆH but **a** not belong to K

We also suppose

K is not a subset of H

bâˆˆK but b does not belong to H

Since, aâˆˆH, bâˆˆK

a,bâˆˆ HUK

abâˆˆ HUK

HUK is a subgroup

abâˆˆH or abâˆˆK

When abâˆˆH

aâˆˆH,a^{-1}âˆˆH H is a subgroup

b=a^{-1}(ab) âˆˆH H is a subgroup

a^{-1} a(b) âˆˆH

ebâˆˆH ,bâˆˆH

when abâˆˆK

bâˆˆK,b^{-1}âˆˆK K is a subgroup

a=(ab)b^{-1}âˆˆK

a=a(bb^{-1}) âˆˆK

a=aeâˆˆK

a âˆˆK

Hence

bâˆˆH or a âˆˆK

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 âˆ— is called a subgroup of G if H also forms a group.

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