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# Theorems of the Subgroup

July 19, 2022
written by Azhar Ejaz

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

## 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 -1âˆˆH –>a-1âˆˆH

by (1)

The inverse of each element exists in H

Let a,bâˆˆH –> a,b-1 âˆˆH

a(b-1)-1âˆˆH

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

Now we prove H is subgroup

Let a,bâˆˆH

a,bâˆˆHi for each iâˆˆ I

Since each Hi is a subgroup of G

ab-1 âˆˆHi  for each iâˆˆ I

ab-1 âˆˆH

so H is a subgroup

Thus

## 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=ah1a-1  ,y=ah2a-1

h1,h2âˆˆH

Now

x,y-1=( ah1a-1  )( ah2a-1)-1

x,y-1= ah1a-1 .(a-1)-1 h2-1 a-1

x,y-1 = ah1a-1ah2-1 a-1

x,y-1=ah1eh2-1 a-1            aa-1=e

x,y-1= ah1h2-1 a-1

H is a subgroup

and h1,h2âˆˆH –>h1h2-1âˆˆH

so

ah1h2-1 a-1 = x,y-1 âˆˆH

aHa-1 is a subgroup of G.

## 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-1âˆˆH        H is a subgroup

b=a-1(ab) âˆˆH        H is a subgroup

a-1 a(b) âˆˆH

ebâˆˆH ,bâˆˆH

when abâˆˆK

bâˆˆK,b-1âˆˆK  K is a subgroup

a=(ab)b-1âˆˆK

a=a(bb-1) âˆˆK

a=aeâˆˆK

a âˆˆK

Hence

bâˆˆH    or      a âˆˆK

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.

### 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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