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

Let   ∩i∈I Hi=H

Now we prove H is subgroup

Let a,b∈H

a,b∈Hi for each i∈ I

Since each Hi is a subgroup of G

ab-1 ∈Hi  for each i∈ I

ab-1 ∈∩i∈I Hi

ab-1 ∈H

so H is a subgroup

Thus

i∈I Hi 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=ah1a-1  ,y=ah2a-1

h1,h2∈H

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,h2∈H –>h1h2-1∈H

so

ah1h2-1 a-1 = x,y-1 ∈H

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

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

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.