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7.Quotient groups III

We know that the kernel of a group homomorphism is a normal subgroup. In fact the opposite is true, every normal subgroup is the kernel of a homomorphism: Theorem 7.1.IfHis a normal subgroup of a groupGthen the map :G!G=Hgiven by (x) =xH; is a homomorphism with kernelH.

Proof.Suppose thatxandy2G. Then

(xy) =xyH =xHyH (x) (y):

Therefore

is a homomorphism. eH=Hplays the role of the identity. The kernel is the inverse image of the identity. (x) =xH=H if and only ifx2H. Therefore the kernel of isH.

If we put all we know together we get:

Theorem 7.2(First isomorphism theorem).Let:G!G0be a group homomorphism with kernelK. Then[G]is a group and :G=H![G]given by(gH) =(g); is an isomorphism. If :G!G=His the map (g) =gHthen(g) = (g). The following triangle summarises the last statement: G -G0 G=H

Example 7.3.Determine the quotient group

Z 3Z7Z 3 f0g Note that the quotient of an abelian group is always abelian. So by the fundamental theorem of nitely generated abelian groups the quotient is a product of abelian groups. 1 Consider the projection map onto the second factorZ7: :Z3Z7!Z7given by (a;b)!b: This map is onto and the kernel isZ3f0g. So by the rst isomorphism theorem the quotient group is isomorphic to the image which isZ7.

Theorem 7.4.LetHbe a subgroup of a groupG.

The following are equivalent:

(1)ghg12Hfor allg2Gandh2H. (2)gHg1=Hfor allg2G. (3)gH=Hg.

Proof.Suppose that (1) holds. Then

gHg

1=fghg1jh2Hg H;

for anyg2G. To prove (2) we have to establish that the RHS is a subset of the

LHS. Pickh2H. Then

g

1hg2H;

as we assuming (1), applied to the elementg12G. Thusg1hg=h1 for someh12H. Multiplying on the left bygand on the right byg1 we get h=gh1g12gHg1: Thus the RHS is a subset of the LHS and (2) holds. Now suppose that (2) holds. We have to show thatgH=Hg. We rst show that the LHS is a subset of the RHS. Pickx2gH. Then x=gh, for someh2H. We have xg

1=ghg12gHg1=H;

so thatxg1=h12H. But thenx=h1g2Hg. Thus the LHS is a subset of the RHS. By symmetry the RHS is a subset of the LHS.

Thus (3) holds.

Finally suppose that (3) holds. Pickg2Gandh2Hand let x=ghg1. Then gh2gH=Hg so thatgh=h1g2Hg, for someh12H. But then x=ghg1=h1gg1=h12H:

Thus (1) holds.

Corollary 7.5.IfGis abelian then every subgroup is normal. 2 Proof.Suppose thatHis a subgroup ofG. Ifh2Handg2Gthen ghg

1=gg1h=h2H:

Thus (1) of (7.4) holds. Therefore (3) holds and soHis normal. Denition-Lemma 7.6.AnautomorphismofGis an isomorphism :G!G.

Fixg2G. Then the map

i g:G!Ggiven bya!gag1 is an automorphism ofG, called aninner automorphism. Proof.We have to check thatigis a group homomorphism and thatig is a one to one correspondence.

Suppose thataandb2G. We have

i g(ab) =g(ab)g1 =ga(g1g)bg1 = (gag1)(gbg1) =ig(a)ig(b):

Thusigis a homomorphism.

There are two ways to check thatigis a one to one correspondence. To check thatigis one to one, we just have to check that the kernel is trivial. Suppose thata2Kerig. Then gag

1=ig(a) =e:

Multiplying on the left byg1and on the right bygwe get a=g1eg=e:

Thus the kernel is trivial andigis one to one.

Now we check thatigis onto. Suppose thatb2G. Leta=g1bg2

G. Then

i g(a) =gag1 =g(g1bg)g1 =b: Thusigis onto. It follows thatigis an automorphism. Here is another way to show thatigis an automorphism. Let's try to write down the inverse map. We guess that the inverse ofigisig1. 3

We check

i g1(ig(a)) =ig1(gag1) =g1(gag1)g = (g1g)a(g1g) =a: Thus the composition one way is the identity. If we replacegbyg1 we see that the composition the other way is the identity. It follows thatigis an automorphism. 4quotesdbs_dbs21.pdfusesText_27