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Math 3230 Abstract Algebra I

Sec 4.3: The fundamental homomorphism theorem

Slides created by M. Macauley, Clemson (Modied by E. Gunawan, UConn) http://egunawan.github.io/algebra

Abstract Algebra I

Sec 4.3The fundamental homomorphism theoremAbstract Algebra I 1/11

Quotients: via Cayley diagrams

Dene the homomorphism:Q8!V4via(i) =vand(j) =h.SinceQ8=hi;ji, we can determine wheresends the remaining elements:(1) =e; (1) =(i2) =(i)2=v2=e; (k) =(ij) =(i)(j) =vh=r; (k) =(ji) =(j)(i) =hv=r;

(i) =(1)(i) =ev=v; (j) =(1)(j) =eh=h:Note that Ker=f1;1g. Let's see what happens when we quotient out by Ker:1i

kj1ikjQ 8Q

8organized by thesubgroupK=h1i1i

kj1ikjK jKiK kKQ

8left cosets ofKare near each otherKiK

jKkKQ

8=Kcollapse cosets

into single nodes

Do you notice any relationship betweenQ8=Ker() and Im()?Sec 4.3The fundamental homomorphism theoremAbstract Algebra I 2/11

The Fundamental Homomorphism Theorem

The following is one of the central results in group theory.Fundamental homomorphism theorem (FHT)

If:G!His a homomorphism, then Im()=G=Ker().The FHT says that every homomorphism can be decomposed into two steps: (i)

quotient out by the kernel, and then (ii) relabel the nodes via.G (KerEG) any homomorphism G

Kergroup of

cosetsImq quotient processi remaining isomorphism (\relabeling") Sec 4.3The fundamental homomorphism theoremAbstract Algebra I 3/11

Proof of the FHT

Fundamental homomorphism theorem

If:G!His a homomorphism, then Im()=G=Ker().Proof

We will construct an explicit mapi:G=Ker()!Im() and prove that it is an isomorphism.LetK:= Ker(), and recall thatG=K:=faK:a2Gg.Dene i:G=K!Im();i:gK7!(g): Showiis well-dened:We must show that ifaK=bK, theni(aK) =i(bK).SupposeaK=bK.We have aK=bK=)b

1aK=K=)b

1a2K:By denition ofb1a2Ker(),

1

H=(b1a)=(b1)(a)=(b)1(a)=)(a) =(b):By denition ofi:i(aK) =(a)=(b)=i(bK).XSec 4.3The fundamental homomorphism theoremAbstract Algebra I 4/11

Proof of FHT (cont.) [Recall:i:G=K!Im();i:gK7!(g)]Proof (cont.)

Showiis a homomorphism:We must show thati(aKbK) =i(aK)i(bK).i(aKbK) =i(abK) (aKbK:=abKfrom Slides 3.5 \quotient groups")=(ab) (denition ofi)=(a)(b) (is a homomorphism)=i(aK)i(bK) (denition ofi)Thus,iis a homomorphism.XShowiis surjective (onto):

This means showing that for any element in the codomain (here, Im()), that some

element in the domain (here,G=K) gets mapped to it byi.Pick any(a)2Im().By dention,i(aK) =(a), henceiis surjective.XSec 4.3The fundamental homomorphism theoremAbstract Algebra I 5/11

Proof of FHT (cont.) [Recall:i:G=K!Im();i:gK7!(g)]Proof (cont.) Showiis injective (1{1):We must show thati(aK) =i(bK) impliesaK=bK.Suppose thati(aK) =i(bK).Then

i(aK) =i(bK) =)(a) =(b) (by denition of the mapi)=)(b)1(a) = 1H=)(b1a) = 1H(is a homom.)=)b1a2K(denition of Ker())=)b1aK=K(aH=H,a2H)=)aK=bK

Thus,iis injective.XIn summary, sincei:G=K!Im() is a well-dened homomorphism that isinjec tive (1{1) and surjective

(onto), it is an isomorphism.Therefore,G=K=Im(), and the FHT is proven.Sec 4.3The fundamental homomorphism theoremAbstract Algebra I 6/11

Consequences of the FHT

An alternative proof of Prop 1 part 3

If:G!His a homomorphism, then Im f1Hg= Im()=G=G:Let's use the FHT to determine all homomorphisms:C4!C3:By the FHT,G=Ker=Im

hencejImj=jG=Kerj 2 f1;2;4g.Thus,jImj= 1, and so theonlyhomomorphism:C4!C3is the trivial one.Sec 4.3The fundamental homomorphism theoremAbstract Algebra I 7/11

What does \well-dened" really mean?

Recall that we've seen the term \well-dened" arise in dierent contexts:a well-denedbina ryop erationon a set G=Nof cosets,a well-denedfunction i:G=N!Hfrom a set (group) of cosets.In both of these cases, well-dened means that:

our denition doesn't depend on our choice of coset representative.Formally: IfNEG, thenaNbN:=abNis aw ell-denedbina ryop erationon the set

G=Nof cosets, because

ifa1N=a2Nandb1N=b2N, thena1b1N=a2b2N:The mapi:G=K!H, wherei(aK) =(a), is aw ell-denedh omomorphism, meaning that

ifaK=bK, theni(aK) =i(bK) (that is,(a) =(b)) holds.Whenever we dene a map and the domain is aquotient, we must show it's

well-dened. Sec 4.3The fundamental homomorphism theoremAbstract Algebra I 8/11

How to show two groups are isomorphic

The standard way to showG=His toconstruct an isomo rphism:G!H.When the domain is a quotient, there is another method, due to the FHT.

Useful technique

Suppose we want to show thatG=N=H. There are two approaches:(i)Dene a map :G=N!Hand prove that it isw ell-dened, ah omomorphism,

and a bijection .(ii)Dene a map :G!Hand prove that it is ahomomo rphism, asurjection (onto), and that Ker =N.Usually, Method (ii) is easier. Showing well-denedness and injectivity can be tricky. For example, each of the following are results for which (ii) works quite well:

Z=hni=Zn;Q

=h1i=Q+;AB=B=A=(A\B) (assumingA;BEG);G=(A\B)=(G=A)(G=B) (assumingG=AB).Sec 4.3The fundamental homomorphism theoremAbstract Algebra I 9/11

Cyclic groups as quotients

Consider the following (normal) subgroup ofZ:

12Z=h12i=f:::;24;12;0;12;24;:::gCZ:Theelementsof thequotient group Z=h12iare thecosets:

0 +h12i;1 +h12i;2 +h12i; ::: ;10 +h12i;11 +h12i:Number theorists call these setscongruen ceclasses mo dulo12 . We say that two

numbers are congruent mo d12 if they a rein the same coset. Recall how to add cosets in the quotient group: (a+h12i) + (b+h12i) := (a+b) +h12i:

\(The coset containinga) + (the coset containingb) = the coset containinga+b."It should be clear thatZ=h12iis isomorphic toZ12.Formally, this is just the FHT

applied to the following homomorphism: :Z!Z12; :k7!k(mod 12);Clearly, Ker() =f:::;24;12;0;12;24;:::g=h12i. By the FHT:Z=Ker() =Z=h12i =Im()=Z12:Sec 4.3The fundamental homomorphism theoremAbstract Algebra I 10/11

A picture of the isomorphismi:Z12!Z=h12i(from the VGT website)Sec 4.3The fundamental homomorphism theoremAbstract Algebra I 11/11

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