[PDF] [PDF] Lecture 43: The fundamental homomorphism theorem - School of

The following result is one of the central results in group theory Fundamental homomorphism theorem (FHT) If φ: G → H is a homomorphism, then Im(φ) ∼ 



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[PDF] Lecture 43: The fundamental homomorphism theorem - School of

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Lecture 4.3: The fundamental homomorphism theorem

Matthew Macauley

Department of Mathematical Sciences

Clemson University

http://www.math.clemson.edu/ ~macaule/

Math 4120, Modern Algebra

M. Macauley (Clemson)Lecture 4.3: The fundamental homomorphism theoremMath 4120, Modern Algebra 1 / 10

Motivating example (from the previous lecture)

Dene the homomorphism:Q4!V4via(i) =vand(j) =h.SinceQ4=hi;ji: (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:Let's quotient out by Ker=f1;1g:1i kj1ikjQ 4Q

4organized by thesubgroupK=h1i1i

kj1ikjK jKiK kKQ

4left cosets ofKare near each otherKiK

jKkKQ

4=Kcollapse cosets

into single nodes

Key observation

Q

4=Ker()=Im().M. Macauley (Clemson)Lecture 4.3: The fundamental homomorphism theoremMath 4120, Modern Algebra 2 / 10

The Fundamental Homomorphism Theorem

The following result 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 (KerCG) any homomorphism G

Kergroup of

cosetsImq quotient processi remaining isomorphism (\relabeling")

M. Macauley (Clemson)Lecture 4.3: The fundamental homomorphism theoremMath 4120, Modern Algebra 3 / 10

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): Show i is 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).XM. Macauley (Clemson)Lecture 4.3: The fundamental homomorphism theoremMath 4120, Modern Algebra 4 / 10

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

Show i is a homomorphism:We must show thati(aKbK) =i(aK)i(bK).i(aKbK) =i(abK) (aKbK:=abK)=(ab) (denition ofi)=(a)(b) (is a homomorphism)=i(aK)i(bK) (denition ofi)Thus,iis a homomorphism.XShow i is 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.XM. Macauley (Clemson)Lecture 4.3: The fundamental homomorphism theoremMath 4120, Modern Algebra 5 / 10

Proof of FHT (cont.) [Recall:i:G=K!Im();i:gK7!(g)]Proof (cont.) Show i is 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)=)(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 isin jective (1{1) and surjective

(onto), it is an isomorphism.Therefore,G=K=Im(), and the FHT is proven.M. Macauley (Clemson)Lecture 4.3: The fundamental homomorphism theoremMath 4120, Modern Algebra 6 / 10

Consequences of the FHT

Corollary

If:G!His a homomorphism, then ImH.A few special cases If:G!His an embedding, then Ker() =f1Gg.The FHT says that Im()=G=f1Gg=G:If:G!His the map(g) = 1Hfor allh2G, then Ker() =G, so the FHT says that

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.M. Macauley (Clemson)Lecture 4.3: The fundamental homomorphism theoremMath 4120, Modern Algebra 7 / 10

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 that we will see very soon, for which (ii) works quite well:Z=hni=Zn;Q

=h1i=Q+;AB=B=A=(A\B) (assumingA;BCG);G=(A\B)=(G=A)(G=B) (assumingG=AB).M. Macauley (Clemson)Lecture 4.3: The fundamental homomorphism theoremMath 4120, Modern Algebra 8 / 10

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:M. Macauley (Clemson)Lecture 4.3: The fundamental homomorphism theoremMath 4120, Modern Algebra 9 / 10

A picture of the isomorphismi:Z12!Z=h12i(from the VGT website)M. Macauley (Clemson)Lecture 4.3: The fundamental homomorphism theoremMath 4120, Modern Algebra 10 / 10

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