[PDF] LECTURE 7 The homomorphism and isomorphism theorems 7.1

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LECTURE 7

The homomorphism and isomorphism theorems

PAVEL R

UZICKA

Abstract.We prove the homomorphism theorem and the three iso- morphism theorems for groups. We show that the alternating group of permutationsAnis simple for alln?= 4.

7.1.The homomorphism theorem.We prove a theorem relating homo-

morphisms, kernels, and normal subgroups. Theorem7.1(The homomorphism theorem).Let?:G→Hbe a group homomorphism andNa normal subgroup ofG. There is a homomorphism ψ:G/N→Hsuch that?=ψ◦πG/Nif and only ifN?ker?. The homomorphismψis necessarily unique. Moreoverψis a group embedding if and only ifN= ker?. N ker? G H G/N

πG/N

Figure 1.The homomorphism theorem

Proof.(?) Suppose that?=ψ◦πG/Nfor someψ:G/N→Hand let n?N. Then?(n) =ψ◦πG/N(n) =ψ(N) =uH, hencen?ker?. It folows thatN?ker?. (?) Suppose thatN?ker?. Iff·N=g·N, for somef,g?G, then g -1·f?Ndue to Lemma 4.2(iii)?(i). SinceN?ker?, we have that u H=?(g-1·f) =?(g)-1·?(f), hence?(g) =?(f). Therefore we can define a mapψ:G/N→Hbyg·N?→?(g). Fromψ((f·N)·(g·N)) = ψ(f·g·N) =?(f·g) =?(f)·?(g) =ψ(f·N)·ψ(g·N), for allf,g?G, we The Lecture and the tutorial took place in Mal´a strana, roomS11, on November 20, 2018.
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infer thatψ:G/N→His a group homomorphism. It is straightforward that?=ψ◦πG/Nand thatψis unique with the required properties. Suppose thatψis a group embedding. Letg?kerφ. We compute that ψ(N) =uH=?(g) =ψ(πG/N(g)) =ψ(N·g), henceN=N·g, whence g?N. It follows that ker??N. SinceN?ker?due to the first part of the theorem, we conclude thatN= ker?. Coversely, suppose thatN= ker?. Letf,g?Gsatisfyψ(f·N) = ψ(g·N). It follow that?(f) =?(g), hence?(g-1·f) =?(g)-1·?(f) =uH, whenceg-1·f?ker?=N. We get thatf·N=g·N, due to Lemma 4.2.

We conclude thatψis an embedding.?

Corollary7.2.A group homomorphism?:G→His an embedding if and only ifker?={uG}.

7.2.The isomorphisms theorems.

Theorem7.3(The 1st isomorphism theorem).Let?:G→Hbe a group homomorphism. Then?(G)is a subgroup ofHisomorphic toG/ker?. ker? ker? G H

G/ker?

πG/ker?

Figure 2.The 1st isomorphism theorem

Proof.Since?is a group homomorphism, we have that

?(g)·?(h)-1=?(g·h-1)??(G), for allg,h?G. Therefore?(G) is a subgroup ofH. It follow from Theorem 7.1 that there is an embeddingψ:G/ker?→?(G) such that ?=ψ◦πG/ker?. Thusψinduces an isomorphism betweenG/ker?and ?(G).?

Lemma7.4.LetN,Kbe subgroups of a groupG.

(i)IfN?GorK?G, thenN·Kis a subgroup ofG. (ii)If bothN?GandK?G, thenN·Kis a normal subgroup of G. Proof.(i) SinceN?GorK?G, the equalityN·K=K·Nholds true.

It follows that

(N·K)·(N·K) =N·N·K·K=N·K,

Group homomorphisms and their kernels 3

henceN·Kis a sub-universe ofG. For alln?Nandk?Kwe have that (n·k)-1=k-1·n-1?K·N=N·K. ThereforeN·Kis a subgroup ofG. (ii) If bothNandKare normal subgroup ofG, theng·N·K=N·g·K= N·K·g, for allg?G. It follows the subgroupN·Kis normal due to

Lemma 4.11.?

Observe that if at least one of subgroupsNandKof a groupGis normal (resp. if both the subgroupsNandKare normal inG),N·Kis the least subgroup (resp. the least normal subgroup) ofGcontaining both the groups NandK. On the other hand the intersectionN∩Kis the largest common subgroup (resp. the largest common normal subgroup if bothNandKare normal inG) ofNandK. Theorem7.5(The 2nd isomorphism theorem).LetGbe a group,Ha subgroup ofG, andNa normal subgroups ofG. Then (N∩H)?H,and(N·H)?N?H?(N∩H).

N·H

N∩HG

0 HN

Figure 3.The 2nd isomorphism theorem

Proof.The productN·His a subgroup ofGdue to Lemma 7.4 (i). Since Nis a normal subgroup ofH, we have thatg·N·g-1?Nfor allg?G(and a fortiorifor allh?H) due to Lemma 5.1. It follows thath·(N∩H)·h-1? N∩H, for allh?H. ThereforeN∩His a normal subgroup ofH. Since clearly g·h-1?N∩Hif and only ifg·h-1?N, for allg,h?H, we have that g·(N∩H) =h·(N∩H) if and only ifg·N=h·N, for allg,h?H. Therefore the maps

H/(N∩H) (N·H)/N

h·(N∩H)h·N

4P. R°UZICKA

are well defined. It is straightforward to verify that they are mutually inverse group isomorphisms.? Theorem7.6(The 3rd isomorphism theorem).LetGbe a group andN,

Knormal subgroups ofG. IfK?N, then

N/K?G/KandG/N?(G/K)?(N/K).

G N K {uG} G/K N/K {uG/K}

Figure 4.The 3rd isomorphism theorem

Proof.SinceK?G, we have that

for alln?Nandg?G. SinceN?G, there isn??Nsuch that g·n·g-1=n?, hence for somen??N. According to Lemma 5.1 we have thatN/K?G/K. From g·h-1?Nif and only ifg·h-1·K?N/K,for allg,h?G, we infer that g·N=h·Nif and only if (g·K)·N/K= (h·K)·N/K, for allg,h?G. Therefore the maps

G/N(G/K)?(N/K)

g·N(g·K)·(N/K) are well defined. It is straightforward to verify that the maps are mutually inverse isomorphisms of the groups.?

Group homomorphisms and their kernels 5

7.3.Simplicity of the alternating groups.A (non-trivial) group has two

trivial subgroups, the singleton subgroup containing onlythe unit element and the group itself. These two subgroups are necessarily normal and they are the only subgroups of finite groups of a prime order due to the Langrange theorem. Infinite groups and finite groups of a composite order have non- trivial subgroups. However it can still happen that they have only trivial normal subgroups. The groups whose only normal subgroups are the trivial ones are called simple. We prove that this is the case of the alternating groups with the only exception ofA4. Theorem7.7.The alternating group of permutationsAnis simple for all n?= 4. Proof.Alternating groupsA2andA3are of a prime order, and so they are that the groupAnis generated by 3-cycles. Using the assumption that n≥5, we show that

Claim1.All 3-cycles are conjugated inAn.

Proof of Claim 1.Letπ= (a,b,c) andρ= (d,e,f) be 3-cycles (with not necessarily disjoint supports). According to formula (5.2),ρis conjugated toπby a permutationσsatisfyingσ(a) =d,σ(b) =eandσ(c) =f. If σis even, we are done. Ifσis odd, we findg,hdistinct fromd,e,fand replaceσwith the even permutation ˆσ= (g,h)·σ. We still have that ˆσ(a) =d, ˆσ(b) =eand ˆσ(c) =f, and soρ=ˆσπ. This is possible since n≥5.?

Claim 1.

Claim2.Avery non-singleton normal subgroup ofAncontains a 3-cycle. Proof of Claim 2.LetNbe a non-singleton normal subgroup of the alter- nating group. Let us denote byπa non-unit permutation fromNwith suppπof the least possible size (among non-unit permutations fromN).

We discus two complementary cases.

First suppose that in the decomposition ofπinto the product of inde- pendent cycles there is a cycle (a,b,c,...) of the length at least 3. Ifπis a

3-cycle, we are done. It this not the case, there ise?suppπnot in{a,b,c}.

Setf:=π(e) and observe thatf /? {b,c}. It follows that the permutation

σ= (a,e)·(b,f) is even andc /?suppσ. We putρ=π-1·σπ=π-1·σ·π·σ-1.

Observe that suppρ?suppπand, sinceN?An, the permutationρbe- longs toN. Applying (5.2) we compute fromf=σ(b) andc=σ(c) that ρ(f) =π-1·σπ(f) =π-1·σ(π(b)) =π-1·σ(c) =π-1(c) =b. Sincef?=b, the permutationρis not the unit. Next we compute that ρ(a) =π-1(σπ(a)) =π-1·σ·π(e) =π-1(b) =a, hence suppρ?suppπ. This contradicts the choice ofπ. The remaining case is whenπ= (a,b)·(c,d)···is a product of inde- pendent transpositions. Sincen≥5, we can picke /? {a,b,c,d}and put

6P. R°UZICKA

σ= (a,b)·(c,e). As in the previous case letρ=π-1·σπ(which here equals toπ·σπ). Observe that suppρ?suppπ? {e}and as aboveρ?N. We easily compute thatρ(a) =a,ρ(b) =bandρ(e) =π-1(d) =c?=e. It follows thatρis a non-unit permutation anda,b /?suppρ. We conclude that|suppρ|<|suppπ|, which is a contradiction.?

Claim 2.

From Claim 1 and Claim 2 we conclude thatNcontains all 3-cycles. It follows thatN=Andue to Lemma 5.7. Therefore the groupAnis simple.? Remark7.8.Note that all the Klein"s VierergrupeVis a non-trivial nor- mal subgroup ofA4(cf. Example 6.9).

Exercises

Exercise7.1.Recall thatVdenotes the Klein"s Vierergrupe

Prove thatS4/V?S3.

Exercise7.2.LetZbe the group of all integers with the operation of addi- tion. For each non-negative integernput n·Z:={n·z|z?Z}={w?Z|wis divisible byn}. Prove that then·Zare subgroups ofZand all subgroups ofZare of this form. Exercise7.3.For a positive integernand an integerzletz(modn)denote the reminder ofzwhen dividing byn. LetZndenote the set{0,1,...,n-1} with the binary operation+ndefined bya+nb=a+b(modn). Prove that Z nis a group isomorphic toZ/n·Z. Exercise7.4.Letm,nbe positive integers such thatn|m. Prove that n·Z? m·Z?Zm n. Exercise7.5.Letm,nbe positive integers. Letddenote their greatest common divisor andltheir least common multiple. (i)Prove that (ii)Prove that n·Z? (m·Z∩n·Z)?(n·Z+m·Z)? m·Z?Zn d=Zlm.quotesdbs_dbs12.pdfusesText_18

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