Answers to Second Midterm

MA441: AlgebraicStructures I

14 December 2003

All questions are worth ten points.In addition to the questions, there will be an additional ten points to be awarded for style and clarity in writing. 1) Deﬁnitions 1. Deﬁnetheindex|G:H|of a subgroupH < G. The index is equal to the number of distinct left cosets ofHinG. The index is also equal to|G|/|H|. 2. Deﬁnetheexternal direct productG1⊕G2of groupsG1andG2. (Specify the set of elements that make up the direct product and its group operation.) The external direct product consists of the set of pairs{(a, b)|a∈ G1, b∈G2}with the group operation (a, b)∙(c, d) = (ab, cd), where acis composed inG1andbdis composed inG2.

2) Fill in the blanks or answer True/False. 29 1. 11≡11 (mod29). 2. Trueor False:Every groupGof order 7 contains an elementa∈G such that|a|True= 7. 3. Trueor False:if Aut(G1)≈Aut(G2) thenG1≈G2. False 4. InS5,|(12)(13)(456)|= 3 (Compose left to right.) 5. Trueor False:LetGbe an arbitrary ﬁnite group of ordern. Ifd|n, then there is anH < Gof orderd. False 1

3) Lagrange’s Theorem 1. Givea clear and complete statement of Lagrange’s Theorem for a sub-groupHof a ﬁnite groupG. LetHbe a subgroup ofGthe order of. ThenHdivides the order of Gnumber of cosets of. TheHinGis|G|/|H|. 2. UseLagrange’s Theorem to prove that the order of an elementa∈G divides the order ofG. The cyclic subgrouphaiis a subgroup ofG, so its order|hai|divides the order ofGorder of. Thea,|a|equals|hai|, so|a|divides|G|. 3. UseLagrange’s Theorem to prove that if|G|is prime, thenGmust be cyclic. If|G|is prime, then the order of any element is either 1 or|G|. There-fore any nonidentity element has order|G|and generatesG. 4. IfHandKare subgroups ofG, where|H|= 33 and|K|= 28, what are the possible orders ofH∩Kyour answer.? Explain Since the intersectionH∩Kis a subgroup of bothHandK, its order must divide both 33 and 28 by Lagrange’s Theorem.However, 2 33 = 3∙11 and 28 = 2∙7 are relatively prime, so the order of the intersection must be 1.

4) List the distinct left cosets ofHinGif 1.G=D4,H=hF Ri, whereRis a rotation andFis a ﬂip. The distinct left cosets can be written asH={e, F R},RH={R, F}, 2 23 33 2 R H={F RR ,},R H={R ,F R}. 2.G=Z/15Z,H=h3i. H={0,3,6,9,12}, 1+H={1,4,7,10,13}, 2+H={2,5,8,11,14}.

5) Supposeφ:G1→G2Prove thatis an isomorphism.G1is abelian if and only ifG2is abelian.

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SupposeG1is abelian.Any pair of elementsx, y∈G2are the images of some elementsa, b∈G1say. Let’sφ(a) =xandφ(b) =y. ThensinceG1is abelian, xy=φ(a)φ(b) =φ(ab) =φ(ba) =φ(b)φ(a) =yx. ThereforeG2is abelian. SupposeG2is abelian.Thenφ(ab) =φ(a)φ(b) =xy. Also,φ(ba) = φ(b)φ(a) =yx. SinceG2is abelian,xy=yx. Becauseφis one-to-one, since abandbahave the same image, thenab=ba. ThereforeG1is abelian.

6) Cosets:Given a subgroupH < G, prove that any two cosets ofHhave the same order, that is, for anya, b∈G,|aH|=|bH|. Letψ:aH→bHviaah7→bh(∀h∈H). Themapψis one-to-one −1 because ifah=bhthena=bbecause we can right multiply byh. The mapψis onto because any element in the cosetbHhas the formbh, which has preimageahunderψ. Thisshows thatψis a bijection on the two cosets aHandbH, which shows|aH|=|bH|.

7) Classiﬁcation:Prove the following: LetGbe a group of order 2p, forp >Suppose2 prime.Gis not cyclic and thata∈Ghas orderp. Ifb6∈ hai, show that|b|= 2. Since|a|=p, the index ofhaiinGis 2, that is, there are exactly two cosets ofhaiinGleft coset. Thebhaimust not be the same ashaibecause thenb∈ haithe only two cosets are. Sohaiandbhai. 2 2 The left cosetbhaimust be one of these two cosets.Ifbhai=bhai, then by cancellingbon the left, we would get thatb∈ hai, so we eliminate that 2 2 possibility. Therefore,bhai=hai, sob∈ hai. 2 Since|hai|=p, the order ofbmust dividep, which means it is either 1 2 2 orp. If|b|=p, thenhbi=hai. Sinceb6∈ hai, thenhbimust be all ofG. 2 But we assumed thatGTherefore,is not cyclic.|b|must be 1 and|b|= 2. (This is part of Theorem 7.2.)

8) Let|x|= 40.List all the elements ofhxithat have order 10.Explain your answer. This is Exercise 46 from Chapter 4, taken from Homework Assignment 6. k The order ofxis 40/gcd(40, k). Toget an element of order 10, choose 4 12 28 36 ksuch that gcd(40, k) = 4.The elements of order 10 are{xx ,x ,x ,}.

9) (extra credit) Prove Lagrange’s Theorem.You may cite basic properties of cosets, such as those listed in Gallian’s Lemma, if you state them accurately. 3

Please refer to the proof on page 137 of Gallian.It suﬃces to note that Gcan be partitioned by its cosets, that all cosets have the same size, and that therefore the order ofGis an even multiple of the order of a coset.

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