5310-6310-test-ch1-answer (1) - Abstract Algebra Chapter I...

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Unformatted text preview: Abstract Algebra Chapter I Name: Ans-Wain Totally 110 points, including 10 bonus ones. Please ﬁnish in 50 minutes. 1. (20 points) Consider the following binaryr algebraic structures. (A) Determine which ones are groups, which ones are not. (B) If a. structure is not a group, then state a reason. a.) (51+) yEg} a group (bl (Zr) No. 0 has no inverse (in ford, all fax-reﬂex orlzer than i! have no {pm/erfe) (c) The set {an i n E Z} with product operation, where a. E R‘. 341:6 I it 15 a ﬁrm/r53 (d) The structure (3Z36, +35), where 3Z35 := {371 j n E Z and 0 _<_ 312 < 36}. \ng it is Q e‘rOUD In fad“, ‘1"!- {S Q SuberMp of 236 I l n J 2. (20 points) Sketch the subgroup diagram of Zn, that is, ﬁnd out all subgroups of Z12 and their inclusion relations. (mi-[Zn— /\ <2) <3} / \<6>/ <4? \/ <¢> Math 2660 Final Exam, Page 2 of 2 September 18, 2006 3. (20 points) Write down the group table of the Klein 4-group V := {6, a, (7.0}, which is isomorphic to 23 X 22. 4. (30 points) A cyclic group G is either isomorphic to Z (when G is of inﬁnite order), or isomorphic to Zm (when G is of order m). Suppose d and m are nonzero integers, and d. divides m. Let H := (a!) be the cyclic subgroup of Zm generated by d E Zm. (A) Construct a. group isomorphism qﬁ : H —+ Zm/d between H and Zm/d. B Prove that the ,on construct is a Group isomorphism. 3 O m A” elements 0'? H “9 OilTlne form ClK £01“ osdk<m. So osk<a— ‘ \ , . \. ‘ [‘4- Deﬁne Zm/d (picnic/ﬁzk for oskcm/q. We anCKTQL d? is an isomorphism.- \ n : (dK ‘- ‘l'len K :k and 30 CiKJZCle G) Inga-Me: it CWde 4’ 22,1 i z (2) SurJ‘ectWe: For ever?- 0\$K< "Md; “there is dke H Svclﬂ "that qbtclk‘=!<_ ® HONWWM PVOPWW= Given d‘k. mi elk; a». dk + dkz_ { e-.H<;-1rdi<L if dk,+cli-<L<-m I m '— dkz-m I“; dk|+dk13m : Mean/dim @m so 4>tdl<i+mdkﬁ 5. ('20 points) Denote the set a b ,5 S.ﬁ{[01]|a€R,bER}. Prove that S with the matrix multiplication forms a nonabelian group. @ T‘ne opera-Han is closed; Gt‘ven [a b] 'C d] e S we product Ol ’ 01 I “ : dl'i“ ‘ .‘ l a b (“q—i : QC a I e S "in aeration :5 C95“. ['0 ‘ILD [J o i J I so T e P 0&1"- +ch+b G) ASSOCI‘aiiviﬁx/: { a :9 [c cl [3 {11 : ac MIND] Fe {13:i‘ace W \i-‘0 i] 0 l] 0 d o l L0 M) 0 i ‘ n m llz‘fil‘: fl): [: iir: r: So it {s ASSoci‘aﬁve. @ IdEn-tE-t : Th ,~ 1 em S[[O]€@S Suchtlnai‘ L; :’1”:‘:]:[: WWW ® Inverse: @vaem [a b e 3 2 Com we find [2 jije 3 Such Hm [‘2 m: “U = [a ‘1] ? By l 91:5" So ‘15an Cwerse of [qb {5 [6‘1 mac] \ O I O ' we alsO have Them‘ﬁ)”, S WH’h +3119. Ma:th meCFUCcUHon “ﬁn/mg QﬁFOl/LF_ It is nonabetfam Sn‘nce [2: f] "L: i 3 2 [02: bcfd] w mmwvﬁ 1 F” WW [2 an: 21: r; ’1'] E: in: Hz? ...
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