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Abstract Algebra I (1St Exam) Solution 2005

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Abstract Algebra I (1St Exam) Solution 2005 Abstract Algebra I (1st exam) Solution 2005. 3. 21 1. Mark each of following true or false. Give some arguments which clarify your answer (if your answer is false, you may give a counterexample). a. (5pt) The order of any element of a ¯nite cyclic group G is a divisor of jGj. Solution.True. Let g be a generator of G and n = jGj. Then any element ge of n G has order gcd(e;n) which is a divisor of n = jGj. b. (5pt) Any non-trivial cyclic group has exactly two generators. Solution. False. For example, any nonzero element of Zp (p is a prime) is a generator. c. (5pt) If a group G is such that every proper subgroup is cyclic, then G is cyclic. Solution. False. For example, any proper subgroup of Klein 4-group V = fe; a; b; cg is cyclic, but V is not cyclic. d. (5pt) If G is an in¯nite group, then any non-trivial subgroup of G is also an in¯nite group. Solution. False. For example, hC¤; ¢i is an in¯nite group, but the subgroup n ¤ Un = fz 2 C ¤ jz = 1g is a ¯nite subgroup of C . 2. (10pt) Show that a group that has only a ¯nite number of subgroups must be a ¯nite group. Solution. We show that if any in¯nite group G has in¯nitely many subgroups. (case 1) Suppose that G is an in¯nite cyclic group with a generator g. Then the n subgroups Hn = hg i (n = 1; 2; 3;:::) are all distinct subgroups of G. Thus G has in¯nitely many subgroups. (case 2). Suppose that G has an in¯nite cyclic subgroup H. By (case 1), H has in¯nitely many subgroups. Since any subgroup of H is also a subgroup of G, G also has in¯nitely many subgroups. (case 3) Finally assume that G has no in¯nite cyclic subgroup, i.e. hgi is a ¯nite subgroup of G for any g 2 G. Choose e 6= g1 2 G, and set H1 = hg1i. By assumption, H1 G. Choose g2 2 G ¡ H1, and set H2 = hg2i. Then H2 6= H1. Note that H1 [ H2 is a ¯nite subset of G. Choose g3 2 G ¡ (H1 [ H2), and set H3 = hg3i. Then 1 2 H3 6= H1;H2. In this way (or inductively) we can choose in¯nitely many subgroups of G. 3. Let S be any subset of a group G. a. (5pt) Show that HS = fx 2 Gjxs = sx for all s 2 Sg is a subgroup of G. ¡1 Solution. It su±ces to show that for any x; y 2 HS, xy 2 HS. Let x; y 2 HS. For any s 2 S, one has xs = sx, ys = sy (equivalently y¡1s = sy¡1), and so (xy¡1)s = x(y¡1s) = x(sy¡1) = (xs)y¡1 = (sx)y¡1 = s(xy¡1): ¡1 Thus xy 2 HS. Therefore HS is a subgroup of G. b. (5pt) Show that HG is an abelian group. Solution. To show that HG is an abelian group, we need to show that xy = yx for any x; y 2 HG. Let x; y 2 HG. By de¯nition of HG, it holds that xy = yx for any x; y 2 HG. 4. (10pt) Show that any ¯nite group G without proper non-trivial subgroups is a cyclic group with a prime number of elements (We assume that jGj > 1). Solution. Let g be any non-identity element of G. Then hgi is a non-trivial subgroup of G. Thus G = hgi, and so G is a cyclic group. Let jGj = n. If n is not a prime, then there exists a positive integer a such that ajn and 1 < a < n. Then the n=a n subgroup hg i has order gcd(n;n=a) = a, which is a contradiction. Therefore n must be a prime. 5. (10pt) Let a and b be positive integers. Find the subgroup of Z generated by fa; bg. Solution. The subgroup of Z generated by fa; bg is fax+byjx; y 2 Zg. We already have known that fax+byjx; y 2 Zg is the cyclic subgroup of Z generated by gcd(a; b) (from Elementary Number Theory course). 6. (10pt) Prove that hQ; +i is not a cyclic group. Solution. Suppose that hQ; +i is a cyclic group. Let x 2 Q be a generator. Then ¡x is also a generator of Q. Thus we may assume that x is a positive rational m number. Write x = n with m; n 2 N; gcd(m; n) = 1. Then any element of Q can be am written as ax = n for some integer a. Choose a su±cient large prime integer p > n. 1 am Then p = ax = n for some integer a (note that a must be positive). But it implies that n = amp > p, which is a contradiction. Therefore Q is not a cyclic group..
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