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Home , Infinite group

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 of any element of a finite cyclic G is a divisor of |G|. Solution.True. Let g be a generator of G and n = |G|. Then any element ge of n G has order gcd(e,n) which is a divisor of n = |G|. b. (5pt) Any non-trivial 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 is cyclic, then G is cyclic. Solution. False. For example, any proper subgroup of Klein 4-group V = {e, a, b, c} is cyclic, but V is not cyclic. d. (5pt) If G is an infinite group, then any non-trivial subgroup of G is also an infinite group. Solution. False. For example, hC∗, ·i is an infinite group, but the subgroup n ∗ Un = {z ∈ C ∗ |z = 1} is a finite subgroup of C .

2. (10pt) Show that a group that has only a finite number of must be a finite group. Solution. We show that if any infinite group G has infinitely many subgroups. (case 1) Suppose that G is an infinite 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 infinitely many subgroups. (case 2). Suppose that G has an infinite cyclic subgroup H. By (case 1), H has infinitely many subgroups. Since any subgroup of H is also a subgroup of G, G also has infinitely many subgroups. (case 3) Finally assume that G has no infinite cyclic subgroup, i.e. hgi is a finite

subgroup of G for any g ∈ G. Choose e 6= g1 ∈ G, and set H1 = hg1i. By assumption,

H1 G. Choose ∈ G − H1, and set H2 = hg2i. Then H2 6= H1. Note that H1 ∪ H2

is a finite subset of G. Choose g3 ∈ G − (H1 ∪ H2), and set H3 = hg3i. Then 1 2

H3 6= H1,H2. In this way (or inductively) we can choose infinitely many subgroups of G.

3. Let S be any subset of a group G.

a. (5pt) Show that HS = {x ∈ G|xs = sx for all s ∈ S} is a subgroup of G. −1 Solution. It suffices to show that for any x, y ∈ HS, xy ∈ HS. Let x, y ∈ HS. For any s ∈ 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 ∈ HS. Therefore HS is a subgroup of G.

b. (5pt) Show that HG is an .

Solution. To show that HG is an abelian group, we need to show that xy = yx

for any x, y ∈ HG. Let x, y ∈ HG. By definition of HG, it holds that xy = yx

for any x, y ∈ HG.

4. (10pt) Show that any finite group G without proper non-trivial subgroups is a cyclic group with a prime number of elements (We assume that |G| > 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 |G| = n. If n is not a prime, then there exists a positive a such that a|n 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 . Find the subgroup of Z generated by {a, b}. Solution. The subgroup of Z generated by {a, b} is {ax+by|x, y ∈ Z}. We already have known that {ax+by|x, y ∈ Z} 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 ∈ 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 ∈ N, gcd(m, n) = 1. Then any element of Q can be am written as ax = n for some integer a. Choose a sufficient 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.