Introduction to Abstract Algebra, Spring 2013 Solutions to Midterm I

Introduction to Abstract Algebra, Spring 2013 Solutions to Midterm I 1. a. What is an equivalence relation on a set? b. How many distinct equivalence relations are there on a set with three elements? c. Let S be a non-empty set and write P(S) for the set of all subsets of S. For X and Y in P(S), define X ∗ Y = X ∩ Y . Is P(S) a group under this operation? a. An equivalence relation ∼ on a set S is a reflexive, symmetric, transitive relation on S. That is, for all x, y, z ∈ S, i. x ∼ x (∼ is reflexive), ii. if x ∼ y then y ∼ x (∼ is symmetric), iii. if x ∼ y and y ∼ z then x ∼ z (∼ is transitive). b. The number of equivalence relations on a set is the number of partitions of the set (that is, the number of ways in which the set can be written as a disjoint union of non-empty subsets). If S = {a, b, c} then the distinct partitions of S are given by {a, b, c}, {a, b} ∪ {c}, {a, c} ∪ {b}, {b, c} ∪ {a}, {a} ∪ {b} ∪ {c}. Thus there are 5 distinct equivalence relations on S. c. For any X ⊂ S, we have X ∩ S = X = S ∩ X. This means that S is an identity element. Suppose X $ S (for example, take X = ∅). Then, for any Y ⊂ S, X ∩ Y ⊂ X 6= S. Thus X does not have an inverse and hence P(S) cannot be a group. 2. Let G be a group and let H be a subset of G. a. What does it mean to say that H is a subgroup of G? b. Suppose that H is a finite non-empty subset of G that is closed under the binary operation in G (i.e., if x, y ∈ H then xy ∈ H). Show that H is a subgroup of G. c. Consider the group Z16 of residues modulo 16 (under addition modulo 16). How many subgroups does this group have? Explain your answer. a. It means that H is a nonempty subset of G that is closed with respect to the binary operation on G (i.e., if x, y ∈ H then xy ∈ H) and such that the induced binary structure on H makes H into a group. Comment. We showed in class that this is the same as saying that H satisfies the conditions: i. if x, y ∈ H then xy ∈ H; ii. 1 ∈ H (i.e., the identity element of G belongs to H) iii. if h ∈ H then also h−1 ∈ H.

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b. By hypothesis, H is closed under the binary operation in G. We’ll show that 1 ∈ H and that if h ∈ H then h−1 ∈ H. By the preceding comment, this suffices to show that H is a subgroup of G. For h ∈ H, the infinite sequence h, h2 , h3 , . . . , hn , . . . consists of elements of H. Since H is finite, we have hk = hl , for distinct positive integers k and l with say k < l. Thus hl−k = 1, and so hm = 1, for some positive integer m. Hence 1 ∈ H and h−1 = hm−1 ∈ H. c. The group Z16 is cyclic. From class, a finite cyclic group of order n has a unique subgroup of order d for each (positive) divisor d of n and no other subgroups. Thus # subgps. of Z16 = # divisors of 16 = 5.

3. a. What does it mean to say a group is (i) cyclic, (ii) abelian? b. Show that every cyclic group is abelian. c. Consider the multiplicative group C× of non-zero complex numbers and let µ∞ = {z ∈ C× : z n = 1 for some n ∈ Z}. i. Show that µ∞ is a subgroup of C× and that µ∞ is not cyclic. ii. Show that every finite subgroup of µ∞ is cyclic. (You may use without proof the fact that the group of N -th roots of unity is cyclic (for any positive integer N ).) a. A group G is cyclic if there is an a ∈ G such that G = {an : n ∈ Z}. In other words, for any g ∈ G, then there is an integer n (depending on g) such that g = an . In this case, a is said to generate G or to be a generator of G. A group G is abelian if gh = hg, for any g, h ∈ G. b. Suppose G is cyclic and that a generates G. Let g, h ∈ G. Thus there are integers k and l such that g = ak and h = al . Hence gh = ak al = ak+l = al+k = al ak = hg, and so G is abelian. c.

i. Clearly, 1 ∈ µ∞ . Let z and w be elements of µ∞ so that z m = 1 and wn = 1 for integers m and n. We have (zw)mn = z mn wmn = (z m )n (wn )m = 1n 1m = 1, and so µ∞ is closed under multiplication. Further, (z −1 )m = (z m )−1 = 1−1 = 1, so µ∞ is also closed under taking inverses. This proves that µ∞ is a subgroup of C× . To see that µ∞ cannot be cyclic, note first that µ∞ is infinite. For example, for any positive integer N , µ∞ contains the cyclic group µN of order N consisting of the complex N -th roots of unity. By definition, each element of µ∞ has finite order and thus the cyclic group generated by any element if µ∞ is finite. In particular, no element of µ∞ is a generator of µ∞ , that is, µ∞ is not cyclic.

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ii. Let ν be a finite subgroup of µ∞ . Note there is a positive integer N such that z N = 1 for all z ∈ ν. (For example, we could take N to be the product of the orders of the elements of ν, or more efficiently we could take N = |ν|, the order of ν (recall from class the order of each element of ν divides |ν|).) Thus ν is a subgroup of µN , the cyclic group of N -th roots of unity in C× . It follows that ν must be cyclic (from class, any subgroup of a cyclic group is cyclic).   a b 4. Let G = { : a, b ∈ Z with a = ±1}. 0 1 a. Show that G is a group under matrix multiplication. b. Is G abelian? Explain your answer. c. Describe all elements of order two in G.     a b c d 0 a. Let g = and g = be elements of G. Then 0 1 0 1      a b c d ac ad + b gg 0 = = 0 1 0 1 0 1 Note ac = ±1 (since a = ±1 and c = ±1) and clearly ad + b ∈ Z. Thus gg 0 ∈ G. In other words, matrix multiplication defines a binary operation on G. This is an associative operation (since matrix multiplicationfor matrices with integer entries is  1 0 associative). The usual 2 × 2 identity matrix I2 = is an identity element (since 0 1   a b AI2 = A = I2 A, for any 2 × 2 matrix A with say integer entries). For any g = 0 1   1/a −b/a in G, the matrix g is invertible with inverse g −1 = . (Check this!) Since 0 1 a = ±1, we see that 1/a = ±1 and −b/a ∈ Z and thus the matrix g −1 belongs to G. This proves that G is a group under matrix multiplication. b. No, G is not abelian. For example,   −1 1 1 0 1 0

  1 −1 = 1 0

 1 0

  1 −1 = 1 0

 2 , 1

while  1 0

1 1



−1 0

so that  −1 0 c. An element

 1 1 1 0

  1 1 6= 1 0

 1 −1 1 0

 1 . 1

  a b in G has order two if and only if it not the identity and 0 1     2    a b a b a ab + b 1 0 = = . 0 1 0 1 0 1 0 1

So we must have a = ±1 and b(a + 1) = 0. Note b 6= 0 implies a = −1. On the other hand, b = 0 and a = 1 gives I2 which has order one, not two. Hence a = −1

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for of order two. Thus the elements of order two are exactly the matrices  all elements  −1 b as b varies through Z. 0 1 5. a. State Lagrange’s theorem. b. Show that every group of prime order is cyclic. c. The set of ordinary integers Z is a subgroup of the additive group of rational numbers Q. Show that Z has infinite index in Q (that is, there are infinitely many (left or right) cosets of Z in Q). a. Let G be a finite group and let H be a subgroup of G. Then the order of H divides the order of G. b. Suppose |G| is prime. Let g ∈ G with g 6= 1. Then hgi = {g n : n ∈ Z} is subgroup of G. By Lagrange’s Theorem, |hgi| divides |G|. Note |hgi| > 1 as g ∈ hgi (and 1 ∈ hgi). Since |G| is prime, it follows that |hgi| = |G|. Thus hgi = G, and G is cyclic. c. Let x, y ∈ Q with 0 ≤ x < 1 and 0 ≤ y < 1. We have x + Z = y + Z if and only if x − y = n, for some n ∈ Z. Since x and y are each in the interval [0, 1), this is only possible if x = y. Thus the cosets x + Z, for 0 ≤ x < 1, form an infinite family of distinct cosets of Z in Q. (In fact, these are all the cosets of Z in Q.)