Math 307 Abstract Algebra Sample Examination 1 with solution There will be 6 questions in the examination. At least 4 comes from the following list. 1. Prove that a group G is Abelian if and only if (ab)−1 = a−1 b−1 . Solution. Let G be Abelian, that is, for any a, b ∈ G, ab = ba. Then (ab)−1 = (ba)−1 = a−1 b−1 . Now suppose (ab)−1 = a−1 b−1 for all a, b ∈ G. Then (ab)(ab)−1 = e and (ba)(ab)−1 = ba(a−1 b−1 ) = e. By cancellation, ab = ba. 2. Prove that in any group, an element and its inverse have the same order. Solution. Let G be a group and let a ∈ G. First suppose |a| = m is finite. Then am = e so that (a−1 )m am = e. Thus, (a−1 )m = e and |a−1 | = n ≤ m because |b| is the smallest positive integer n such that bn = e. If n < m, then am−n = am (a−1 )n = e, contradicting the fact that |a| = m. So, m = n. Next, suppose |a| is infinite. If |a−1 | = n is finite, then by the above argument |a| = |(a−1 )−1 |, which is a contradiction. 3. Suppose that H is a proper subgroup of Z under addition and H contains 18, 30 and 40, Determine H. Solution. Since gcd(18, 30, 40) = 2, there exists an x, y, z ∈ Z such that 18x + 30y + 40z = 2. In fact, one easily checks that 2 = 2 ∗ 40 − 2 ∗ 30 − 1 ∗ 18 ∈ H. So, H contains 2Z, which is the set of all even numbers. If H contains any additional element a, it will be of the form 2k + 1. Then 1 = (2k + 1) − 2k ∈ H and H = Z. Hence, H cannot contain other elements, and H = 2Z. 4. Let H and K be subgroups of a group G. Show that H ∪ K ≤ G if and only if H ≤ K or K ≤ H. Solution. Let G be a group and let H, K ≤ G. Assume without loss of generality that H ≤ K, that is H ⊆ K, which implies that H ∪ K = K ≤ G. Conversely, assume that H 6≤ K and K 6≤ H, that is H 6⊆ K and K 6⊆ H, which implies that H ∪ K 6= K and H ∪ K 6= H. Then, there exists an h ∈ H\K and a k ∈ K\H such that h, k ∈ H ∪ K, Suppose, H ∪ K were a subgroup of G. Then hk ∈ H ∪ K. Case 1. If hk ∈ H, then h−1 ∈ H and hence k = h−1 (hk) ∈ H, which is a contradiction. Case 2. If hk ∈ K, then k −1 ∈ K and hence h = (hk)k −1 ∈ K, which is a contradiction. Thus, H ∪ K cannot be a subgroup. 5. Suppose a and b are elements in a group such that |a| = 4, |b| = 2, and a3 b = ba. Find |ab|. Solution. We prove that |ab| = 2. Note that (ab)(ab) = a(ba)b = a(a3 b)b = a4 b2 = e. So, |ab| = 1 or 2. If |ab| = 1, then a is the inverse of b so that 4 = |a| = |b| = 2, which is absurd. So, |ab| = 2. 6. Let a and b belong to a group. If |a| and |b| are relatively prime, show that hai ∩ hbi = {e}. Solution. Suppose H = hai = {a, a2 , . . . , am } and K = hbi = {b, b2 , . . . , bn }, where am = bn = e, such that gcd(m, n) = 1. Clearly, e ∈ H ∩ K. Suppose c ∈ H ∩ K and |c| = k. Then k is factor of m and also a factor of n. Thus, k = 1 and c = e. 1 7. Suppose G is a set equipped with an associative binary operation ∗. Furthermore, assume that G has an left identity e, i.e., eg = g for all g ∈ G, and that every g ∈ G has an left inverse g 0 , i.e., g 0 ∗ g = e. Show that G is a group. Solution. Let g ∈ G. We first show that the left inverse g 0 of g is also the right inverse. To see this, let gˆ be the left inverse of g 0 . Then (ˆ g ) = (ˆ g )(g 0 g) = (ˆ g g 0 )g = eg = g. So, gˆ = g 0 0 satisfies e = gˆg = gg . Now, because gg 0 = g 0 g = e, we have ge = g(g 0 g) = (gg 0 )g = g. 8. Suppose x is an element of a cyclic group of order 15 and exactly two of x3 , x5 , and x9 are equal. Determine |x13 |. Solution. Let x ∈ G = hai = {a, . . . , a15 }. Clearly, |x| > 1, else e = x3 = x5 = x9 . Note also that |x| is a factor of |G| = 15. Thus, |x| ∈ {3, 5, 15}. Consider 3 cases. 1. x3 = x5 6= x9 . Then x2 = x5−3 = e. So, |x| = 2, a contradiction. 2. x3 6= x5 = x9 . Then x4 = x9−5 = e so that |x| ∈ {2, 4}, a contradiction. 3. x3 = x9 6= x5 . Then x6 = x9−3 = e so that |x| ∈ {2, 3, 6}. Thus, |x| = 3 and |x13 | = |x| = 3. 9. Consider σ = (13256)(23)(46512). (a) Express σ as a product of disjoint cycles. Solution. σ = (1, 2, 4)(3, 5). (b) Express σ as a product of transpositions. Solution. σ = (1, 4)(1, 2)(3, 5). (c) Express σ as a product minimum number of transpositions. (Prove that the number is minimum!) Solution. σ moves more than 5 numbers in {1, . . . , 6}. So, we need at least three transpositions. 10. (a) Let α = (1, 3, 5, 7, 9, 8, 6)(2, 4, 10). What is the smallest positive integer n such that αn = α−5 ? Solution. We need to find the smallest n such that αn+5 = ε. Since |α| = lcm(5, 3) = 21, we see that n = 16. (b) Let β = (1, 3, 5, 7, 9)(2, 4, 6)(8, 10). If β m is a 5-cycle, what can you say about m? Solution. Note that β m is a 5-cycle if and only if (2, 4, 6)m = (8, 6)m = ε and (1, 3, 5, 7, 9)m is a five cycle. This happen if and only if m is a multiple of 6 = lcm(3, 2) and m is not a multiple of 5. That is m = 6k and k is not a multiple of 5. 11. In S7 show that x2 = (1, 2, 3, 4) has no solutions, but x3 = (1, 2, 3, 4) has at least two. Solution. Note that (x2 )4 = ε. So, |x| = 1, 2, 4. Clearly, |x| 6= 1, 2, else x2 6= (1, 2, 3, 4). If |x| = 4, then x is a 4-cycle, or the product of a 4-cycle and a 2-cycle; in either case, x2 6= (1, 2, 3, 4). A shorter proof is to observe that x2 = (1, 2, 3, 4) = (1, 4)(1, 3)(1, 2) is an odd permutation. But x2 must be an even permutation for any x ∈ Sn . Let x ∈ {(1, 4, 3, 2), (1, 4, 3, 2)(5, 6, 7)}. Then x3 = (1, 2, 3, 4). 2 12. Let H ≤ Sn . (a) Show that either H ≤ An or |H ∩ An | = |H|/2. Solution. Suppose H ≤ Sn . Let S1 = H ∩ An , and S2 = H − S1 . Case 1. If S2 = ∅, then H ≤ An . Case 2. If S2 6= ∅ and g ∈ S2 is an odd permutation. Then define f : S1 → S2 by f (x) = gx. It is well defined because for every even permutation x ∈ H, gx ∈ H is an odd permutation and will be in S2 . It is 1-1 because f (x1 ) = f (x2 ) implies gx1 = gx2 so that x1 = x2 by cancellation. If is onto because for every y ∈ S2 , we can let x = g −1 y ∈ H ∩ An = S1 so that f (x) = y. Since there is a bijection from S1 to S2 , we see that |S1 | = |S2 |, and |H ∩ An | = |H|/2 as asserted. (b) If |H| is odd, show that H ≤ An . Solution. Since |H| is odd, it cannot be the case that |H ∩ An | = |H|/2. So, H ≤ An . 13. Let G be a group. Show that φ : G → G defined by φ(g) = g −1 is an isomorphism if and only if G is Abelian. Solution. Suppose G is Abelian. First, we show that φ is bijective. Clearly, if φ(a) = φ(b), then a−1 = b−1 . Taking inverse on both sides, we see that a = b; so φ is 1-1. If a ∈ G, then φ(a−1 ) = a; so φ is onto. Now, by commutativity, for any a, b ∈ G. φ(ab) = (ab)−1 = b−1 a−1 = a−1 b−1 = φ(a)φ(b). Thus, φ is a group isomorphism. Conversely, suppose φ is an isomorphism. Then for any a, b ∈ G, a−1 b−1 = φ(a)φ(b) = φ(ab) = (ab)−1 = b−1 a−1 . Taking inverse on both sides, we see that ba = ab. 14. Let G be a group with |G| = pq, where p, q are primes. Prove that every proper subgroup of G is cyclic. But the whole group may not be cyclic. Solution. Let H be a proper subgroup of G. Then |H| ∈ {1, p, q}. By Homework 2, or a corollary of Lagrange theorem, H has prime order or order 1 is cyclic. Consider S3 of order 6. Every proper subgroup is cyclic, but S3 is not. 15. Suppose G is a group of order n, and k ∈ N is relatively prime to n. Show that g : G → G defined by g(x) = xk is one-one. If G is Abelian, show that g is an automorphism. Solution. Let H = hxi and K = hyi. Then H ∩ K contains xk = y k . Note that |H| and |K| are factors of n by Lagrange Theorem, k is relatively prime to |H| and also to |K|. So, xk and y k are generators of H and K. Thus, H = K = hxk i = hy k i. Hence, y ∈ H so that y = xr for some 1 ≤ r ≤ m = |H|. Note that m = |H| is a factor of n = |G| is relatively prime to k. If r 6= 1, then xk = y k = xrk so that xk(r−1) = e. So, k(r − 1) is a multiple of m. Since k and m are relatively prime, m is a factor of 1 ≤ r − 1 < m, which is impossible. So, r = 1, i.e., x = y. If G is Abelian, then (xy)k = xk y k so that the map x 7→ xk is an isomorphism. 3