Math 307 Abstract Algebra Homework 2 Sample Solution 1. Prove that the set of all 2 × 2 matrices with entries from R and determinant 1 is a group under matrix multiplication. Let M be the set of all 2 × 2 matrices with real entries such that for A ∈ M , det(A) = 1. (G0) Let A, B ∈ M . Then AB is a 2 × 2 matrix, and det(AB) = det(A) det(B) = 1 · 1 = 1. Thus, AB ∈ M . (G1) We know that matrix multiplication is associative. (G2) Since det(I2 ) = 1, I2 ∈ M and satisfies I2 A = AI2 = A for each A ∈ M . (G3) Suppose A ∈ M . Then det(A) = 1 6= 0, so an inverse, A−1 exists such that det(A−1 ) = det(A)−1 = 1−1 = 1. We have AA−1 = A−1 A = I2 . Combining (G0) – (G3), we see that M is a group under matrix multiplication. 2. Prove that the set of all rational numbers of the form 3m 6n , where m, n ∈ Z, is a group under multiplication. Let G = {3m 6n : m, n ∈ Z}. (G0) If a = 3m 6n and b = 3r bs with m, n, r, s ∈ Z, then ab = 3m+r 6n+s ∈ G. (G1) Integer multiplication is associative. (G2) Note that 1 = 30 60 ∈ G satisfies 1a = a1 = a in G. (G3) If a = 3m 6n ∈ G, then b = 3−m 6−n ∈ G satisfies ab = ba = 1. Combining (G0) – (G3), we see that G is a group under matrix multiplication. 3. Prove that a group G is Abelian if and only if (ab)−1 = a−1 b−1 . 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. 4. Prove that a group G is Abelian if and only if (ab)2 = a2 b2 for any a, b ∈ G. Let G be Abelian and let a, b ∈ G, then (ab)2 = (ab)(ab) = a(ba)b = a(ab)b = a2 b2 . Now let abab = (ab)2 = a2 b2 for any a, b ∈ G. Then by cancellation bab = ab2 and by cancellation again, ab = ba. 5. Prove that in any group, an element and its inverse have the same order. 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 1 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. 6. Suppose that H is a proper subgroup of Z under addition and H contains 18, 30 and 40, Determine H. 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. 7. Let H and K be subgroups of a group G. Show that H ∩ K ≤ G. [Extra 2 points for showing intersection of any (finite or infinite) collection of subgroups of a group is a subgroup.] We prove the general case. One can specialize the proof to the case when Γ = {1, 2}. Let Hα ≤ G for all α ∈ Γ for a non-empty set Γ. T T Then since for each α, e ∈ Hα , e ∈ α Hα and α Hα 6= ∅. T T Now let a, b ∈ α Hα , so for each α, a, b ∈ Hα so ab−1 ∈ Hα . Hence ab−1 ∈ α Hα . T By the One-Step Test, α Hα ≤ G. 8. (Extra credit) Let H and K be subgroups of a group G. Show that H ∪ K ≤ G if and only if H ≤ K or K ≤ H. [Extra 2 points for proving the result of any (finite or infinite) collection of subgroups.] 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. Extension of the result to 3 subgroups or more could be problematic. Try it in the next homework! 2