Solutions to the Algebra problems on the Comprehensive Examination 2018

Solutions to the Algebra problems on theComprehensive Examination of January 26, 2018 1. (25 points) Let G1 and G2 be groups, let H1 ⊆ G1 be a subgroup, and let φ : G1 → G2 be a homomorphism. The set H2 = {φ(x) : x ∈ H1 } is called the image of H1 under φ, sometimes denoted φ(H1 ). Prove that H2 is a subgroup of G2 . [This is a standard theorem in Math 350. You must actually prove it, not just quote it.] Solution: It suffices to show that H2 is nonempty, closed under ·, and closed under −1 . Nonempty : Let e1 be the identity in G1 . Then e1 is also the identity in H1 (e.g. e1 ∈ H1 ), so that φ(e1 ) ∈ H2 , hence H2 is nonempty. (In fact, because φ is a homomorphism, φ(e1 ) = e2 , where e2 is the identity in G2 .) Closed under · : Let z, w ∈ H2 . Then there exists x, y ∈ H1 such that z = φ(x), w = φ(y). Since H1 is a subgroup, xy ∈ H1 , hence φ(xy) ∈ H2 . But φ is a homomorphism, so that φ(xy) = φ(x)φ(y). Hence, zw = φ(x)φ(y) ∈ H2 , so that H2 is closed under ·. Closed under −1 : Let z ∈ H2 . Then there exists x ∈ H1 such that z = φ(x). Since H1 is a subgroup, x−1 ∈ H1 , hence φ(x−1 ) ∈ H2 . But φ is a homomorphism, so that φ(x−1 ) = φ(x)−1 . Hence, z −1 = φ(x)−1 ∈ H2 , so that H2 is closed under −1 . 2. (25 points) Let G be a group, let N ⊆ G be a normal subgroup, and let m ≥ 1 be an integer. Suppose that for every element y ∈ G/N , the order of y divides m. Prove that for all x ∈ G, we have xm ∈ N . Solution: Let x ∈ G. Then xN ∈ G/N . Let n be the order of xN in G/N . By hypotheses, n divides m, so there is some integer k such that nk = m. Then (xN )n = N (the identity in G/N ), so that xm N = (xN )m = (xN )nk = ((xN )n )k = N k = N . That is, xm N = N , which implies xm ∈ N as claimed. 3. (25 points) Consider the group S8 of permutations of the set {1, 2, 3, . . . , 8}. Let σ, τ ∈ S8 be the permutations σ = (1, 2, 3)(4, 5, 6) and τ = (3, 5)(1, 7, 8, 4). (a) (7 points) Write σ 2 τ as a product of disjoint cycles. Solution: σ 2 τ = (1, 2, 3)(4, 5, 6)(1, 2, 3)(4, 5, 6)(3, 5)(1, 7, 8, 4) = (1, 7, 8, 6, 5, 2)(3, 4) (b) (9 points) Compute the order of each of σ, τ , and σ 2 τ . Solution: σ is given as the product of two disjoint 3-cycles, so the order of σ is lcm(3, 3) = 3. Similarly, τ is given as the product of a 2-cycle and a 4-cycle which are disjoint, so the order of σ is lcm(2, 4) = 4. Finally, from part (a), σ 2 τ is the product of a 6-cycle and a 2-cycle which are disjoint, so the order of σ 2 τ is lcm(6, 2) = 6. (c) (9 points) Decide whether each of σ, τ , and σ 2 τ is an even or odd permutation; don’t forget to justify. Solution: σ is a product of two 3-cycles (which are both even) so is even + even = even. τ is a product of a 2-cycle (odd) and a 4-cycle (odd), so is odd + odd = even. σ 2 τ is a product of a 6-cycle (odd) and a 2-cycle (odd), so is odd + odd = even. 4. (25 points) Let R be a ring. (a) (6 points) Define what it means for a subset I ⊆ R to be an ideal of R. If you use any other technical terms like “closed,” “subring,” “group,” “subgroup,” etc., you must fully define those terms as well. Solution: A subset I ⊆ R is an ideal of R if • I 6= ∅, • x, y ∈ I implies x − y ∈ I, • x ∈ I and r ∈ R implies xr ∈ I and rx ∈ I. (b) (19 points) Let I, J ⊆ R be ideals, and define I + J = {x + y : x ∈ I and y ∈ J}. Prove that I + J is an ideal of R. Solution: We will show I + J satisfies the three conditions given in part (a). • Since I and J are ideals, they are non-empty. In particular, 0R ∈ I and 0R ∈ J (where 0R is the zero element in R). Then 0R + 0R = 0R ∈ I + J. Hence, I + J 6= ∅. • Let z1 , z2 ∈ I + J. Then there exist x1 , x2 ∈ I and y1 , y2 ∈ J such that z1 = x1 + y1 and z2 = x2 + y2 . Since I is an ideal, x1 − x2 ∈ I. Similarly, since J is an ideal, y1 − y2 ∈ J. Thus, z1 − z2 = (x1 + y1 ) − (x2 + y2 ) = (x1 − x2 ) + (y1 − y2 ) ∈ I + J. • Let r ∈ R and z ∈ I + J. Then there exists x ∈ I, y ∈ J such that z = x + y. Since I is an ideal, rx and xr are in I. Similarly, since J is an ideal, ry and yr are in J. Thus, rz = r(x + y) = rx + ry ∈ I + J, and zr = (x + y)r = xr + yr ∈ I + J. 2