- July 31, 2020

Math 4281 Final Exam Practice Summer 2020 1. Prove that if I1 ⊆ I2 ⊆ · · · are ideals in a ring R, then I = ⋃∞ n=1 In is an ideal of R Proof. Since 0 ∈ I1 ⊂ I, I 6= ∅. Let a, b ∈ I. Then there exist n,m ∈ Z+ such that a ∈ In and b ∈ Im. WLOG, say m ≥ n. Then a ∈ In ⊆ Im, so a, b ∈ Im, which is an ideal, so a− b ∈ Im ⊆ I. Further, let r ∈ R. Then ra, ar ∈ In ⊆ I because In is an ideal. Hence, I is an ideal by the Ideal Test. 2. Show that n element has order 2 in Sn if and only if its cycle decomposition if a product of commuting 2-cycles. Proof. Suppose first that σ ∈ Sn is a (disjoint) product of 2-cycles. But since disjoint cycles commute, σ2 would be a product of squared 2-cycles. Since (a b)2 = 1 for any a, b, we conclude that σ is of order 2. Suppose now that σ ∈ Sn isn’t a product of 2-cycles, but instead has some other n-cycle (n 6= 2) in its decomposition. Stil, the product of disjoint cycles commutes. But every n- cycle has order n. Therefore σ2 would contain the square of an n-cycle, but that cannot be the identity. Therefore, it still permutes some elements, and thus σ2 itself cannot be the identity. 3. Let G = {z ∈ Z | zn = 1 for some n ∈ Z+}. (a) Prove that G is a group under multiplication (called the groups of roots of unity in C). Proof. For (a), let x, y ∈ G. Then by definition, xn = 1 and ym = 1 for some n,m ∈ Z+. Then (xy)nm = xnmynm = (xn)m(ym)n = 1m1n = 1, so G is closed under multiplication. Since G ⊆ C, and C is a field, associativity holds in G. Since 11 = 1 and 1x = x = x1 for all x ∈ G, G has an identity element. Let x ∈ G. Then xn = 1 for some n ∈ Z+. Note that this implies that x 6= 0, so x ∈ C is a unit and there exists x−1 ∈ C. Now to see that x−1 ∈ G, note that (x−1)n = (xn)−1 = 1−1 = 1. Thus by definition, G is a group under multiplication. (b) Prove that G is not a group under addition. Proof. note that 1 ∈ G, but 1 + 1 = 2 /∈ G. Thus G is not closed under addition, so it cannot be a group under the operation. Page 1 Math 4281 Final Exam Practice Summer 2020 4. Construct a field with 343 elements. Justify your work. Proof. Note that the polynomial f(x) = x3 + x+ [1] ∈ Z7[x] has no roots in Z7, so it is irreducible. Thus by Theorem 15.11, F = Z7[x]/〈f(x)〉 if a field, and the elements in F can be uniquely written as ax2 + bx+ c with a, b, c ∈ Z7, so F has 73 = 343 elements. 5. Suppose G is a finite group and x, g ∈ G. (a) Prove that |x| = |g−1xg|. Proof. For (a), since G is finite, |x| = n < ∞ and |g−1xg| = m < ∞ for some n,m ∈ Z+. let e denote the identity element of G. Using xn = e, and multiplying both sides by g−1 on the left and g on the right to get g−1xng = g−1eg = e. But then e = g−1xng = g−1xexex · · ·xexg = g−1xgg−1xgg−1x · · ·xgg−1xg = (g−1xg)n. Thus |g−1xg| = m ≤ n. Now e = (g−1xg)m = g−1xgg−1xg · · · g−1xg = g−1xmg. Multiplying both sides on the left by g and the right by g−1, we obtain xm = geg−1 = e, so n ≤ m. Hence we have shown that n = m, or |x| = |g−1xg|. (b) Use part (a) to deduce that for all a, b ∈ G, |ab| = |ba|. Proof. let x = ab and g = a. Then |ab| = |x| = |g−1xg| = |a−1aba| = |ba|. 6. Find all of the elements in the subgroup K = 〈(12)(34), (125)〉 ≤ S5. Proof. We compute the powers of the given elements and all products to obtain K = {I, (12)(34), (125), (152), (25)(34), (15)(34)} 7. (a) Let G = 〈g〉 be a cyclic group of order 32. Compute the order of g24 in G, justifying your answer. Proof. By Theorem 23.6.ii, |g24| = 32 gcd(24,32) = 32 8 = 4. (b) Let G′ be a group, and let a ∈ G′ be an element of order 21. Let H = 〈a〉 and K = 〈a7〉. Find all of the distinct left cosets of K in H and provide their elements. Proof. We compute the sets gK for g ∈ H = 〈a〉, so all g = an for n ∈ Z. The six distinct left cosets of K in H are: Page 2 Math 4281 Final Exam Practice Summer 2020 K = 〈a7〉 = {1, a7, a14}, aK = a〈a7〉 = {a, a8, a15}, a2K = a2〈a7〉 = {a2, a9, a16}, a3K = a3〈a7〉 = {a3, a10, a17}, a4K = a4〈a7〉 = {a4, a11, a18}, a5K = a5〈a7〉 = {a5, a12, a19}, and a6K = a6〈a7〉 = {a6, a13, a20}. (c) Compute the distinct left cosets of 〈R2〉 in D6 and provide their elements. Proof. The group D6 = {I, R,R2, R3, R4, R5, r, rR, rR2, rR3, rR4, rR5} is the group of symmetries of a regular hexagon. The subgroup 〈R2〉 = {I, R2, R4}, so its left cosets are: 〈R2〉 = {I, R2, R4}, R〈R2〉 = {R,R3, R5}, r〈R2〉 = {r, rR2, rR4}, and rR〈R2〉 = {rR, rR3, rR5}. 8. Find all odd prime integers p so that x+ [2] divides f(x) = x4 + x+ [1] ∈ Zp[x]. Proof. x+ [2] divides f(x) is −[2] is a root of f(x). Note that f(−[2]) = (−[2])4 − [2] + [1] = [15] which will be zero in Zp if and only if p = 3 or 5. Page 3