Solutions to the Algebra problems on the Comprehensive Examination of February 1, 2008

Solutions to the Algebra problems on the Comprehensive Examination of February 1, 2008 1. Let G be a group, let H ⊆ G be a subgroup, and define the normalizer of H to be N(H) = {x ∈ G : x−1 Hx = H}. (a) Prove that N(H) is a subgroup of G. Solution: (N(H) nonempty) Trivially e−1 He = H, so e ∈ N(H).X (N(H) closed under group operation) Given x, y ∈ N(H), using associativity and the property of elements of N(H): (xy)−1H(xy) = y −1x−1 Hxy = y −1 Hy = H so xy ∈ N(H).X (N(H) closed under −1 ) Given a ∈ N(H), a−1 Ha = H, so H = a(a−1 Ha)a−1 = aha−1 = (a−1 )−1 Ha−1 , so a−1 ∈ N(H).X Thus N(H) is a subgroup of G as desired. QED (b) Prove that H is a subgroup of N(H). Solution: It is given that H is a subgroup of G, so we already know that H forms a group under the group operation. Thus, it is sufficient to show that H ⊆ N(H). Given a ∈ H (note that since H is closed under inverses, a−1 ∈ H), we show that a−1 Ha = H. ⊆: Given a−1 ha ∈ a−1 Ha, since H is closed under group operation, a−1 ha ∈ H.X ⊇: Given h ∈ H, since H is closed under group operation, aha−1 ∈ H so h = a−1 (aha−1 )a ∈ a−1 Ha.X Thus a ∈ H ⇒ a ∈ N(H) so H ⊆ N(H), so H is a subgroup of N(H) as desired. QED (c) Prove that H is a normal subgroup of N(H). Solution: Given x ∈ N(H), h ∈ H: by the property of N(H), x−1 Hx ⊆ H. Thus since x−1 hx ∈ x−1 Hx, we have x−1 hx ∈ H as desired. QED 2. Recall that Sn denotes the group of permutations on n symbols. (a) Find an element of S10 of order 21. Solution: The order of a permutation σ is the lcm of the orders of each individual disjoint cycle comprising σ (and the order of an n-cycle is n). Since 21 = lcm(3, 7), the element σ = (1 2 3)(4 5 6 7 8 9 10) has order 21. (b) Prove that no element of S10 has order 11. Solution: Suppose there exists a permutation σ of order 11. Then σ must have a disjoint cycle structure such that the lcm of the cycle lengths is 11. If Xσ = {x1 , . . . , xn } is the set of cycle lengths of σ, this means that 11 = lcm(x1 , . . . , xn ) = x1 × · · · × xn gcd(x1 , . . . , xn ) But since 11 is prime, this means that 11 xi for some i; for that i, therefore, xi ≥ 11. Thus σ has a cycle that is (at least) an 11-cycle. But that’s impossible, since σ is a permutation on only 10 symbols! ⇒⇐ Thus no element of S10 has order 11. QED 3. Let R be a ring. (a) Define what it means for a subset I ⊆ R to be an ideal of R. Solution: I ⊆ R is an ideal of R if (I, +) is a subgroup of (R, +) and ∀x ∈ I, r ∈ R, xr ∈ I and rx ∈ I. (b) Let S be another ring, and let φ : R → S be a ring homomorphism. Let I ⊆ R be an ideal of R, and set J = {x ∈ I : φ(x) = 0S }, where 0S denotes the zero element of S. Prove that J is an ideal of R. Solution: First, we show that (J, +) is a subgroup of (R, +): (J nonempty) Pick some x ∈ I (such an x must exists because I ideal ⇒ J 6= ∅). Then since I is an ideal, 0R = 0R x ∈ I. Since φ is a ring homomorphism, φ(0R ) = 0S so 0R ∈ J.X (J closed under +) Given x, y ∈ J ⊆ I, x + y ∈ I because I is closed under +. Also, since φ is a homomorphism, φ(x + y) = φ(x) + φ(y) = 0S + 0S = 0S . Thus, x + y ∈ J.X (J closed under negatives) Given x ∈ J ⊆ I, −x ∈ I because I is closed under negatives. Also, since φ is a homomorphism, φ(−x) = −φ(x) = −0S = 0S . Thus, −x ∈ J.X Thus (J, +) is a subgroup of (R, +) X Now given r ∈ R, x ∈ J ⊆ I: rx, xr ∈ I since I is an ideal. Now since φ is a homomorphism, φ(rx) = φ(r)φ(x) = φ(r)0S = 0S and φ(xr) = φ(x)φ(r) = 0S φ(r) = 0S , so xr, rx ∈ J.X Thus, J is an ideal of R. QED 4. Let R = R[x] be the ring of polynomials with coefficients in the field R of real numbers. Let I ⊆ R be the subset I = {f ∈ R[x] : f (1) = f (2) = 0}. (a) Prove that I is an ideal of R. Solution: First, we show that (I, +) is a subgroup of (R, +): (I nonempty:) Let 0(x) be the zero polynomial (all coefficients are zero). Then 0(1) = 0(2) = 0 so 0(x) ∈ I.X (I closed under +) Given f, g ∈ I, (f + g)(1) = f (1) + g(1) = 0 + 0 = 0 and (f + g)(2) = f (2) + g(2) = 0 + 0 = 0, so (f + g) ∈ I.X (I closed under negatives) Given f ∈ I, (−f )(1) = −f (1) = −0 = 0 and 2 (−f )(2) = −f (2) = −0 = 0 so −f ∈ I.X Thus (I, +) is a subgroup of (R, +) X Now given r ∈ R, f ∈ I: (rf )(1) = r(1)f (1) = r(1)(0) = 0 and (rf (2) = r(2)f (2) = r(2)(0) = 0. Similarly, (f r)(1) = 0 and (f r)(2) = 0. Thus, rf, f r ∈ I.X Thus, I is an ideal of R. QED (b) Prove that R/I has zero-divisors. That is, show that there are two nonzero elements of R/I whose product is zero in R/I. Solution: Let f (X) = X − 1 and g(X) = X − 2. Then f (2) = 1 6= 0 and g(1) = −1 6= 0, so f, g ∈ / I. Thus, I + f and I + g are nonzero. However, (I + f )(I + g) = I + h, where h(X) = (X − 1)(X − 2). Since h(1) = h(2) = 0, h ∈ I which means that I + h = I + 0 (0 here is the zero polynomial 0(X)). Thus, f and g are zero-divisors as desired. QED 3