Solutions to the Algebra Problems on the Comprehensive Examination

Solutions to the Algebra problems on the Comprehensive Examination of January 27, 2012 1. (25 points). Let G be a group, and let H, K ⊆ G be subgroups of G. (a) Prove the following standard theorem about subgroups: that H ∩ K is a subgroup of G. Solution: (H ∩ K nonempty) Since H and K are subgroups of G, e ∈ H and e ∈ K so e ∈ H ∩ K, which means that H ∩ K 6= ∅.X (H ∩ K closed under group operation) Given a, b ∈ H ∩ K, ab ∈ H and ab ∈ K since H and K are closed under ∗, so ab ∈ H ∩ K as desired.X (H ∩ K closed under −1 ) Given a ∈ H ∩ K, a−1 ∈ H and a−1 ∈ K since H and K are closed under inverses, so ab ∈ H ∩ K as desired.X Thus H ∩ K is a subgroup of G as desired. QED (b) If H and K are both normal subgroups of G, prove that H ∩ K is also a normal subgroup of G. Solution: First of all, we know from part (a) that H ∩ K is a subgroup of G. Now, given a ∈ H ∩ K, g ∈ G, gag −1 ∈ H and gag −1 ∈ K since H and K are normal. Thus, gag −1 ∈ H ∩ K, so H ∩ K is normal in G as desired. QED 2. (25 points). Let G and H be groups. Recall that a homomorphism φ : G → H is said to be trivial if φ(g) = eH for all g ∈ G. (a) If |G| = 144 and |H| = 25, prove that any homomorphism φ : G → H is trivial. Solution: Given g ∈ G, since o(g) | |G| = 144, we have g 144 = eG ; note that similarly, (φ(g))25 = eH . Since φ is a homomorphism, we have eH = φ(eG ) = φ(g 144 ) = (φ(g))144 . 25 144 In any group,
hm = e implies o(h) = eH
| m. Applying this to (φ(g)) =4 (φ(g)) 2 2 gives o φ(g) | 144
and o φ(g) | 25. Since gcd(144, 25) = gcd(2 3 , 5 ) = 1, we must have o φ(g) = 1, which means φ(g) = eH as desired. QED (b) Let G be the cyclic group of order 2, and let H be the cyclic group of order 6. Give an example of a nontrivial homomorphism φ : G → H. Solution: Define φ : G → H by 0 7→ 0, 1 7→ 3. G is small enough that we can confirm this is a homomorphism by hand: φ(0 ⊕ 0) = 0 = 0 ⊕ 0 = φ(0) ⊕ φ(0), φ(0 ⊕ 1) = 3 = 0 ⊕ 3 = φ(0) ⊕ φ(1), and φ(1 ⊕ 1) = 0 = 3 ⊕ 3 = φ(1) ⊕ φ(1).X 3. (25 points). (a) List all elements of A4 , the alternating group of degree four, expressing each such element as a product of disjoint cycles. Solution: A4 consists of e, 3-cycles, and ”double 2-cycles”: A4 = {e, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3), (1 2 3), (1 3 2), (1 2 4), (1 4 2), (1 3 4), (1 4 3), (2 3 4), (2 4 3)} Note that |A| = 12 = 4!/2 as it should be. (b) For each element you listed, say what its order is. Solution: o(e) = 1, the order of each 3-cycle is 3, and the order of the double 2-cycles ((1 2)(3 4), (1 3)(2 4), and (1 4)(2 3)) is 2. 4. (25 points). 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, we have xr ∈ I and rx ∈ I.    a b (b) Let R = : a, b, c ∈ R . You may assume that R is a ring under the 0 c operations of matrix addition and matrix multiplication.    a b : a, b ∈ R . Prove that I is an ideal of R. Let I = 0 0 Solution: First,  we show that (I, +) is a subgroup of (R, +): 0 0 (I non-empty) ∈ I, so I 6= ∅.X 0 0       a b c d a+c b+d (I closed under +) Given x = and y = , x+y = ∈ 0 0 0 0 0 0 I.X     −a −b a b ∈ I.X , −x = (I closed under negatives) Given x = 0 0 0 0 Thus (I, +) is a subgroup     of (R, +).X   ax by x y a b ∈ I and xr = , rx = ∈ R and x = Now, given r = 0 0 0 0 0 c   ax bx + cy ∈ I.X 0 0 Thus I is an ideal of R. QED 2