Some of the problems are very easy, some are harder. 1. Let F : Z Z be a function defined as F (x) = 10x. (a) Prove that F is a group homomorphism. (b) Find Ker(F ) Solution: Ker(F ) = {0}. Proof: Let x Ker(F ). Then F (x) = 0 by definition of the kernel of a homomorphism. Therefore 10x = 0. In Z product of two numbers can be 0 only if one of them is 0. Therefore x = 0. So Ker(F ) = {0}. (c) Find Im(F ) Im(F ) = {10i i any integer} = {all integer multiples of 10} = {all integers of the form 10i, whereiis an {..., 30, 20, 10, 0, 10, 20, 30,...} 2. Let f : Z 6 D 3 be the homomorphism (!) defined as f(1) = ρ. Solution: Z 6 = ({0, 1, 2, 3, 4, 5}, + 6 ) and D 3 = ({e, ρ, ρ 2, t, tρ, tρ 2 ), multiplication) (a) Find f(2) = f(2) = f(1 + 1) = f(1)f(1) = ρρ = ρ 2 (b) Find f(3) = (c) Find f(4) = f(3) = f(2 + 1) = f(2)f(1) = ρ 2 ρ = ρ 3 = e f(4) = f(3 + 1) = f(3)f(1) = eρ = ρ (d) Find f(5) = f(5) = f(3 + 2) = f(3)f(2) = eρ 2 = ρ 2 (e) Find f(0) = f(0) = e (f) Find Ker(f) Ker(f) = {x Z 6 f(x) = e} is the definition of the kerenel Ker(f) = {0, 3} by the above computation. (g) Find Im(f) Im(f) = {y D 3 there exist x Z 6 so that f(x) = y} defn of image Im(f) = {e, ρ, ρ 2 } by the above computation. (h) Find Z 6 /Ker(f) These are (left) cosets of Ker(f) = {0, 3} in Z 6 : 0 + {0, 3} = {0, 3} 1 + {0, 3} = {1, 4} 2 + {0, 3} = {2, 5} 3 + {0, 3} = {3, 0} = {0, 3} 4 + {0, 3} = {4, 1} = {1, 4} 5 + {0, 3} = {5, 2} = {2, 5} 1
(i) Describe an isomorphism Z 6 /Ker(f) Im(f) f : Z 6 /Ker(f) Im(f) is defined as f(x + Ker(f)) := f(x) f({1, 4}) = f(1 + {0, 3}) = f(1) = ρ f({2, 5}) = f(2 + {0, 3}) = f(2) = ρ 2 f({0, 3}) = f(3 + {0, 3}) = f(3) = e 3. Let G and H be two groups and G H the external direct product of G and H. (a) Prove that the map f : G H H G defined as f(g, h) = (h, g) is a group homomorphism. i. Find Ker(f) Ker(f) = {(e G, e H )}. You have to prove this! Proof: First notice that: elements of G H are all pairs {(g, h) g G, h H} operation is componentwise identity of G H is the pair (e G, e H )! Similarly: elements of H G are all pairs {(h, g) g G, h H} operation is componentwise identity of H G is the pair (e H, e G )! This was just set-up and reminder of the definitions! Now the actual proof: Let (a, b) Ker(f). Then f(a, b) is the identity in H G. So f(a, b) = (e H, e G ). But f(a, b) = (b, a) be definition of f. Therefore (e H, e G ) = (b, a). Therefore e H = b and e G = a. So (a, b) = (e G, e H ). Hence Ker(f) {(e G, e H )} and since this is only one element in G H, it follows that Ker(f) = {(e G, e H )} ii. Find Im(f) Im(f) = H G. You have to prove this! (b) Prove that the map i : G G H defined as i(g) = (g, e h ) is a group homomorphism (e H is the identity element in H). i. Find Ker(i) ii. Find Im(i) Im(i) = {(g, e H ) g G} You need to prove this! (c) Prove that the map p : G H G defined as p(g, h) = g is a group homomorphism. i. Find Ker(p) ii. Find Im(p) Im(p) = G. You need to prove this! 4. Describe all Abelian groups G, up to isomorphism, such that: 2
(a) G = 16 (b) G = 16 and G has no elements of order 16. (c) G = 16 and G has no elements of order 8. (d) G = 16 and all elements of G have order 2. (e) G = 360 (f) G = 12 but G does not have elements of order 12. (g) G = 18 but G is not cyclic. Solution: G = 18 = 2 3 2 So G is isomorphic to a product of an Abelian group G (2) of order 2 and an Abelian group G (3) of order 3 2. The only possibility for G (2) is: Z 2. The only possibilities for G (3) are: Z 9 and Z 3 Z 3. So G is isomorphic to one of: Z 2 Z 9 or Z 2 Z 3 Z 3. External product of cyclic groups of relatively prime orders is isomorphic to a cyclic group with order equal to the product of orders (done in class). So Z 2 Z 9 = Z18 which is cyclic. Therefore there is only one Abelian non-cyclic group of order 18 (up to isomorphism!): Z 2 Z 3 Z 3. 5. Let G be a group acting on a set X. Solutions: Main formulas and properties to use: X = disjoint O x O x O x divides G O x = 1 if and only if x is a fixed point of the action of G on X. (a) If G = 11 and X = 12 prove that the action has at least one fixed point. Proof: Since G = 11, the orbits can have 1 or 11 elements. Since X = 12 we have 12 = X = disjoint O x O x. So, we have either 12 = 11+1 or 12 = 1+1+1+1+1+1+1+1+1+1+1+1. These are the only possibilities and in both cases there is at least one x X such that the orbit O x = 1 has only one element, hence G fixes that point, i.e. x is a fixed point of the action. (b) If G = 11 and X = 10 prove that G(x) = x for all x X. Proof: Since G = 11, the orbits can have 1 or 11 elements. Since X = 10 we have 10 = X = disjoint O x O x. So, we have 10 = 1+1+1+1+1+1+1+1+1+1. So for each x X the orbit O x = 1 has only one element, hence G fixes that point, i.e. gx = x for all g G, or G(x) = x. Hence for each x X we have G(x) = x. 3
(c) If G = 12 and X = 11 prove that the action has at least 3 orbits. (d) If G = 16 and X = 155 prove that the action has at least one fixed point. (e) If G = p 2 where p is a prime and X = q, q a prime q > p then G has at least one fixed point. 6. Let G X ϕ X be an action of group G on the set X = G given by conjugation, i.e. ϕ(g, x) = gxg 1. Suppose x is a fixed point of this action. Prove that x Z(G), i.e. x is in the center of G. Proof: Let x is a fixed point of the action of the group G on the set X = G, given by ϕ(g, x) = gxg 1. If x is fixed point then ϕ(g, x) = gxg 1 = x for all g G. So gxg 1 = x for all g G. Multiply on the right by g and get: So gx = xg for all g G. Therefore x Z(G), the center of G. (Definition Z(G) = {x G gx = xg for all g G}.) 7. Let G X ϕ X be an action of group G on the set X = {subgroups of G} given by conjugation, i.e. ϕ(g, H) = ghg 1. Suppose K is a fixed point of this action. Prove that K is a normal subgroup of G. 8. Let f : G G be a group homomorphism. Prove: If Ker(f) = {e} then f is a one-to-one map. Proof: Assume Ker(f) = {e}. WTS f is one-to-one. Suppose f(x 1 ) = f(x 2 ). WTS x 1 = x 2. Use two facts about homomorphisms, which we did in class: f(ab) = f(a)f(b) this is just definition f(a 1 ) = (f(a)) 1 mentioned many times in class (for homomorphisms) f(x 1 x 1 2 ) = f(x 1 ) f(x 1 2 ) by the first property = f(x 1 ) f(x 1 2 ) = f(x 1 ) (f(x 2 )) 1 by the second property = f(x 1 ) (f(x 1 )) 1 by assumption that f(x 1 ) = f(x 2 ) = e G since f(x 1 ) and (f(x 1 )) 1 are inverses of each other. Therefore f(x 1 x 1 2 ) = e G Therefore x 1 x 1 2 Ker(f) = {e}. So x 1 x 1 2 = e. Now multiply the equation by x 2 on the right. Therefore (x 1 x 1 2 ) x 2 = e x 2. Apply associative law, inverse and identity property. So x 1 = x 2. Therefore f is one-to-one. 9. Let f : G G be a group homomorphism. Prove: Ker(f) is a normal subgroup of G. 10. If G = 18 and a G, what are all possible orders of a? Answer: 1,2,3,6,9,18. 4
11. If G = 17 and a G, what are all possible orders of a? Answer: 1,17. 12. If G = 23 and a G, a e, what are all possible orders of a? Answer: 23. Reamark - Syllow theory will be on the next quiz. 5