S11MTH 3175 Group Theory (Prof.Todorov) Quiz 6 (PracticeSolutions) Name: 1. Let G and H be two groups and G H the external direct product of G and H.


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1 Some of the problems are very easy, some are harder. 1. 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. Proof: Elements of G H are all pairs {(g, h) g G, h H} operation is componentwise elements of H G are all pairs {(h, g) g G, h H} operation is componentwise f((g 1, h 1 ) + (g 2, h 2 )) = f(g 1 + g 2, h 1 + h 2 ) = (h 1 + h 2, g 1 + g 2 ) f(g 1, h 1 ) + f(g 2, h 2 ) = (h 1, g 1 ) + (h 2, g 2 ) = (h 1 + h 2, g 1 + g 2 ) Therefore f((g 1, h 1 ) + (g 2, h 2 )) = f(g 1, h 1 ) + f(g 2, h 2 ) hence f is a homomorphism. i. Find Ker(f). identity of G H is the pair (e G, e H )! identity of H G is the pair (e H, e G )! 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) by 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 )} 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(g 1 + g 2 ) = (g 1 + g 2, e H ) i(g 1 ) + i(g 2 ) = (g 1, e H ) + (g 2, e H ) = (g 1 + g 2, e H + e H ) = (g 1 + g 2, e H ) Therefore i(g 1 + g 2 ) = i(g 1 ) + i(g 2 ) i. Find Ker(i). Ker(i) = {e G }. You have to prove this! ii. Find Im(i). Im(i) = G {e H } = {(g, e H ) g G} < G H 1
2 (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). 2. Describe all Abelian groups G, up to isomorphism, such that: (a) G = 16 Z 16, Z 8 Z 2, Z 4 Z 4, Z 4 Z 2 Z 2, Z 2 Z 2 Z 2 Z 2 (b) G = 16 and G has no elements of order Z 16 has order 16. (a, b) = lcm( a, b ) for (a, b) A B a Z 8 = a = 1, 2, 4, 8 a Z 4 = a = 1, 2, 4. a Z 2 = a = 1, 2. lcm(8, 2) = 8 lcm(4, 2) = lcm(4, 4) = lcm(2, 4) = 4 lcm(2, 2) = lcm(2, 2, 2) = 2 lcm(k, 1) = k The only possible orders of the elements in the following groups are: 8,4,2,1. Z 8 Z 2, Z 4 Z 4, Z 4 Z 2 Z 2, Z 2 Z 2 Z 2 Z 2 (c) G = 16 and G has no elements of order 8. 2 Z 16 has order 8 (1, 0) Z 8 Z 2 has order 8. The only possible orders of the elements in the following groups are: 4,2,1. Z 4 Z 4, Z 4 Z 2 Z 2, Z 2 Z 2 Z 2 Z 2 (d) G = 16 and all nonidentity elements of G have order 2.. Z 2 Z 2 Z 2 Z 2 2
3 (e) G = = G (2) = Z 8 or Z 4 Z 2 or Z 2 Z 2 Z 2 G (3) = Z 9 or Z 3 Z 3 G (5) = Z 5 G is isomorphic to one of the following 6 groups: Z 8 Z 9 Z 5 Z 4 Z 2 Z 9 Z 5 Z 2 Z 2 Z 2 Z 9 Z 5 Z 8 Z 3 Z 3 Z 5 Z 4 Z 2 Z 3 Z 3 Z 5 Z 2 Z 2 Z 2 Z 3 Z 3 Z 5 (f) G = 12 but G does not have elements of order 12. If G = 12 = and G is Abelian, then G is isomorphic to one of the following: Z 4 Z 3, Z 2 Z 2 Z 3 1 Z 4 has order 4. 1 Z 3 has order 3. (1, 1) Z 4 Z 3 has order (1, 1) = lcm(4, 3) = 12 G = Z 2 Z 2 Z 3 (g) G = 18 but G is not cyclic. G = 18 = 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 noncyclic group of order 18 (up to isomorphism!): Z 2 Z 3 Z Let G be a group acting on a set X. 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. 3
4 (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 = 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 = 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. (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. 4. 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 be 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. (Def. Z(G) = {x G gx = xg for all g G}.) 5. 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. 6. Let f : G G be a group homomorphism. Prove: If Ker(f) = e then f is a onetoone map. Proof: Assume Ker(f) = {e}. WTS f is onetoone. Suppose f(x 1 ) = f(x 2 ). WTS x 1 = x 2. 4
5 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 onetoone. 7. Let f : G G be a group homomorphism. Prove: Ker(f) is a normal subgroup of G. 8. If G = 18 and a G, what are all possible orders of a? Answer: 1,2,3,6,9, If G = 17 and a G, what are all possible orders of a? Answer: 1, If G = 23 and a G, a e, what are all possible orders of a? Answer: Let G be a group of order G = 35. (a) How many 5Sylow subgroups does G have? r 5 = #{5Sylow subgroups} G = 35 = r 5 = 1, 5, 7, 35 r 5 = #{5Sylow subgroups} = 1(mod5) = r 5 = 1, 6, 11, 16, 21, 26, 31 r 5 = 1. Therefore there is exactly one 5Sylow subgroup. (b) Let P 5 be a 5Sylow subgroup of G. How many elements does P 5 have? P 5 = 5 (c) How many 7Sylow subgroups does G have? r 7 = #{7Sylow subgroups} G = 35 = r 7 = 1, 5, 7, 35 r 7 = #{7Sylow subgroups} = 1(mod7) = r 7 = 1, 8, 15, 22, 29 r 7 = 1. Therefore there is exactly one 7Sylow subgroup. (d) Let P 7 be a 7Sylow subgroup of G. Find P 7. P 7 = 7 5
6 12. Find ALL Sylow subgroups of Z 12. Prove that these are all. r 2 = #{2Sylow subgroups} G = 12 = r 2 = 1, 2, 3, 4, 6, 12 r 2 = #{2Sylow subgroups} = 1(mod2) = r 2 = 1, 3, 5, 7, 9, 11 r 2 = 1 or 3. Notice, since Z 12 is Abelian, all subgroups are normal and since all 2Sylow subgroups are conjugate to each other, it follows that gp 2 g 1 = P 2 for all g G. Hence there is only one 2Sylow subgroup P 2. P 2 = 4 and P 2 = {3, 6, 9, 0}. Similarly, there is exactly one 3Sylow subgroup P 3 = {4, 8, 0}. 13. Determine the Sylow subgroups of the alternating group A 4 (the even permutations of {1, 2, 3, 4}. A 4 = 12 = A 4 is not Abelian, so we really need to find numbers of psylow subgroups for p = 2 and p = 3 r 2 = #{2Sylow subgroups} G = 12 = r 2 = 1, 2, 3, 4, 6, 12 r 2 = #{2Sylow subgroups} = 1(mod 2) = r 2 = 1, 3, 5, 7, 9, 11 r 2 = 1 or 3. P 2 = 4 and P 2 = {(12)(34), (13)(24), (14)(23), (1)} r 3 = #{3Sylow subgroups} G = 12 = r 3 = 1, 2, 3, 4, 6, 12 r 3 = #{3Sylow subgroups} = 1(mod 3) = r 3 = 1, 4, 7, 10 r 2 = 1 or 4. {P (1) 3 = (123), P (2) 3 = (124), P (3) 3 = (134) P (4) 3 = (234) } 6
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