MATH 3005 ABSTRACT ALGEBRA I FINAL SOLUTION


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1 MATH 3005 ABSTRACT ALGEBRA I FINAL SOLUTION SPRING MOON Write your answer neatly and show steps. Any electronic devices including calculators, cell phones are not allowed. (1) Write the definition. (a) (3 pts) Cyclic group. A group G is cyclic if it is generated by an element a G, i.e., G = a. Equivalently, G is cyclic if there is a G such that G = {a n n Z}. (b) (3 pts) The external direct product of two groups. The external direct product G H of two groups G and H is a group defined as G H = {(a, b) a G, b H}. The binary operation is defined by (a 1, b 1 ) (a 2, b 2 ) = (a 1 a 2, b 1 b 2 ). (c) (3 pts) For a group G and a normal subgroup N, the quotient group G/N. (Describe it as a set and provide the definition of its binary operation.) As a set, the quotient group is G/N = {an a G}, the set of all distinct cosets. The binary operation is defined by an bn = (ab)n. (d) (3 pts) The kernel of a group homomorphism. For a group homomorphism φ : G G, ker φ = {g G φ(g) = ē} where ē is the identity of G. (e) (3 pts) For a permutation group G on a set S and a S, the stabilizer of a. The stabilizer is stab(a) = {g G g(a) = a}. Date: May 7,
2 (2) (6 pts) For a group G and its subgroup H, the normalizer of H is defined as N(H) = {g G ghg 1 H for all h H}. Show that N(H) is a subgroup of G. For the identity e G, ehe 1 = h H for all h H. So e N(H) and N(H). If g 1, g 2 N(H), then for any h H, g 1 hg1 1 = h H. Then g 1 g 2 h(g 1 g 2 ) 1 = g 1 g 2 hg2 1 g1 1 = g 1 h g1 1. Because h H and g 1 N(H), g 1 h g1 1 H. Therefore g 1 g 2 N(H). Finally, if g N(H) and h H, then ghg 1 = h for some h H, so gh = h g. Then g 1 h(g 1 ) 1 = g 1 hg = g 1 gh = h H. Therefore g 1 N(H) and N(H) G. Showing the nonemptyness: 1 pt. Proving the closedness: 3 pts. Showing the existence of inverse: 2 pts. 2
3 (3) (a) (5 pts) In S 7, what is the maximum order of an element? If σ = σ 1 σ 2 σ k S n is a product of disjoint cycles σ 1, σ 2,, σ k, then σ = lcm( σ 1, σ 2,, σ k ). For σ S 7, if we describe σ as a product of disjoint cycles, then the length distributions of them are one of 7, 6 + 1, 5 + 2, , 4 + 3, , , , , , , , , , In each case the least common multiple is 7, 6, 10, 5, 12, 4, 4, 3, 6, 6, 3, 2, 2, 2, 1. Therefore the maximum order is 12. Indicating the fact that the order of σ is the least common multiple of the length of disjoint cycles: 2 pts. Getting the answer 12: +3 pts. If there is no explanation of the answer, 2 pts. (b) (4 pts) Find an element σ S 7 with the maximum order. Write σ as a product of disjoint cycles and a product of 2cycles. By (a), when σ S 7 is a product of two disjoint cycles σ 1 and σ 2 of length 4, 3 respectively, the length is 12. One example is σ = (1234)(567). If we write it as a product of 2cycles, then σ = (14)(13)(12)(57)(56). Finding an element of order 12 and writing it as a product of disjoint cycles: 2 pts. Describing it as a product of 2cycles: 2 pts. (c) (3 pts) Write σ 1 as a product of disjoint cycles. σ 1 = (1432)(576) (d) (4 pts) In A 7, what is the maximum order of an element? A cycle of odd (resp. even) length is even (resp. odd). So a permutation of type 4 + 3, is odd, but length 7 cycle is even. It has the maximum order 7. 3
4 (4) (6 pts) Let G be a cyclic group of order 36. Draw the subgroup lattice of G. Let G = a. Then any subgroup of G is cyclic and generated by a k for some positive divisor k of 36. Therefore the subgroups of G are G = a, a 2, a 3, a 4, a 6, a 9, a 12, a 18, {e} = a 36. Note that a m a n if and only if n m. Therefore the subgroup lattice of G is G a 2 a 3 a 4 a 6 a 9 a 12 a 18 {e}. Listing all subgroups: 2 pts. Drawing incorrect paths or missing paths/nodes: 1 pt per 2 paths. Using incorrect names for nodes: 2 pts. Flipping the diagram vertically: 1 pt. 4
5 (5) (7 pts) Prove that S 4 S 3 Z 4. In S 4, if one write σ S 4 as a product of disjoint cycles, then its length can be decomposed 4, 3 + 1, 2 + 2, , So the maximum order of an element is 4. On the other hand, in S 3 there is an element τ of order 3. In Z 4, 1 = 4. Then (τ, 1) = lcm( τ, 1 ) = 12. Therefore S 2 S 3 Z 4. Explaining an idea to proof: 1 pt. Finding the maximum order of one group: 3 pts each. Alternatively, counting the number of elements of fixed order in a group: 3 pts each. (6) (7 pts) Show that a group of prime order is cyclic. Let G be a group of prime order p. If a G is nonidentity element, a G = p. Because a e, a 1 and a = a = p. Therefore G = a and G is cyclic. Stating the definition of cyclic group: 1 pt. Stating Lagrange s theorem correctly: 2 pts. Proving the result correctly: 7 pts. (7) (9 pts) Let G be a group of order 55. Show that G has a normal subgroup of order 11. Because 11 is a prime divisor of G = 55, there is an element a G with a = 11. Then a is a subgroup of order 11. Let H and K be two subgroups of order 11. Then H K is a subgroup of H, so H K H = 11. Therefore H K is 1 or 11. But 55 = G HK = H K H K = 121 H K. Thus H K = 1 is impossible. So H K = 11 and H = K. In other words, there is a unique subgroup of order 11 in G. Let H be the subgroup. For any element g G, ghg 1 is a subgroup of G and ghg 1 = H = 11. From above observation, ghg 1 = H. This implies that H G. Showing the existence of order 11 subgroup: 3 pts. Proving that an order 11 subgroup is unique: 3 pts. Showing that the unique order 11 subgroup is normal: 3 pts. 5
6 (8) Let p be a prime and let G = Z p 2 Z p 2. (a) (6 pts) Determine the number of elements of order p in G. Note that a Z p 2 has order 1, p, or p 2. a = p if and only if gcd(a, p 2 ) = p 2 /p = p. Therefore there are p 1 such elements, p, 2p, 3p,, (p 1)p. Let H be the set of such elements. For any (a, b) Z p 2 Z p 2, (a, b) = lcm( a, b ). Therefore there are three possibilities of (a, b) = p. Case 1. a = p, b = 1. Then b = 0. So there are p 1 such elements. {(a, 0) a H} is the set of such elements. Case 2. a = 1, b = p. Similarly, there are p 1 such elements. {(0, b) b H} is the set of such elements. Case 3. a = b = p. In this case, {(a, b) a, b H} is the set of such elements. So the number of elements is (p 1) 2. In summary, there are (p 1) 2 + 2(p 1) = p 2 1 elements of order p in G. Listing all possibilities to get an order p elements: 3 pts. Counting the number of elements of oder p in G: 3 pts. (b) (6 pts) Find the number of subgroups of order p in G. Note that for any order p element a G, a is a subgroup of order p. Therefore every order p element is in an order p subgroup. Conversely, any nonidentity element in an order p subgroup has order p. So an order p subgroup has exactly p 1 order p elements. Moreover, if H and K are two distinct order p subgroups, then H K H = p and H K p. Therefore H K = 1 and H K = {e}. Therefore for each order p element, there is a unique subgroup of order p containing it. In summary, the number of subgroups of order p is (p 2 1)/(p 1) = p
7 (9) (a) (6 pts) Show that Z 5 Z 5 / (1, 1) Z 5. Because (1, 1) = lcm( 1, 1 ) = 5, (1, 1). Therefore Z 5 Z 5 / (1, 1) = Z 5 Z 5 (1, 1) = 25 5 = 5. A group of prime order is cyclic. Therefore it is a cyclic group of order 5, which is isomorphic to Z 5. Computing the order of Z 5 Z 5 / (1, 1) : 3 pts. Concluding that it is isomorphic to Z 5 : 3 pts. (b) (6 pts) By using the first isomorphism theorem, show that GL(2, R)/SL(2, R) R. Define a map φ : GL(2, R) R as φ(a) = det A. Then φ(ab) = det(ab) = det A det B = φ(a)φ(b), so it is a homomorphism. For any c R, [ c 0 φ( 0 1 ] ) = c. Therefore φ is onto and φ(gl(2, R)) = R. Furthermore, A ker φ det A = 1 A SL(2, R). So ker φ = SL(2, R). By the first fundamental theorem, GL(2, R)/SL(2, R) = GL(2, R)/ ker φ φ(gl(2, R)) = R. Stating the first isomorphism theorem: 1 pt. Giving the homomorphism φ: 2 pts. Showing that φ is a homomorphism: 1 pt. Proving that φ is onto: 1 pt. Indicating that ker φ = SL(2, R): 1 pt. 7
8 (10) (10 pts) Determine the number of ways in which the vertices of a square can be colored with four colors. Two colorings are equivalent if one is obtained from another by applying a permutation induced by the symmetry group D 4 of the square. Let D 4 = {e, R 90, R 180, R 270, H, V, D, D } where R θ is the rotation by θ and H, V are horizontal/vertical flips, D, D are two diagonal flips. Let S be the set of all colorings. Then S = 4 4 = 256. For e, fix(e) = S = 256. For R 90, if a coloring is in fix(r 90 ), then all vertices must be the same color. So fix(r 90 ) = 4. By the same idea, fix(r 270 ) = 4, too. For R 180, a coloring is in fix(r 180 ) exactly when the opposite vertices have the same color. Therefore fix(r 180 ) = 4 2 = 16. A coloring is fixed by H when two adjacent vertices connected by a vertical edge have the same color. So fix(h) = 4 2 = 16. By the same token, we have fix(v ) = 16. Finally, for two diagonal flips D and D, there are two fixed vertices on the flipping axis and a pair of vertices which share the same color. Therefore fix(d) = fix(d ) = 4 3 = 64. By Burnside s theorem, the number of orbits is 1 D 4 g D 4 fix(g) = 1 8 ( ) = = 55. Stating Burnside s theorem: 2 pts. Getting the number of all possible colorings before identification: 2 pts. Counting the number of fixed colorings for each element of D 4 : 1 pt each. 8
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