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S10MTH 3175 Group Theory (Prof.Todorov) Quiz 5 (Practice-Some Solutions) Name: 1. Let Sn be the group of permutations on n elements f1; 2; 3; : : : ; ng. Let An be the subgroup of even permutations. Prove that An is normal subgroup of Sn. 2. Let S15 be the group of permutations on 15 elements f1; 2; 3;:::; 15g. Prove that S15 has elements of order 56 but does not have any elements of orders 49 or 50. Answer: Use the following facts: (a) Order of a k-cycle is k, i.e. jαj = k for α a k-cycle. (b) jαβj = lcm(jαj; jβj) if α and β are disjoint permutations. • S15 has elements of order 56: Let σ = αβ, where α is an 8-cycle and β is a 7-cycle. Then jσj = jαβj = lcm(jαj; jβj) = lcm(8; 7) = 56. • S15 does not have any elements of orders 49: In order for lcm(m; n) = 49, either m = 49 or n = 49. Therefore we would need a cycle of order 49, hence need a 49-cycle in S15, which is impossible. • S15 does not have any elements of orders 50: In order for lcm(m; n) = 50, either m or n 2 must be a multiple of 5 = 25. Therefore we would need a 25-cycle or 50-cycle in S15, which is impossible. 3. Prove that a factor group of a cyclic group is cyclic. Answer: Recall: A group G is cyclic if it can be generated by one element, i.e. if there exists an element a 2 G such that G =< a > (this means that all elements of G are of the form ai for some integer i.) Recall: Elements of a factor group G=H are left cosets fgH j g 2 G. Proof: Suppose G =< a >. Let G=H be any factor group of G. WTS: G=H is cyclic. Any element of G=H is of the form gH for some g 2 G. Since G is cyclic, there is an integer i such that g = ai. So gH = aiH. aiH = (aH)i by definition of multiplication in factor groups. Therefore gH = (aH)i for any coset gH. So G=H is cyclic, by definition of cyclic groups. 4. Prove that a factor group of an Abelian group is Abelian. Answer: Recall: A group G is Abelian if ab = ba for all a; b 2 G. Proof: Suppose G is Abelian. Let G=H be any factor group of G. WTS: G=H is Abelian. Let aH; bH be any elements of G=H. Then (aH)(bH) = (ab)H by definition of multiplication in factor groups. Then (ab)H = (ba)H since G is abelian. (ba)H = (bH)(aH) again definition of multiplication in factor group. Therefore (aH)(bH) = (bH)(aH). Therefore G=H is Abelian. 1 S10MTH 3175 Group Theory (Prof.Todorov) Quiz 5 (Practice-Some Solutions) Name: 5. Prove that any subgroup of an Abelian group is normal subgroup. Answer: Recall: A subgroup H of a group G is called normal if gH = Hg for every g 2 G. Proof: Suppose G is Abelian. Let H be a subgroup of G. WTS: gH = Hg for all g 2 G. Let g 2 G. Then gH = fgh j h 2 Hg by definition of left coset. gh = hg for all h since G is Abelian. Therefore fgh j h 2 Hg = fhg j h 2 Hg = Hg by definition of right coset Hg. Therefore gH = Hg for all g 2 G. Therefore H is normal subgroup of G. 6. Let H =< 5 > be the subgroup of G = Z generated by 5. (a) List the elements of H = h5i. Answer: Notice: Operation in G = Z is addition. H = h5i = f0; ±5; ±10;::: g = f5k j k 2 Zg (b) Prove that H is normal subgroup of G. Answer: Addition in Z is commutative. So G = (Z; +) is Abelian group and by previous problem every subgroup of an Abelian group is normal. (c) List the elements of the factor group G=H = Z= < 5 >. Answer: Elements of G=H are left cosets of H in G: 0 + H = 0 + h5i = f0 + 0; 0 ± 5; 0 ± 10;::: g = f0; ±5; ±10;::: g = H 1 + H = 1 + h5i = f1 + 0; 1 ± 5; 1 ± 10;::: g = f:::; −9; −4; 1; 6; 11;::: g 2 + H = 2 + h5i = f2 + 0; 2 ± 5; 2 ± 10;::: g = f:::; −8; −3; 2; 7; 12;::: g 3 + H = 3 + h5i = f3 + 0; 3 ± 5; 3 ± 10;::: g = f:::; −7; −2; 3; 8; 13;::: g 4 + H = 4 + h5i = f4 + 0; 4 ± 5; 4 ± 10;::: g = f:::; −6; −1; 4; 9; 14;::: g (d) Write the Cayley table of G=H = Z= h5i. G=H = Z= h5i 0 + h5i 1 + h5i 2 + h5i 3 + h5i 4 + h5i 0 + h5i 0 + h5i 1 + h5i 2 + h5i 3 + h5i 4 + h5i 1 + h5i 1 + h5i 2 + h5i 3 + h5i 4 + h5i 0 + h5i 2 + h5i 2 + h5i 3 + h5i 4 + h5i 0 + h5i 1 + h5i 3 + h5i 3 + h5i 4 + h5i 0 + h5i 1 + h5i 2 + h5i 4 + h5i 4 + h5i 0 + h5i 1 + h5i 2 + h5i 3 + h5i (e) What is the order of 2 + h5i in Z= h5i? Answer: j2 + h5i j = 5 as 2 + h5i 2 Z= h5i. 2 S10MTH 3175 Group Theory (Prof.Todorov) Quiz 5 (Practice-Some Solutions) Name: 7. Let K = h15i be the subgroup of G = Z generated by 15. (a) List the elements of K = h15i. Answer: K = h15i = f15k j k 2 Zg (b) Prove that K is normal subgroup of G. Proof: (Z +) is Abelian group and any subgroup of an Abelian group is normal (from 5). (c) List the elements of the factor group G=K = Z= h15i. Answer: G=K = Z= h15i = fi + h15i j 0 ≤ i ≤ 14g. (There are 15 elements.) !!! This is just one way of expressing these cosets. Notice that there are many ways of expressing the same coset, eg. (2+h15i) = (17+h15i) = (32+h15i) = (−13+h15i) = ::: . (d) Write the Cayley table of G=K = Z= h15i. Answer: Don't have to actually write it, but make sure that you know what it would look like. (e) What is the order of 3 + K = 3 + h15i in Z= h15i? Answer: j3 + h15i j = 5 in Z= h15i, since (3 + h15i) + (3 + h15i) + (3 + h15i) + (3 + h15i) + (3 + h15i) = (15 + h15i) = (0 + h15i). (f) What is the order of 4 + K = 4 + h15i in Z= h15i? Answer: j4 + h15i j = 15 in Z= h15i. (g) What is the order of 5 + K = 5 + h15i in Z= h15i? Answer: j5 + h15i j = 3 in Z= h15i. (h) What is the order of 6 + K = 6 + h15i in Z= h15i? Answer: j6 + h15i j = 5 in Z= h15i, (i) Prove that G=K is cyclic. Answer: Z= h15i is generated by 1 + h15i, hence it is cyclic. underlineAnswer 2: Z= h15i is cyclic since it is factor of the cyclic group (Z; +) (this group is generated by 1). 3 S10MTH 3175 Group Theory (Prof.Todorov) Quiz 5 (Practice-Some Solutions) Name: (j) Prove that G=K = Z= h15i is isomorphic to Z15. Answer: • Elements of Z= h15i are left cosets, and operation is addition of cosets (as defined for factor groups!) • Elements of Z15: f0; 1; 2; 3; 4; 5; 6; 7; 8; 9; 10; 11; 12; 13; 14g and addition is modulo 15. • Define the map: f : Z= h15i ! Z15 by f(i + h15i) = ri, where ri is the remainder after dividing i by 15. { Notice that f, as defined, is a mapping from Z= h15i to Z15 since the values of f are the remainders after dividing by 15, hence they are non-negative integers between 0 and 14. { Prove that f is well-defined: Suppose the same coset (i + h15i) is given by another integer j, i.e. (i + h15i) = (j + h15i). Then f(i + h15i) = ri 2 Z15 and f(j + h15i) = rj 2 Z15. To show that f is well defined we must show that f(i + h15i) = f(j + h15i), i.e. WTS ri = rj 2 Z15. From (i+h15i) = (j +h15i) it follows that j 2 (i+h15i) and therefore j = i+k15 for some integer k 2 Z. By definition of remainders: i = n·15+ri and j = m·15+rj, with 0 ≤ ri; rj < 15. Therefore from j = i + k15, we have m · 15 + rj = n · 15 + ri + k15. Hence rj − ri = (k + n − m) ·15 which is an integer multiple of 15. Since the remainders are 0 ≤ ri; rj < 15, it follows that rj = ri. Therefore f is a well defined function. { Let f : Z= h15i ! Z15 defined by f(i+h15i) = ri, where ri is the remainder after dividing i by 15. Then f is "one-to-one" function. Proof: Suppose there are a; b 2 Z= h15i such that f(a) = f(b). WTS a = b. From a; b 2 Z= h15i, it follows that a = i + h15i and b = j + h15i. From f(a) = f(b) it follows f(i + h15i) = f(j + h15i), and by definition of f it follows that ri = rj. So i = n · 15 + ri and j = m · 15 + rj. Therefore i − j is a multiple of 15, hence and element of the subgroup h15i. Therefore a = i + h15i = j + h15i = b. Therefore f is a one-to-one function. { f : Z= h15i ! Z15 is onto. Proof: Let y 2 Z15. WTS: there is an x 2 Z= h15i such that f(x) = y.
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