S11MTH 3175 Group Theory (Prof.Todorov) Quiz 5 (Practice) Name: Some problems are really easy, some are harder, some are repetitions. 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. Answer: • One way: ◦ jSnj = n! and jAnj = n!=2 ◦ # (left cosets of An in Sn)= [Sn : An] = jSnj=jAnj = n!=(n!=2) = 2 ◦ # (right cosets of An in Sn)=# (left cosets of An in Sn)=2 ◦ Want to show: aAn = Ana for all a 2 Sn - If a 2 An then aAn = An = Ana. - If a2 = An then aAn 6= An 6= Ana Sn = An [ aAn disjoint union of 2 distinct left cosets. Sn = An [ Ana disjoint union of 2 distinct right cosets. An [ aAn = An [ Ana disjoint unions implies aAn = Ana for all a2 = An ◦ An is normal subgroup in Sn since aAn = Ana for all a 2 Sn • Another way: ◦ An = feven permutations on n elementsg ◦ A subgroup H in a group G is normal subgroup, if and only if gHg−1 ⊂ H. Propo- sition proved in class. −1 ◦ WTS σAnσ ⊂ An Let α 2 An. Then α is an even permutation (by definition of An). −1 Then σασ is even permutation for all σ 2 Sn (by a problem from a previous quiz). −1 Therefore σασ 2 An −1 Therefore σAnσ ⊂ An Therefore An is normal subgroup in 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. 1 S11MTH 3175 Group Theory (Prof.Todorov) Quiz 5 (Practice) Name: • 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. ( Factor Groups - Quotient Groups - G=H ) 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 generated by aH, hence G=H is cyclic, by the 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. 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. 2 S11MTH 3175 Group Theory (Prof.Todorov) Quiz 5 (Practice) Name: 6. Let H =< 5 > be the subgroup of G = Z generated by 5. (a) List the elements of H =< 5 >. 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. 3 S11MTH 3175 Group Theory (Prof.Todorov) Quiz 5 (Practice) 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). (j) Prove that G=K = Z= h15i is isomorphic to Z15. Answer: • One way - using the First Isomorphism Theorem: { Define a group homomorphism: f : Z ! Z15: ∗ Since Z is cyclic it is enough to define homomorphism on a generator, and extend to all other elements. ∗ Define f(1) := 1(mod15) and f(n1) := n1 (mod15) (i.e. f(n) := n (mod15) ∗ Since j1j = 1 for 1 2 Z and jf(1)j = 15 for f(1) = 1 2 Z= h15i we have jf(1)j j j1j. { Claim 1: f is onto (this is quite clear, but here is a detailed proof). 4 S11MTH 3175 Group Theory (Prof.Todorov) Quiz 5 (Practice) Name: ∗ Elements of Z15 are integers f0; 1; 2;:::; 14g ∗ For each n 2 f0; 1; 2;:::; 14g = Z15 consider n 2 Z. ∗ Then f(n) = n 2 Z15. Hence f is onto. { Im(f) = Z15 (As stated in class this is equivalent to f being onto.) { Claim 2: Ker(f) = h15i < Z (this is quite clear, but here is a detailed proof). ∗ Ker(f) = fn 2 Z j f(n) = 0 2 Z15g ∗ n (mod15) = r 2 f0; 1;:::; 14g, where r is the remainder after dividing n by 15, i.e.