Assignment 8: Selected Solutions


 Marlene Joseph
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1 Section 4.1 Assignment 8: Selected Solutions 1. and 2. Express each permutation as a product of disjoint cycles, and identify their parity. (1) (1,9,2,3)(1,9,6,5)(1,4,8,7)=(1,4,8,7,2,3)(5,9,6), odd; (2) (1,2,9)(3,4)(5,6,7,8,9)(4,9)=(1,2,9,3,4,5,6,7,8), even; (3) (1,4,8,7)(1,9,6,5)(1,5,3,2,9)=(1,4,8,7)(2,6,5,3), even; (4) (1,4,2,3,5)(1,3,4,5)=(1,5,4)(2,3), odd; (5) (1,3,5,4,2)(1,4,3,5)=(1,2)(3,4,5), odd; (6) (1,9,2,4)(1,7,6,5,9)(1,2,3,8)=(1,4)(2,3,8,7,6,5), even; (7) (2,3,7)(1,2)(3,5,7,6,4)(1,4)=(1,7,6,4,3,5,2), even; (8) (4,9,6,7,8)(2,6,4)(1,8,7)(3,5)=(1,4,2,7)(3,5)(6,9), odd. 14. List the elements of A 3 in cycle notation. A 3 = (1, 2, 3) = {(1), (123), (321)} S Let G be a group. For each element a G, let k a : G G be the map k a (g) = ga 1. (1) Prove that each k a is a permutation of G; (2) Prove that K = {k a a G} is a group under composition of maps; (3) Define ϕ : G K by ϕ(a) = k a. Determine if ϕ is an isomorphism. (1) Observe that k a is surjective, since given g G, k a (ga) = (ga)a 1 = g. Also, k a is injective since, if ga 1 = k a (g) = k a (h) = ha 1, then g = (ga 1 )a = (ha 1 )a = h. (2) We show that K Sym(G) is a subgroup (I am using the notation Sym(G) for the group of permutations of G). Indeed, k 1 : G G is the identity map, so K contains the identity. Next, one calculates k 1 a = k a 1: k a k a 1(g) = k a (ga) = (ga)a 1 = g and k a 1k a (g) = k a 1(ga 1 ) = (ga 1 )a = g. Finally, we check that k a k b = k ab : k a k b (g) = k a (gb 1 ) = gb 1 a 1 = g(ab) 1 = k ab (g). (3) First note that ϕ is obviously bijective. The computation in the previous line shows that ϕ(ab) = k ab = k a k b = ϕ(a)ϕ(b) 1
2 2 so ϕ is an isomorphism Let G be the group of the rigid motions of the tetrahetron: Determine G Actually, we can show that G = A 4. Indeed, G is generated by the 3 cycles (123), (124), (134), (234). Therefore, we just need to show that A n, n 3 is generated by 3cycles. To this end we prove Claim: Any even permutation can be written as the product of 3cycles. Indeed, any even permutation can be written as product of an even number of transpositions, so we just need to show that any pair of transpositions can be written as a product of 3cycles. To this end, we consider the two possible cases: (1) Product of disjoint transpositions: (a, b)(c, d) = (a, b)(b, c)(b, c)(c, d) = (a, b, c)(b, c, d); (2) Product of nondisjoint transpositions: (a, b)(b, c) = (a, b, c). This proves the claim. Therefore, G = A 4, so G = A 4, since S 4 / A 4 = [S 4 : A 4 ] = 2, it follows that A 4 = 4!/2 = Let H be the subgroup (1, 2) S 3. (1) Find the distinct left cosets of H in S 3. (2) Find the distinct right cosets of H in S 3. (1) The left cosets are: (123)H = {(123), (13)}. H = {1, (12)} (23)H = {(23), (321)}
3 3 (2) The right cosets are: H = {1, (12)} H(23) = {(23), (123)} H(321) = {(321), (13)}. 7. Let H be a subgroup of the group G. Prove that if two right cosets Ha and Hb are not disjoint, then Ha = Hb. Assume x Ha Hb. Then x = h 1 a = h 2 b for some h 1, h 2 H. Therefore a = h 1 1 x = h 1 1 h 2 b, showing that a Hb and therefore Ha Hb. By a symmetric argument H b H a, so equality holds. 22. Let G be a group of order pq where p and q are primes. Show that any proper nontrivial subgroup of G is cyclic. Let H G be a proper nontrivial subgroup. Then, by Lagrange s theorem, H is either p or q. In particular, H has prime order, so H is cyclic Show that every subgroup of an abelian group is normal. Let G be an abelian group and H a subgroup of G. Then, for h H and g G, ghg 1 = gg 1 h = 1h = h H. Hence, H is normal. 15. If {H λ } λ L is a collection of normal subgroups, for some index set L, then λ L H λ is a normal subgroup. Let h H λ and g G. Then, h H λ for all λ. Since each H λ is normal, ghg 1 H λ for each λ. Thus, ghg 1 H λ as required. 21. Prove that if H and K are normal subgroups of a group G such that H K = {1}, then hk = kh for all h H and k K. Note that hk = kh if, and only if hkh 1 k 1 = 1. But, (since K is normal) (hkh 1 k 1 K, and (since H is normal) h(kh 1 k 1 ) H. Therefore, hkh 1 k 1 H K = {1} proving the result. 32. Let H be a subgroup of G of index 2. (1) Prove that H is normal in G. (2) Prove that g 2 H for all g G. (1) To prove that H is normal, we first observe that G/H = {H, xh} where x is any element of G that is NOT in H. Now, assume g G and h H. If
4 4 g H, then of course ghg 1 H as H is a subgroup. We may therefore consider the case where g / H. To show that ghg 1 H, we show that ghg 1 H = H. Indeed, suppose not. Then ghg 1 H = gh (as g / H is one such x as above). But this means that g 1 (ghg 1 ) = hg 1 = h H. It now follows that g = (h ) 1 h H, a contradiction. We must therefore have ghg 1 H. (2) Since H is normal, G/H is a group of order 2. This means that for every g G, g 2 H = (gh) 2 = H. But this simply says that g 2 H Let H be a normal subgroup of a finite group G. If the order of the quotient group G/H is m, prove that g m H for every g G. This is a generalization of the last problem. Indeed, as G/H = m, gh divides m for every g G. This just means that g m H = (gh) m = H, so g m H. 22. Let H be a normal subgroup of the group G. Prove that G/H is abelian if, and only if a 1 b 1 ab H for every a, b G. The group G/H is abelian if, and only if, abh = ahbh = bhah = bah. This is equivalent to saying ab H ba, or a 1 b 1 ab = (ba) 1 ab H. 32. Let Inn(G) be the group of inner automorphisms of G. It consists of the automorphisms t a : G G (a G) defined by t a (g) = aga 1. Prove that Inn(G) is a normal subgroup of Aut(G). Let a G and β Aut(G). We show that βt a β 1 = t β(a). Indeed, suppose that g G. Then (βt a β 1 )(g) = (βt a )(β 1 (g)) = β(aβ 1 (g)a 1 ) = β(a)β(β 1 (g))β(a 1 ) = β(a)gβ(a) 1 = t β(a) (g). 34. If H and K are normal subgroups of a group G such that G = HK and H K = {1}, then G is said to be the internal direct product of H and K, written G = H K. If G = H K, prove that ϕ : H G/K defined by ϕ(h) = hk is an isomorphism.
5 First, note that ϕ(h 1 h 2 ) = h 1 h 2 K = h 1 Kh 2 K = ϕ(h 1 )ϕ(h 2 ), so ϕ is a homomorphism. Note that if h ker ϕ, then ϕ(h) = hk = K, so h H K = {1} showing that ϕ is injective. To see that ϕ is surjective, note that any g G can be written as g = hk for some h H and k K. Therefore, gk = hkk = hk = ϕ(h). This proves that ϕ is an isomorphism. 5
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