# Adv Alegbra HW Solutions 56 - ϕ G → X be a bijection...

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56 (iii) The inclusion Z R is a homomorphism of additive groups. Solution. True. (iv) The subgroup { 0 } of Z is isomorphic to the subgroup { ( 1 ) } of S 5 . Solution. True. (v) Any two f nite groups of the same order are isomorphic. Solution. False. (vi) If p is a prime, any two groups of order p are isomorphic. Solution. True. (vii) The subgroup h ( 12 ) i is a normal subgroup of S 3 . Solution. False. (viii) The subgroup h ( 123 ) i is a normal subgroup of S 3 . Solution. True. (ix) If G is a group, then Z ( G ) = G if and only if G is abelian. Solution. True. (x) The 3-cycles ( 765 ) and ( 52634 ) are conjugate in S 100 . Solution. True. 2.65 If there is a bijection f : X Y (that is, if X and Y have the same number of elements), prove that there is an isomorphism ϕ : S X S Y . Solution. De f ne ϕ : S X S Y by ϕ : α 7→ f α f 1 . It is routine to check that ϕ is a homomorphism; it is an isomorphism, for its inverse is the function β 7→ f 1 β f . 2.66 Let G be a group, let X be a set, and let
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Unformatted text preview: ϕ : G → X be a bijection. Prove that there is an operation on X which makes X into a group such that ϕ : G → X is an isomorphism. Solution. Absent. 2.67 (i) Prove that the composite of homomorphisms is itself a homomor-phism. Solution. If f : G → H and g : H → K are homomorphisms, then ( g ◦ f )( ab ) = g ( f ( ab )) = g ( f a f b ) = g ( f a ) g ( f b ) = ( g ◦ f )( a )( g ◦ f )( b ). (ii) Prove that the inverse of an isomorphism is an isomorphism. Solution. If f is an isomorphism, then f − 1 ( x ) = a if and only if x = f ( a ) . Hence, if f − 1 ( y ) = b , then f − 1 ( xy ) = ab (since f ( ab ) = f ( a ) f ( b ) = xy ), and so f − 1 ( xy ) = ab = f − 1 ( x ) f − 1 ( y ) ....
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