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Unformatted text preview: THE UNIVERSITY OF SYDNEY Math2968 Algebra (Advanced) Semester 2 Tutorial Solutions Week 11 2008 1. (for general discussion) Let G be a group and N G . (a) Verify that if G is cyclic then G/N is cyclic. (b) Verify that if G is abelian then G/N is abelian. (c) Verify that G/N = G when G is the group of complex numbers of modulus 1 under multiplication and N = { 1 } . (d) Find an example where G/N is not isomorphic to any subgroup of G . Solution (a) If G is cyclic then G = ( x ) for some x G , so that G = { x i  i Z } , giving G/N = { Nx i  i Z } = { ( Nx ) i  i Z } = ( Nx ) , which verifies that G/N is cyclic. (b) If G is abelian and x,y G then ( Nx )( Ny ) = Nxy = Nyx = ( Ny )( Nx ) , which verifies that G/N is abelian. (c) Let : G G where x = x 2 for all x G . Then is easily seen to be a homomorphism with image G and kernel N . By the Fundamental Homomorphism Theorem, G/N = G/ ker = im G = G . (d) Let G = Q 8 and N = { 1 } . Then g 2 N for each g G , so that G/N = C 2 C 2 . However, G has exactly one element of order 2 whilst C 2 C 2 has 3 elements of order 2. It is impossible therefore to find a subgroup of G isomorphic to C 2 C 2 . 2. Let G be an abelian group and put T = { x G  x has finite order } . (a) Verify that T G , called the torsion subgroup of G . (b) Verify that G/T has no nontrivial elements of finite order. (c) Calculate T when G is the group of nonzero reals under multiplication and show that G is the internal direct product of T with H where H is the group of positive reals under multiplication. Deduce that G = ( R , +) Z 2 . Solution (a) Certainly 1 T . Let x,y T so that  x  = n and  y  = m for some n,m < . Then ( xy 1 ) mn = ( x n ) m ( y m ) n = 1 , so that xy 1 has order mn < , giving...
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