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Please Solve This Question 1. If f : G - G' is a homomorphism then show that f (G) = (f (a) | a E
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1. If f : G - G' is a homomorphism then show that f (G) = (f (a) | a E G} is a subgroup of G'. We also write f(G) = Imf (Image f). Show further that if H is normal in G then f (H) is normal in f (G), i.e., homomorphic image of a normal subgroup is normal. 2. For a fixed element a in a group G, define fa : G - G, s.t., fa(x) = al xa, x E G Show that fa is an isomorphism. 3. Let f, g be homomorphisms from G - G'. Show that H = {x E G |f(x) = g(x); is a subgroup of G. 4. Let G be a finite abelian group. Suppose o(G) and n are co-prime. Show that ( : G -+ G, s.t., p(x) = x" is an isomorphism (in other words, any g e G can be expressed as g = x" where x E G). 5. Show that the relation of isomorphism in groups is an equivalence relation. 6. Prove that the group G = {1, -1} under multiplication is isomorphic to G' = {0, 1} under addition modulo 2. 7. Show that 2Z = 3Z by considering the mapping 2x - 3x. Generalise. 8. Let G be the group of real numbers under addition. Show that 0 : G - G, s.t., O(x) = [x] is not a homomorphism, where [x] is the greatest integer not greater than X. 9. Show that f : C - C, s.t. f(z) = z is an automorphism where C = complex numbers. 10. Let G be the group of 2 x 2 matrices over reals of the type [ a ] s.t. ad - bc #0, under matrix multiplication and G' be the group of non zero real numbers under multiplication. Show that the map b - ad - bc is an onto homomorphism. 11. Show that homomorphic image of (a) an abelian group is abelian. (b) a cyclic group is cyclic. (c) a finite group is finite. 12. Show that converse does not hold in all the cases of the previous problem. (See example 9, page 149). 13. Let R be the set of real numbers. For a, be R, (a # 0) define Tat : R -> R, s.t., Tab (x) = ax + b. Let G be set of all such maps and let N = {], E G}. Show that G is a group and N is a normal subgroup of G and that ~ G is isomorphic to the group of non zero real numbers under multiplicationStep by Step Solution
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