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Solutions to exercises

Chapter 1

1.   Note that all subgroups of G are normal, since G is abelian; and G ≠ {1} since G is simple. Let g be a non-identity element of G. Then 〈g〉 is a normal subgroup of G, so 〈g〉 = G. If G were infinite, then 〈g2〉 would be a normal subgroup different from G and {1}; hence G is finite. Let p be a prime number which divides |G|. Then 〈gp〉 is a normal subgroup of G which is not equal to G. Therefore gp = 1, and so G is cyclic of prime order.

2.   Since G is simple and Ker ϑ G, either Ker ϑ = {1} or Ker ϑ = G. If Ker ϑ = {1} then ϑ is an isomorphism; and if Ker ϑ = G then H = {1}.

3.   First, GAn = {gG: g is even}, ...

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