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, G ∩ An = {g ∈ G: g is even}, so G ∩ An ◁ G. Since G ∩ An ≠ G, we may choose h ∈ G with h ∉ An. For all odd g in G, we have g = (gh−1)h ∈ (G ∩ An)h. Therefore G ∩ An and (G ∩ An)h are the only right cosets of G ∩ An in G, and G/(G ∩ An) ≅ C2.

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