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 〈*g*^{2}〉 would be a normal subgroup different from *G* and {1}; hence *G* is finite. Let *p* be a prime number which divides |*G*|. Then 〈*g ^{p}*〉 is a normal subgroup of

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* ∩ *A*_{n} = {*g* ∈ *G*: *g* is even}, ...

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