**1)(a) If G is an abelian group and then is a subgroup of G.
(b) If , then and
Proof-** a) Since we can write we have that is nonempty. Let which means we can write and for some . Then we have so that it is a subgroup by the subgroup test.

b) Assume that so that we can write for some g, To show the next part we define by . Then this function is onto and we have it is a homomorphism by property of addition of cosets. Thus f is an epimorphism. Now let this gives that which tells us that for each i. This means that for all i so that . So that . Then using the first isomorphism theorem we have

**4) A free abelian group is a free group iff it is cyclic.
Proof-** Let G be a free abelian group that is also a free group. We know that any free group on a set X with is nonabelian so we must have that |X| = 1. Every element in a free group has infinite order and .

Suppose that G is a free abelian group which is cyclic. Then for some . Then each element of G has the form for some integer n. Define by is a group isomorphism so that G is free since is free.

**10)(a) Show that the additive group is not finitely generated.
(b)Show that is not free
Proof- **Assume instead that is finitely generated by some set . Denote the product . Let so that we can write for some . Then we can write for some . This tells us that . Thus , but obviously must also contain so that . Let such that . Then for some integer b. Finally we get that which contradicts gcd(a,n) = 1. Thus the rational numbers are not finitely generated.

b) Let be any two rationals. If , then we have and we can have and . This shows x and y are not linearly independent. Thus if, was a free group, it would have to be of rank 1, which would make it cyclic, but from part a) we know that it is not cyclic (because it is not finitely generated). Thus the rationals are not a free group.