Proof 1. Let H be a subgroup of the abelian group G. Let x,y be elements of H. Since x,y \in G and G is abelian, we have that xy = yx. As the elements x,y are taken arbitrarily, we have that the subgroup H is abelian too.

Proof 2. Let H is a subgroup of the abelian group G. Assume by contradiction that if x,y \in H then xy \neq yx. Since x,y \in G, we have that xy \neq yx in G too, which means that G is not abelian, a contradiction. So H is indeed abelian.

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