A question on Abelian Groups

Question

Prove that every subgroup of an Abelian group is Abelian but the converse is not true. I recently stumbled onto this question , but not able to solve it . Please help me out!

Answer 1

Let $a_k$ be the group elements of your abelian group $A$. Since all elements of $A$ commute among each other, this includes every subgroup of $A$.

Otherwise there is a subgroup of a non-abelian group $G$, which contains only elements that commute among each other, called the center. This might be trivially the identity element ...

Answer 2

it is obvious that every subgroup of every abelian group is abelian. for the inverse $S_3$(group of permutations of 3 objects) is not an abelian group but it's subgroups(except $S_3$) are abelian(beacuse subgroups are of order $1$ or $2$ or $3$ that are abelian).

Answer 3

let,G be group.h is subgroup of G. given that G is abelian i.e. ab=ba for all element of G. let,x,y be any element of H. H is subgroup of G means H is also subset of G. therefoer x,y are also elt. of G therefore xy=yx for all elt. of H so H is abelian. For converse look at S3.