Suppose is a finite abelian group. Then is the quotient group where denotes the set of 2 torsion elements in (that is, the elements such that is the identity) ? I feel this should be easy to prove if it were true but haven't managed to get anywhere.
Yes, this was my first thought, but then I couldn't think of a homomorphism with kenel and image . Intuitively I think we want to consider "odd and even orders" of the elements (elements with even order are sent to identity) but I didn't feel like I got anywhere by using the FIT.
Actually, have you come across the fundamental theorem of finite abelian groups? This will make your job a lot easier! It means you can split A off into a direct product of groups of order a power of 2 (2-groups), and everything else. Everything else `disappears', so you are left with your 2-groups.
This means (whith a bit of work) you can induct on n, where A is a cyclic group of order . Which is easy, if you keep in mind the isomorphism you are trying to prove.
EDIT: I can't think of a way of doing it without using the FTFAG to `throw away' the subgroup of elements of odd order. Basically, you can use the FTFAG to write every element (multiplicatively) in the form,
where has odd order and has order (a power of 2). Then your homomorphism is,
With a bit of work you can show that your kernel is what you want, and that image is also what you want. (You would also need to show it is a homomorphism though!)