# Thread: A/2A isomorphic to A[2]?

1. ## A/2A isomorphic to A[2]?

Suppose $A$ is a finite abelian group. Then is the quotient group $A/2A \cong A[2]$ where $A[2]$ denotes the set of 2 torsion elements in $A$ (that is, the elements $a \in A$ such that $2a$ is the identity) ? I feel this should be easy to prove if it were true but haven't managed to get anywhere.

2. Originally Posted by amanda19
Suppose $A$ is a finite abelian group. Then is the quotient group $A/2A \cong A[2]$ where $A[2]$ denotes the set of 2 torsion elements in $A$ (that is, the elements $a \in A$ such that $2a$ is the identity) ? I feel this should be easy to prove if it were true but haven't managed to get anywhere.
I believe the hard bit is working out what you need to prove. My hind is "The First Isomorphism Theorem"...

3. Yes, this was my first thought, but then I couldn't think of a homomorphism with kenel $2A$ and image $A[2]$. 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.

4. 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 $2^n$. 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,

$g=a_1^{\alpha_1}\ldots a_n^{\alpha_n}b_1^{\beta_1}\ldots b_m^{\beta_m}$

where $a_i$ has odd order and $b_j$ has order $2^{x_j}$ (a power of 2). Then your homomorphism is,

$a_1^{\alpha_1}\ldots a_n^{\alpha_n}b_1^{\beta_1}\ldots b_m^{\beta_m}\mapsto b_1^{\beta_12^{x_1-1}}\ldots b_m^{\beta_m2^{x_m-1}}$

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!)

5. something tells me this is over-kill. why not consider 2 cases: |A| odd, and even.

i could be wrong here, but i believe if |A| is odd, A[2] = {0} so a-->0 should work.