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Math Help - A/2A isomorphic to A[2]?

  1. #1
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    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.
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by amanda19 View Post
    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"...
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  3. #3
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    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.
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  4. #4
    MHF Contributor Swlabr's Avatar
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    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!)
    Last edited by Swlabr; May 12th 2011 at 03:22 AM.
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  5. #5
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    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.

    if |A| is even (will think about this more later).
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  6. #6
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Deveno View Post
    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.

    if |A| is even (will think about this more later).
    Yes - if the order is odd 2A=A and everything vanishes. What I was doing was using the FTFAB to write A=BxC, B of odd order C of order a power of 2, and then you can chuck away B, so you just need to think about when |A| is a power of 2...
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