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Isomorphism question

  1. Nov 27, 2011 #1
    All groups are finite abelian

    if K⊕K ≅ N⊕N, prove that K≅N

    I'm thinking of constructing bijection, but I don't know if my argument makes sense!

    since K⊕K ≅ N⊕N, there exists a bij between the two

    assume ψ: K⊕K ----> N⊕N
    (k,k') |---> (n,n') where n = f(k) for some fct f and n' = g(k')

    since ψ is a bij it is invertible, hence f is invertible and then it is a bijection


    since ψ is a homomorphism, then so is f b/c ψ(kk',k"k'")=(nn',n"n"')=(f(k),f(k'),g(k'')g(k'''))=(f(kk'),g(k"k'''))

    and f is onto inherited from ψ. ker ψ = ψ(e,e)=(e,e) in (N⊕N) = ker(f,g)

    I don't like it but it's the only thing I can think of.

    any ideas. thanks
  2. jcsd
  3. Nov 27, 2011 #2


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    do you know the decomposition theorem for finite abelian groups?
  4. Nov 27, 2011 #3
    in other words, the fundamental thm of finite abelian groups,

    every fin. ab. grp G is the direct sum of cyclic groups, each of prime power order.

    let G = K⊕K ≅ N⊕N = G'

    by thm K and N are both cyclic. hence K= <k> and N = <n> with |<k>| = |<n>|

    hence K ≅ N.
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