Proving K≅N from K⊕K ≅ N⊕N

[Bachelier]

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

or:

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

 

[mathwonk]

do you know the decomposition theorem for finite abelian groups?

 

[Bachelier]

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.