# Isomorphism question

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
Homework Helper
do you know the decomposition theorem for finite abelian groups?

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.