- #1
Bachelier
- 376
- 0
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
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