Is G_nm Isomorphic to G_n x G_m?

[barbiemathgurl]

Let k be a positive integer.

define G_k = {x| 1<= x <= k with gcd(x,k)=1}

prove that:
a)G_k is a group under multiplication modulos k (i can do that).

b)G_nm = G_n x G_m be defining an isomorphism.

 

[matt grime]

What have you done for b)? There is only one possible way you can think of to write out a map from G_nm to G_n x G_m, so prove it is an isomorphism. Remember, G_n x G_m looks like pars (x,y)...

 

[Kummer]

We can use the Chinese Remainder Theorem on this one.

Define the mapping,
\phi: G_{nm}\mapsto G_n\times G_m
As,
\phi(x) = (x\bmod{n} , x\bmod{m})

1)The homomorphism part is trivial.
2)The bijection part is covered by Chinese Remainder Theorem.

 

[mathwonk]

but the point is to prove that theorem.