# Direct product

1. Jul 29, 2007

### 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.

2. Jul 29, 2007

### 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)....

3. Jul 29, 2007

### 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.

4. Jul 30, 2007

### mathwonk

but the point is to prove that theorem.