How can i prove that a cartesian product is isomorphic to another cartesian product

Question

$\def\<#1>{\left<#1\right>}\def\Z{\mathbb Z}\<\Z_6, \oplus> \times \<\Z_{10},\oplus>$ is isomorphic to $\<\Z_2, \oplus> \times \<\Z_{30}, \oplus>$

i know i have to prove that with the chinese remainder theorem but i cant see how.

I denote the group operation by $*$. $(x,y)*(z,w)=(a,b)$ from the first cartesian product. and $(t,s)*(u,v)=(a', b')$ from the second direct product now $F(a,b)=(a',b')$ but what is $F$ here?

Thanks for any help!!!

Answer 1

This is the direct sum of two abelian groups (integers modulo n under addition).

$$\mathbb Z_m \oplus \mathbb Z_n \cong \mathbb Z_{mn}\text{ if and only if } \gcd(m, n) = 1$$

$\gcd(2, 3)$, so $\mathbb Z_6 \cong \mathbb Z_2\oplus \mathbb Z_3$ and $\gcd (3, 10) = 1$, so $\mathbb Z_{30} \cong \mathbb Z_3\oplus \mathbb Z_{10}.$

That gives us $$\mathbb Z_6\oplus \mathbb Z_{10} \cong \mathbb Z_2\oplus \mathbb Z_3 \oplus \mathbb Z_{10} \cong \mathbb Z_2 \oplus \mathbb Z_{30}$$