Are $G(30,30)$ and $G(6,10,15)$ isomorphic? Here is a little problem if anyone could help me with it, it would be a lot appreciated

Let $G(k_1,k_2,...,k_r)$ the abelian group define as: $$\mathbb Z /k_1\mathbb Z \,\times \mathbb Z /k_2\mathbb Z \,\times ...\mathbb Z  /k_r\mathbb Z$$

We define: $G(30,30)$ and $G(6,10,15)$.
Can we prove using the Chinese reminder theorem that there is an isomorphism between this two groups?
Thanks in advance for your help.
 A: The Chinese remainder theorem can be stated as
$$\mathbb{Z}/nm\mathbb{Z}\simeq\mathbb{Z}/n\mathbb{Z}\times \mathbb{Z}/m\mathbb{Z}$$
whenever $n,m$ are coprime. Or equivalently, in your notation
$$G(n\cdot m)\simeq G(n,m)$$
With that we have
$$G(30,30)=G(2\cdot 3\cdot 5,2\cdot 3\cdot 5)\simeq G(2,2,3,3,5,5)$$
$$G(6, 10, 15)= G(2\cdot 3, 2\cdot 5, 3\cdot 5)\simeq G(2,2,3,3,5,5)$$
and so they are clearly isomorphic.
A: If you mean the following statement by Chinese Remainder Theorem, then yes, we can solve it by using it:

Theorem(Chinese Remainder Theorem): Let $m,n \in \mathbb{Z}^+$ such that $\gcd(m,n) = 1$. Then, the map $\phi: \mathbb{Z}_{mn} \to \mathbb{Z}_{m} \times \mathbb{Z}_{n}$ defined by $\phi([x]_{mn}) = ([x]_m, [x]_n)$ is an isomorphism.
Then, we have
$$G(6,10,15) = \mathbb{Z}_6 \times \mathbb{Z}_{10} \times \mathbb{Z}_{15} \cong \mathbb{Z}_{2} \times \mathbb{Z}_{3} \times \mathbb{Z}_{10} \times \mathbb{Z}_{15} \cong \mathbb{Z}_{2} \times \mathbb{Z}_{15} \times \mathbb{Z}_{3} \times \mathbb{Z}_{10} \cong \mathbb{Z}_{30} \times \mathbb{Z}_{30}$$
Note that, we are also using the fact that $G \times H \cong H \times G$ to interchange the terms of the product.
