I was trying to solve the problem of finding the number of elements of $GL_n(\mathbb{Z}_m)$ for some $m\in\mathbb{N}$. The case where $m=p$ prime was easy because of elementary linear algebra. The case $m=p^r$ was quite harder but using some group theory I was able to compute the elements of $GL_n(\mathbb{Z}_{p^{k+1}})$ from the numbers of element of $GL_n(\mathbb{Z}_{p^k})$. But now I am stuck in the general case, I'm not sure whether this is right or not, but my intuition tells me that I can generalize the chinese remainder theorem ( $\mathbb{Z}_{pq}\cong\mathbb{Z}_p\times\mathbb{Z}_q$ if $\gcd(p,q)=1$) to $GL_n(\mathbb{Z}_{pq})\cong GL_n(\mathbb{Z}_{p})\times GL_n(\mathbb{Z}_{q})$ if $\gcd(p,q)=1$. If this where true I would be able to finish the exercise, however I don't know how to do it.
Any hints or help will be thanked.