Chinese remainder theorem for groups of matrices over rings Let $N=p_1^{k_1}p_2^{k_2}$ and consider the group $GL_2(\mathbb{Z}/N\mathbb{Z})$.
I would like to use Chinese remainder map to show that the group is isomorphic to $GL_2(\mathbb{Z}/p_1^{k_1}\mathbb{Z}) \times GL_2(\mathbb{Z}/p_2^{k_2}\mathbb{Z})$. Does the following map gives an isomorphism:
$$A \mapsto (A \bmod{p_1^{k_1}}, A \bmod{p_2^{k_2}})$$
where $A \bmod{p_i^{k_i}}$ just means all entries of $A$ taking modulo $p_i^{k_i}$.
This is clearly a well defined map since if the determinant of $A$ is coprime to $N$, then each modulo $p_i^{k_i}$ is also a unit in each $\mathbb{Z}/p_i^{k_i}\mathbb{Z}$. Moreover, this is one-to-one correspondence by CRT. However, I am a bit lost of showing the subjectivity (both of their sizes are equal). Any thought?
 A: This is a situation where viewing the problem in greater generality makes things a lot less complicated. Suppose you have two commutative rings $R$ and $S$, and let $\varphi \colon R \to S$ be a (unital) ring homomorphism. Try to prove the following:

*

*For any natural number $n$, there is an induced ring homomorphism $M_{n}(\varphi) \colon M_{n}(R) \to M_{n}(S)$ given by applying $\varphi$ element-wise to each matrix entry.

*The ring morphism $M_{n}(\varphi)$ induces a group homomorphism $\operatorname{GL}_{n}(\varphi) \colon \operatorname{GL}_{n}(R) \to \operatorname{GL}_{n}(S)$.

*If $R = S$ and $\varphi = \operatorname{Id}_{R}$, then $\operatorname{GL}_{n}(\varphi) = \operatorname{Id}_{\operatorname{GL}_{n}(R)}$.

*If $\psi \colon S \to T$ is a ring homomorphism, then $\operatorname{GL}_{n}(\psi \circ \varphi) = \operatorname{GL}_{n}(\psi) \circ \operatorname{GL}_{n}(\varphi)$.

(You might recognize that you are actually demonstrating that the association $R \mapsto \operatorname{GL}_{n}(R)$ defines a functor from the category of commutative rings to the category of groups.)
Once you've proven these facts, you may deduce that $\operatorname{GL}_{n}(\varphi)$ is an isomorphism if $\varphi$ is an isomorphism. Your problem then doesn't require any specific require any specific knowledge, other than that the Chinese Remainder Theorem furnishes a ring isomorphism $\mathbb{Z}/N\mathbb{Z} \to \mathbb{Z}/p_{1}^{k_{1}}\mathbb{Z} \times \mathbb{Z}/p_{2}^{k_{2}}\mathbb{Z}$ defined as you suggest.
