Proof of Chinese remainder theorem by isomorphism

Question

My note states that we can prove Chinese remainder theorem as the way shown: Let $m$ and $n$ be coprime natural numbers. Then $C_{mn}$ is isomorphic to $C_m \times C_n$ (where $C_m$ is cyclic group with order $m$)

I understand how to prove they are isomorphic, but I don’t understand how that prove the Chinese remainder theorem.

For example,let $m=5,n=7$. The isomorphism shows that there is a bijective map between adding the result we get after applying $\pmod {35}$ and adding $(a,b)$, where $a$ is the result after $\pmod 5$ and $b$ is the result after $\pmod 7$. But how does it show that we could know what is $a\pmod {35}$ if we know $(a\pmod 7,a\pmod 5)$ for positive integer $a$?

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Edit: I know the chinese remainder theorem as

If one knows the remainders of the Euclidean division of an integer n by several integers, then one can determine uniquely the remainder of the division of n by the product of these integers, under the condition that the divisors are pairwise coprime (no two divisors share a common factor other than 1)

Answer 1

Let $m$ and $n$ be coprime natural numbers. Then $C_{mn}$ is isomorphic to $C_m×C_n$ (where $C_m$ is cyclic group with order $m$)

Note how the chinese remainder theorem is just a generalization of that: Let $\{n_i\}$ be pairwise coprime. Then $C_N$ is isomorphic to $C_{n_1}\times\dots\times C_{n_k}$ via the map $[x]_N\mapsto ([x_{n_1}],\dots,[x_{n_k}])$, where $N=\prod n_i$.

So maybe try to generalize your proof for only two coprime numbers to work for $k$-many pairwise coprime numbers.

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Edit: OPs version of the CRT is

If one knows the remainders of the Euclidean division of an integer $n$ by several integers, then one can determine uniquely the remainder of the division of $n$ by the product of these integers, under the condition that the divisors are pairwise coprime (no two divisors share a common factor other than 1)

Here is how you see that this is implied by the version of the CRT above:

Let $n\in \mathbb N$ be some integer and let $q_1,\dots, q_k$ be pairwise coprime integers. Suppose I know the remainders ($n$ mod $q_i$) for every $i$. All these remainders together give an element in $C_{q_1}\times\dots\times C_{q_k}$. Now using the CRT (as stated by me above) we get one element in $C_Q$ by the isomorphism, where $Q=q_1\cdot \dots\cdot q_k$. This element is precisely the remainder ($n$ mod $Q$)

Answer 2

The thing to note is not just that there is a group isomorphism $f : C_{nm} \to C_n \times C_m$.

We have canonical group homomorphisms $g : \mathbb{Z} \to C_{nm}$ and $h : \mathbb{Z} \to C_n \times C_m$. You denote these maps by $g(a) = (a \mod nm)$ and $h(a) = ((a \mod n), (a \mod m))$ (though this is not standard notation). The key here is that $f \circ g = h$. In other words, if we know $a \mod nm$ - that is, if we know $g(a)$ - we can apply $f$ to compute $h(a)$ - that is, to compute $(a \mod n, a \mod m)$. Because $f$ is an isomorphism, we can also go the other way - we have $f^{-1} \circ h = f^{-1} \circ f \circ g = g$. Thus, if we know $(a \mod n, a \mod m)$, we can apply $f^{-1}$ to compute $a \mod nm$.