# Chinese Remainder Theorem (Groups)

## Theorem

Let $G$ be a group.

Let $N_1, \ldots, N_n$ for some $n \ge 1$ be normal subgroups of $G$.

Let $\pi_i: G \rightarrow G / N_i$ be the canonical projections.

Then the homomorphism $\pi: G \to G / N_1 \times \cdots \times G / N_n$ defined as:

$\map \pi x = \tuple {\map {\pi_1} x, \ldots, \map {\pi_n} x}$

has the kernel $\ds N := \bigcap_{i \mathop = 1}^n N_i$.

Further, $\pi$ is surjective if and only if the normal subgroups have the following property:

$\ds \forall k \le n - 1: N_{k + 1} \bigcap_{i \mathop = 1}^k N_i = G$

## Proof

The mapping $\pi$ is indeed a group homomorphism, because each canonical projection $\pi_i: G \to G / N_i$ is a group homomorphism.

The kernel of $\pi$ is given by:

$\ds \ker \pi = \set {x \in G: \forall i, 1 \le i \le n: \map {\pi_i} x = \map {\pi_i} e} = \set {x \in G: \forall i, 1 \le i \le n: x \in N_i} = \bigcap_{1 \mathop \le i \mathop \le n} N_i =: N$

We will now establish the necessity and sufficiency of the condition for surjectivity of $\pi$.

Indeed, we will demonstrate that if $\pi$ is surjective, then:

$\ds \forall k: N_k \bigcap_{j \mathop \ne k} N_j = G$

which will complete the proof together with a deduction of surjectivity from the condition that:

$\ds \forall k \le n - 1: N_{k + 1} \bigcap_{i \mathop = 1}^k N_i = G$

### Necessary Condition

We will start by showing the condition is necessary for surjectivity.

So suppose $\pi$ is surjective.

Let $x \in G$ be arbitrary.

Then for each $i$, there is $u_i \in G$ such that $\map {\pi_i} {u_i} = \map {\pi_i} x$ and $\map {\pi_j} {u_i} = \map {\pi_j} e$ for $j \ne i$.

Note that $\map {\pi_i} {u_i^{-1} x} = \map {\pi_i} {u_i}^{-1} \map {\pi_i} x = \map {\pi_i} e$.

Therefore, $u_i \in N_j$ for $j \neq i$ and $u_i^{-1} x \in N_i$.

Hence for all $i \ne j$, we find:

$\ds x = u_i \paren {u_i^{-1} x} \in N_j \bigcap_{m \mathop \ne i} N_m$

As $x$ was arbitrary, this completes the proof that $\ds N_j \bigcap_{m \mathop \ne i} N_m = G$.

$\Box$

### Sufficient Condition

We will use induction on $n$, where the base case $n = 1$ is trivial.

By induction, assume the result for the case $n$.

Now take any $x_1, \ldots, x_n, y \in G$.

We will produce an element $g$ with $\map \pi g = \tuple {\map {\pi_1} {x_1}, \ldots, \map {\pi_n} {x_n}, \map {\pi_{n + 1} } y}$.

First, by the induction hypothesis, we can find an $x \in G$ such that we have $\map {\pi_i} x = \map {\pi_i} {x_i}$ for all $i \le n$.

Recall our assumption that $\ds N_{n + 1} \bigcap_{i \mathop = 1}^n N_i = G$.

In particular, we can choose $a \in N_{n + 1}$ and $\ds b \in \bigcap_{i \mathop = 1}^n N_i$ such that $y^{-1} x = a b$.

Equivalently, we have $y a = x b^{-1}$.

Thus for $i \le n$, we have $\map {\pi_i} {y a} = \map {\pi_i} {x b^{-1} } = \map {\pi_i} x = \map {\pi_i} {x_i}$.

Further, we have $\map {\pi_{n + 1} } {y a} = \map {\pi_{n + 1} }y$.

This means that $g = y a$ is the desired element.

This concludes the proof of the sufficiency.

$\blacksquare$