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 $\ds N := \bigcap_{i \mathop = 1}^n N_i$, and is surjective the normal subgroups have the following property:
 * $\forall i \ne j: N_i N_j = G$

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

The kernel of $\phi$ is given by:
 * $\ds \ker \pi = \set {x \in G: \forall i, 1 \le i \le n: \map {\pi_i} x = \map {\pi_i} e} = \set {x \in G: \forall i, 1 \le i \le n: x \in N_i} = \bigcap_{1 \mathop \le i \mathop \le n} N_i =: N$

It remains then to be proved that $\pi$ is surjective the subgroups pairwise generate all of $G$.

Stated explicitly, we will show that the statement:
 * $\forall \sequence {x_i}_{i = 1}^n \in G^n: \exists x \in G: \map {\pi_i} x = \map {\pi_i} {x_i}$

holds :
 * $\forall i \ne j: N_i N_j = G$

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 $j \ne i$.

Clearly, $\map {\pi_j} {a_i} = 0$ for $j \ne i$ while $\map {\pi_i} {u_i^{-1} x} = \map {\pi_i} {u_i}^{-1} \map {\pi_i} {x} = \map {\pi_i} {e}$.

Hence for all $i \ne j$, we find:
 * $x = u_i \paren {u_i^{-1} x} \in N_j N_i$

As $x$ was arbitrary, this completes the proof that $N_i N_j = G$.

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 $x_1, ..., x_n, y \in G$.

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

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 $i \le n$.

Our assumption implies 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.