Prove that if a, b are relatively prime integers, then $\Bbb{Z}/ab\Bbb{Z}$ is isomorphic to $\mathbb{Z}/a\Bbb{Z} \times \Bbb{Z}/b\Bbb{Z}$. I know this is related to the Chinese remainder theorem but I'm having trouble showing there is an isomorphism between the mapping $\mathbb{Z}/ab\mathbb{Z}$ to $\mathbb{Z}/a\mathbb{Z} \times \mathbb{Z}/b\mathbb{Z}$.
Thank You.
 A: It's rather more than "related to" the Chinese Remainder Theorem, in fact it really is the CRT, just stated in more abstract terms.
Let $a,b$ be relatively prime and consider the map
$$\phi:{\Bbb Z}/ab{\Bbb Z}\to({\Bbb Z}/a{\Bbb Z})\times({\Bbb Z}/b{\Bbb Z})
  \quad\hbox{where}\quad \phi(x)=(x\bmod a,x\bmod b)\ .$$
First, this map is well defined: if $x$ has a given value modulo $ab$ then its values modulo $a$ and modulo $b$ are determined, so $\phi(x)$ is uniquely determined.  Also, the CRT tells us that if $(s,t)$ in $({\Bbb Z}/a{\Bbb Z})\times({\Bbb Z}/b{\Bbb Z})$ is given then the system
$$x\equiv s\pmod a\ ,\quad x\equiv t\pmod b$$
has a solution, so $\phi$ is onto (surjective).  Since the domain and codomain have the same size, $\phi$ is a bijection.  Finally, it is easy to check (try it) that
$$\phi(x+y)=\phi(x)+\phi(y)\ ,$$
and since we are talking about finite sets, this is all we need to show that $\phi$ is an isomorphism.
A: Define the map $f: \mathbb{Z}/ab\mathbb{Z} \Longrightarrow \mathbb{Z}/a\mathbb{Z}*\mathbb{Z}/b\mathbb{Z}$ as follows:
\begin{equation*}
                 f(x\   mod\  ab ) = (x \ mod \ a, x\  mod\  b). 
\end{equation*}
Observve that $\mathbb{Z}/ab\mathbb{Z}$ , and $\mathbb{Z}/a\mathbb{Z}*\mathbb{Z}/b\mathbb{Z}$ both have $ab$ elements. So it suffices to prove that $f$ is one to one, hence it will be onto. So if ($x$  mod $a$, $x$ mod $b$) = ( $y$ mod $a$, $y$ mod $b$) $\Rightarrow$
$a$ divides $x - y$, and $b$ divides $x - y$. Since $(a,b) = 1$, $ab$ divides $x - y$. This means $x$ mod $ab = y$ mod $ab$. So $f$ is one to one.
A: The isomorphism is $\phi(ab+N\mathbb{Z})=(a+N\mathbb{Z},b+N\mathbb{Z})$
First show it is a homomorphism, then show it is injective and surjective - this proof is an application of the Chinese Remainder Theorem.
We note that an isomorphism is a bijective homomorphism, so we need to show (1) that $\phi$ is a homomorphism and (2) that $\phi$ is bijective.
(1) $\phi$ is trivially a homomorphism because $\mathbb{Z/nZ} \times \mathbb{Z/mZ}$ is a group, and $\phi(a+b)=\phi(a)+\phi(b) \forall a,b \in \mathbb{Z/mnZ}$ 
(2) a. $\phi$ is injective because $\phi(a) = \phi(b)$ if and only if $a=b$. An ordered pair is only equal if their components are equal.
b. $\phi$ is surjective if and only if $\forall a+\mathbb{Z/n},b+\mathbb{Z/mZ}$, $\exists c$ such that $\phi(c+\mathbb{Z/nmZ})=(a+\mathbb{Z/nZ},b+\mathbb{Z/mZ}).$ Note that this statement is equivalent to:
$$c \equiv a \pmod{n}$$
$$c \equiv b \pmod{m}$$
By the Chinese Remainder Theorem, this system has a unique solution c if m and n are coprime. 
