Inverse of $f: \mathbb{Z}_{21} \times \mathbb{Z}_{10} \to \mathbb{Z}_{210}$ such that $f(a,b)= 10a +21b$ Let $f: \mathbb{Z}_{21} \times \mathbb{Z}_{10} \to \mathbb{Z}_{210}$ such that $$f(a,b)= 10a +21b.$$ 
We have that $f$ is an isomorphism, but how does one go about finding explicitly the inverse $f^{-1}$ of $f$?
 A: Since both are cyclic, both have a generator. You need only find $(a,b)$ such that $f(a,b)=1$, which amounts to the division algorithm. Then 
$$f^{-1}(c) = (ca, cb)$$
In this case we have $a=19, b=1$.
Further explanation:
Suppose you have $f(a,b)=1$, i.e. $10a+21b=1$. If $g(c)=(ca,cb)$ then
\begin{align*}
g(f(x,y)) = g(10x+21y) &= ((10x+21y)a, (10x+21y)b)\\
&=  (10ax,21by)\\
&=  ((1-21b)x,(1-10a)y)\\
&=  (x,y)
\end{align*}
A: To construct a homomorphism (say, $h$) from a cyclic group $< a >$ (i.e, a group generated by the element $a$) to some group $G$, all we have to do is to assign an appropriate image of $a$ (i.e, we will assign some value to $h(a) \in G$ appropriately).
Since the group $\mathbb{Z}_{210}$ is generated by $1$, to find $f^{-1}$, all we have to do is to assign an appropriate element to $f^{-1}(1) \in \mathbb{Z}_{21}\times \mathbb{Z}_{10}$.
Since $f^{-1}$ is the inverse of $f$, to assign a value for $f^{-1}(1)$, all we have to do is to find which element in $\mathbb{Z}_{21} \times \mathbb{Z}_{10}$ that $f$ maps to $1$ in $\mathbb{Z}_{210}$, then assign $f^{-1}(1)$ to be that element, and the element we're talking about is $(19, 1)$. This element can be easily found by noticing the fact that: $(-2) \times 10 + 1 \times 21 = 1$, and the element $-2$ in $\mathbb{Z}_{21}$ is the same as the element $19$ in $\mathbb{Z}_{21}$.
So $f^{-1}: \mathbb{Z}_{210} \to \mathbb{Z}_{21} \times \mathbb{Z}_{10}$ is constructed so that $f^{-1}(1) = (19, 1) \in \mathbb{Z}_{21} \times \mathbb{Z}_{10}$, so, take any element $c \in \mathbb{Z}_{210}$, we'll have:
$$f^{-1}(c) = f^{-1}(\underbrace{1 + 1 + ... + 1}_{c \text{ elements } 1'\text{s}}) = \underbrace{f^{-1}(1) + f^{-1}(1) + ... + f^{-1}(1)}_{c \text{ elements } f^{-1}(1)'\text{s}} = (19c; c)$$
A: It is well known, and easy to check, that if $m$ and $n$ are coprime, and $um+vn=1$ is  a Bézout's relation between $m$ and $n$ the inverse isomorphism from $\mathbf Z/m\mathbf Z\times\mathbf Z/n\mathbf Z$ to $\mathbf Z/mn\mathbf Z$ is given by
$$(x\bmod m,y\bmod n)\longmapsto yum+xvn\bmod mn.$$
Indeed $z=yum+xvn$ satisfies $$z\bmod m=xvn\bmod m=x(um+vn)\bmod m=x\cdot 1\bmod m=x\bmod m,$$
and similarly $\bmod n$.
