Visual proof that $\mathbb Z_m × \mathbb Z_n = \mathbb Z_{mn} \iff \gcd(m,n) = 1$ Is there a way of visually proving or demonstrating that $\mathbb Z_m × \mathbb Z_n = \mathbb Z_{mn} \iff \gcd(m,n) = 1$?
Take $Z_4 × Z_3$ for example:

It's not obvious from the Cayley diagram that that group is cyclic, even if we rearrange it into concentric squares.
What is a good way of showing visually that a direct product is cyclic?  Is there a visual way of showing this in general?
 A: Think of two circles, one of circumference $4$, the other of circumference $3$. 
Choose one shiny dot as the base point on each circle. 
Now let each shiny dot rotate around each circle at the same constant speed of 1 unit per second (not angular speed, but actual speed). 
If you were to animate this, you would see that the first moment that the two shiny dots simultaneously return to their original positions is $3 \cdot 4 = 12$ seconds after they start.
A: In the above case, if we want to represent the red arrows as elements of $\mathbb Z_{12},$ you need $x\equiv 0\pmod{4}$ and $x\equiv 1\pmod 3$ for $x$ being the red line. So the red arrows corresponds to $x=4\pmod{12}$. The blue arrows corresponds to $x\equiv 0\pmod{3}$ and $x\equiv 1\pmod{4}$ which is $x\equiv 9\pmod{12}$.
So the labeling of this Cayley diagram to get the Cayley diagram for $\mathbb Z_{12}$ is essentially due to Chinese Remainder Theorem.
If you went the other way, drawing $\mathbb Z_{12}$ as a regular $12$-gon first, then lines for red would trace out four equilateral triangles in the circle, and the lines for blue would trace out three squares. Any red triangle and blue square would meet at exactly one point (not initially obvious, that is a result of $\gcd(3,4)=1$.)
More generally, if we have a circle of $mn$ points and we draw $n$ regular $m$-gons as red and $m$ regular $n$-gons (with arrows all going, say, counter-clockwise.)

Claim 1: If $\gcd(m,n)=1$, then any regular $m$-gon and any regular $n$-gon inscribed in the same circle share in at most one corner.
Proof: Sharing two corners means that for some $0\leq i<n,0\leq j<m$ you have that $\frac{2\pi i}{n}=\frac{2\pi j}{m}$, which means $nj=mi<mn$ is a common multiple of $m$ and $n$, which is not possible when $n,m$ are relatively prime because it would mean:
$$1<\frac{mn}{\operatorname{lcm}(m,n)}$$ would be a common divisor of $m,n$.

Now, each of the $nm$ points on our circle is on a single $m$-gon and a single $n$-gon. And there are $nm$ pairs of $m$-gon and $n$-gon, so this means that every pair must contain one common corner. That's the real trick. Visually, that isn't obvious, but by a counting argument, using the above claim, it has to be true.
So, if we remove the circle, we see that we can re-arrange our diagram to get your diagram. (You have to argue from commutativity that following a red arrow then a blue arrow is the same as following a blue arrow then a red arrow, to ensure that you get a grid like above.)
Aside: Indeed, the fact that pair of $n$-gon and $m$-gon have a single common points is Chinese Remainder Theorem. The set of points $x+mn\mathbb Z$ such that $x\equiv a\pmod n$ is one $m$-gon, and the set of points $x+mn\mathbb Z$ such that $x\equiv b\pmod m$ is the set of points of one $n$-gon. Having a point in common means finding a common $x$.
