When are the product of minimal maps also minimal?

Question

I am struggling to answer this question from Intro to Dyn Systems, Brin:

Exercise 2.1.10 Assume that f and g are minimal. Find necessary and sufficient conditions for f × g to be minimal.

My attempt:

Let f: X → X and g: Y →Y be continuous maps of compact metric spaces. f × g is minimal if every orbit is dense in X × Y.

From a previous result, the closure of the forward orbit of f × g is equal to the closure of the forward orbit of f times the closure of the forward orbit of g if, and only if, (x, g(y)) is inside the closure f×g (x, y).

I'm now stuck on how to use this, but I think it might be related to ensuring (x, g(y)) is inside the closure of f×g(x, y).

Could anyone also suggest an example that would illustrate why the condition is necessary?

Thank you!

Answer 1

Let $f:x\mapsto (x+\alpha\bmod 1)$ be a map with dense orbits on the unit circle, that is, $α$ is irrational. Let $g=f^{\circ2}$ be the map for the still irrational $2α$. Then $f\times g$ will move along the lines $2x-y=2x_0-y_0=const. \pmod 1$ on the product torus, meaning it does not have dense orbits. $2x_0-g(y_0)=2x_0-y_0-2α \pmod 1$ does not lie on this same line.