Is there a group on which the maps

$x \mapsto x^2, x \mapsto x^3, x \mapsto x^4, \dots$

are injective precisely if they are surjective? For example, $\mathbb{Z}$ is *not* such a group, since the map $x \mapsto x + x$ is injective, but not surjective.

You can use your knowledge of group theory to construct such a group explicitly (easy to think of Abelian examples, so let's make it non-Abelian), or you can appeal to compactness. Consider the first-order theory over the language of group theory that has the axioms of group theory (again, you can make it non-Abelian), along with axioms asserting that each of the functions above is injective precisely if it is surjective.

On a finite set, every function is injective precisely if it is surjective, so every finite group (every finite non-Abelian group) is a model of this theory. Therefore, there must be an infinite group that is a model of this theory as well.

One can, of course, immediately start asking less trivial questions along the same lines (until directly constructing the groups gets difficult enough): this eventually leads to the theory of *pseudofinite* structures, structures that satisfy all sentences that every finite structure of the same kind satisfies (if you thought of $(\mathbb{Q},+)$ in the Abelian case: that is actually a pesudofinite group, but proving that is much harder).