An exam question has asked me to show that there is a normal subgroup of size 4 of the group of rotational symmetries of the cube.
A trick I've seen before is considering the action of a group on its set of left cosets, but I don't think it works here.
Ok, I'll try to piece together all the pieces that ahulpke gave me (thanks) into a single answer.
Let $G$ be the group of rotational symmetries of a cube. Then consider the action of $G$ on the pairs of opposite faces of the cube, $X$. This action induces a homomorphism $$\phi:G \rightarrow \text{Sym}(X)$$ for which $\text{ker} \phi$ is a normal subgroup of $G$. By the first isomorphism theorem, $$G/\text{Ker}\phi \cong \text{Im}\phi<\text{Sym}(X),$$and by observing that $\text{Im}\phi$ has an element of order 2 and 3 by considering certain rotations, we can conclude that $\text{Im}\phi$ must have size six as it is a subgroup of $\text{Sym}(X) \cong S_3$ which has six elements. As $G$ has 24 elements, $\text{Ker}\phi$ has $24/6=4$ elements, and we're done.