For a recent project (which I have since completed) I needed to derive the automorphism group of the cube graph, and I wanted to do so with some reasonable degree of rigor. I defined a group action of $S_4$ on the four pairs of distance three vertices and noted that the kernel of the induced homomorphism is the identity and the permutation that flips every pair of vertices in the block. This is $C_2$, so by the first isomorphism theorem: $$\text{Aut}(\text{Cube})/C_2 \cong S_4 $$ I also know that Aut(Cube) $\cong C_2 \times S_4$, so it seems natural to simply "multiply" by the quotiented group in the above equation. However, I know that this is not a legitimate step, because, for instance $C_4/C_2 \cong C_2$ while $C_4 \not \cong C_2 \times C_2$.
So, my question is two-fold: one, is there a way to derive Aut(Cube) $\cong C_2 \times S_4$ from my argument, and two, are there certain nice conditions under which a direct product of a quotient group and the quotient is isomorphic to the whole group? $$N \times G/N \cong G$$