Finding if the group of isometries of a cube is a normal subgroup of $S_{12}$

Question

Let $G$ be the group of isometries of a cube and let X be the set of edges of the cube. This action defines an injective homomorphism $\phi:G \rightarrow S_{12} $, injective since stabilisers of adjacent edges overlap only at ${e}$. I am struggling to determine if $Im(\phi)$ is a normal subgroup of $S_{12}$. I have tried considering the action of G on its left cosets and a few other ideas but nothing has yielded useful progress.

Thanks a lot in advance.

Answer 1

The unique non-identity proper normal subgroup of $\mathfrak S_{12}$ is the alternating group $\mathfrak A_{12}$.

$G$ is isomorphic to $\mathfrak S_{4} \times \mathbb Z / 2\mathbb Z$ and therefore has $2 \times 4!= 48$ elements. As $ \vert \mathfrak A_{12} \vert \gt 48$, $\text{Im}(\phi)$ isn't normal in $\mathfrak S_{12}$.