Prove that $G_3$ acts on the set of opposite edges of a tetrahedron Problem Statement

Edges $e,e'$ of a tetrahedron $T$ are said to be opposite if they are disjoint; that is, they do not share a vertex. The $6$ edges can be partitioned into a set, $X$, of $3$ pairs of opposite edges. Prove that $G_3$, the group of symmetries of $T$, acts on $X$ and the kernel $K < G_3$ is a normal subgroup of order $4$.


For this problem I've included a picture of a tetrahedron below with labeling for the edges. The perspective I'm taking for this picture is viewing the tetrahedron from above as if it were sitting on a table where the triangle defined by the edges $B = \{ e_1, e_2, e_3 \}$ constitute the base. The edges $H = \{ e_4, e_5, e_6 \}$ are the edges which give the shape depth and thus are pointing towards a common vertex $V_0$ which is pointing directly at the viewer. I've color coded the edges so that pairs of opposite edges are the same color. Hence our pairs of opposite edges are $$X = \{ (e_1, e_4), (e_2, e_5), (e_3, e_6) \}$$ An observation is that a pair of opposite edges always includes one edge that makes up the base (and hence is part of $B$) and one that makes up the "depth" or "vertical" portion of the shape (and hence is an element of $H$)

As a note, I will focus my questioning in on the first part of the problem: proving that $G_3$ acts on $X$, so as to not broaden my scope too much.
I really have no idea where to start here. I know, from a proposition in the text that $G_3 \cong S_4$ But since $S_4$ is the group of symmetries of $4$ identical objects I'm guessing that the elements of $S_4$ are referring to the permutations of the vertices of the tetrahedron. Since we have $6$ edges ($3$ pairs of $6$ distinct elements) we can't really permute them according to the symmetries in $S_4$ without leaving $2$ permanently fixed. Which, doesn't even make sense geometrically, let alone the arbitrary nature of deciding which edges are "left out." To be more specific:
What exactly is meant by this? I'm guessing they want me to find a homomorphism $\eta: G_3 \to \text{Sym}(X)$ hence, showing that an action exists between the group and the set in question? How can this be done with the set of opposite edges given my concerns above?

 A: $S_4$ is the group of tetrahedral symmetry if these symmetries are not orientation preserving (you can take the "mirror image" of the tetrahedron). Since by taking the correct "mirror image", you can intervert any pair of vertices of the tetrahedron, $S_4$ (which we identify in the following as the permutations of $\{1,2,3,4\}$) indeed acts on the set of vertices $V=\{v_1,v_2,v_3,v_4\}$ through the following group action:
\begin{equation}\begin{aligned} \phi: S_4\times V &\rightarrow V \\
(\sigma,v_i)&\mapsto v_{\sigma(i)}
\end{aligned}
\end{equation}
One can indeed verify the identity axiom:
$$\forall v_i\in V,\, \phi(e,v_i)=v_i$$
and the compatibility axiom:
\begin{equation}\begin{aligned}\forall \sigma_1,\sigma_2\in S_4, \forall v_i\in V, \,\phi(\sigma_2,\phi(\sigma_1,v_i)) &= \phi(\sigma_2,v_{\sigma_1(i)}) \\
&= v_{\sigma_2\circ\sigma_1(i)}\\
&= \phi(\sigma_2\circ\sigma_1,v_i)
\end{aligned}
\end{equation}
We now denote $E=\{e_1,\dots,e_6\}\subset V\times V$ the set of edges of the tetrahedron (where $V\times V$ denotes the set of unordered pairs of vertices), and $X\subset E \times E$ the set of unordered pairs of opposite edges:
\begin{equation}\begin{aligned}
X &= \{x_1,x_2,x_3\}\\
&= \{\{e_1,e_4\},\{e_2,e_5\},\{e_3,e_6\}\} \\
&=\{\{\{v_1,v_2\},\{v_3,v_4\}\},\{\{v_1,v_3\},\{v_2,v_4\}\},\{\{v_1,v_4\},\{v_2,v_3\}\}\}
\end{aligned}
\end{equation}
Since all $X$ contains all the possible associations of two sets of same size $e_i,e_j$ such that $e_i\cup e_j = V$, it is clear that the map:
\begin{equation}\begin{aligned} \psi: S_4\times X &\rightarrow X \\
(\sigma,\{\{v_i,v_j\},\{v_k,v_l\}\})&\mapsto \{\{\phi(\sigma,v_i),\phi(\sigma,v_j)\},\{\phi(\sigma,v_k),\phi(\sigma,v_l)\}\}
\end{aligned}
\end{equation}
is stable. Moreover, it is a group action as well. The identity and compatibility axiom are deduced from the fact that $\phi$ is a group action.
