Why do the symmetries of a tetrahedron contain no odd permutations?

Question

This video "Symmetries of a Tetrahedron" shows that the rotational group of a tetrahedron is isomorphic to the alternating group $A_4$.

At time 0:42, it shows that it is geometrically impossible to obtain an odd permutation like $(1 \; 2)$ by rotations. However, how to formally prove it? My main concern here is that how to argue that the symmetries of a tetrahedron do not contain $(1 \; 2)$ does not because we may miss some rotations.

Then, at time 0:48, it shows how to obtain odd permutations by turning the tetrahedron inside out. I don't understand how the tetrahedron is turned inside out. Could you please explain it?

Answer 1

Let $ABCD$ be the tetrahedron and consider

$$K=\det\pmatrix{A-B\\A-C\\A-D}$$

Swapping two vertices from among $B,C$ and $D$ amounts to swapping two rows in the determinant, so it changes the sign of $K$.
Swapping $A$ with another vertex $X$ amounts to subtracting the row $A-X$ from the other two rows and multiplying the row $A-X$ by $-1$. The end result is once again the sign of $K$ changes.

It follows that an odd permutation -- as a composition of an odd number of swaps -- must change the sign of $K$.

Now, we need only show that a rotation leaves the sign of $K$ unchanged. Do you think you can take it from here?