= Solution
A rigid <rotation> of a regular <tetrahedron> permutes its four vertices, giving an injective <group homomorphism> from its <rotation> <group> to $S_4$: fixing every vertex <forces> the <rotation> to be the identity. Its seven nontrivial <rotation> axes are four lines through a vertex and the opposite face centre, and three lines through midpoints of opposite edges.
Each vertex-face axis admits <rotations> by $\pm2\pi/3$, giving eight three-cycles on the vertices. Each opposite-edge axis admits a half-turn, giving three double <transpositions>. Along with the identity there are twelve <rotations>. All their induced <permutations> are even, so the injective image is a twelve-element <subgroup> of $A_4$, hence the whole <group>. Therefore
$$
\boxed{\text{the tetrahedron rotation group is isomorphic to }A_4.}
$$
Its action on $\mathbb R^3$ supplies a three-dimensional <representation>. To prove irreducibility even after complexification, choose vertices
$$
(1,1,1),\quad(1,-1,-1),\quad(-1,1,-1),\quad(-1,-1,1).
$$
The <group> contains the half-turns $D_x=\operatorname{diag}(1,-1,-1)$ and their cyclic variants, as well as the coordinate cycle $(x,y,z)\mapsto(y,z,x)$. If a nonzero <invariant subspace> over the complex numbers contains a <vector> with nonzero $x$ component, applying $(I+D_x)/2$ shows it contains the $x$ axis. If another component is nonzero, use the corresponding projection instead. The coordinate cycle then supplies the other two axes. Every nonzero <invariant subspace> is therefore the full space, proving
$$
\boxed{A_4\text{ has an irreducible representation of dimension }3.}
$$
This also gives an elementary proof without presupposing the <character table>.
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