Past exam of the options II course of the University of Oxford / 2004 / A / A2

The permutation inverse is ${\rho }_{g^{-1}}$: applying the two maps successively sends $\omega$ to $(\omega g)g^{-1}=\omega$. With the paper's right-action convention, permutation products also act on the right, andThus $\boxed{\rho (gh)=\rho (g)\rho (h)}$. If function composition is written on the left instead, the same right action is represented by $g\mapsto {\rho }_{g^{-1}}$; keeping these conventions consistent avoids an artificial reversal. The kernel consists precisely of elements fixing every point, so

The rotational tetrahedral group acts transitively on its four vertices. A rotation fixing a vertex must preserve its opposite face and rotate its three vertices cyclically; its stabilizer has order three. The orbit-stabilizer theorem gives $|G|=4\cdot 3=12$. The rotations are the identity, two nontrivial rotations about each of the four vertex-to-opposite-face axes, and a half-turn about each of the three opposite-edge-midpoint axes. Therefore the counts by element order areA rotation fixing all vertices fixes three linearly independent vertex vectors and is the identity. The vertex permutation action is therefore faithful. The listed rotations induce the identity, 3-cycles and double transpositions, all even permutations. Its injective image has order 12, the full order of $A_4$. Hence $\boxed{G\cong A_4}$.