The symmetries of a tetrahedron are isomorphic to $A_4$

One of my favorite results from Math 412 is the following:

Claim.  The symmetry group of a regular tetrahedron is isomorphic to the alternating group $A_4$.

Proof sketch.   Given a group action of $G$ on $X$, we can fix $g\in G$ and denote $x\mapsto g\cdot x$ by $\mathrm{ad}(g):X\to X$. Using basic properties of group actions, we can show that $\mathrm{ad}$ is a homomorphism from $G$ to the group of bijections on $X$, denoted $\mathrm{Bij}(X)$.

Importantly, we can show $\mathrm{ad}$ is injective iff the action is faithful. This follows because

Now we consider the canonical action of the symmetries of a tetrahedron $\mathrm{Sym}(T)$ on the faces of a tetrahedron $X = \{f_1, f_2, f_3, f_4\}$.

Clearly the action is faithful, so we know immediately that $\mathrm{ad}:\mathrm{Sym}(T)\to\mathrm{Bij}(X)\cong S_4$ is injective. This implies $\mathrm{im}\ \mathrm{ad}\cong \mathrm{Sym}(T)$.

But $\mathrm{im}\ \mathrm{ad}$ contains the set of 3-cycles (corresponding to a cycle of any three faces about their shared vertex), which is a generating set for $A_4$. So $A_4\subseteq\mathrm{im}\ \mathrm{ad}$, implying that there exists a surjection from $\mathrm{Sym}(T)$ to $A_4$.

Finally, since $|\mathrm{Sym}(T)| = 12 = |A_4|$, we conclude $\mathrm{Sym}(T)\cong A_4$.