Exercise 1.7.20 (Group of rigid motions of a regular tetrahedron)

Question

Show that the group of rigid motions of a tetrahedron is isomorphic to a subgroup of $S_4$.

Richard Ganaye · Accepted Answer

\begin{proof} 
Let $(G,\circ)$ be the group of rigid motions of a regular tetrahedron $T = \{A,B,C,D\}$. Then $G$ acts on $T$ by $g\cdot M = g(M) \ (g \in G, M \in T)$.

The permutation representation $\varphi$ is defined by 
$$
\varphi
\left\{
\begin{array}{ccl}
G & 	o &S_T\
g & \mapsto &\varphi_g,
\end{array}
ight.
\qquad 	ext{where} \qquad \varphi_g
\left\{
\begin{array}{ccl}
T & 	o &T\
M & \mapsto &\varphi_g(M) = g\cdot M = g(M),
\end{array}
ight.
$$
in other words
$$
\varphi
\left\{
\begin{array}{ccl}
G & 	o &S_T\
g & \mapsto &\varphi_g = \begin{pmatrix} A & B & C & D\g(A)& g(B) & g(C) & g(D) \end{pmatrix}.
\end{array}
ight.
$$
The action of $G$ on $T$ is faithful: if $g \in \ker(\varphi)$, then $g(A) = A,\ g(B) = B,\ g(C) = C,\ g(D) = D$, where $A,B,C,D$ are not coplanar. Hence $g = e$ (this is true for every affine transformation).

Since $\varphi$ is injective, $G \simeq \varphi(G)$, where $\varphi(G)$ is a subgroup of $S_T$. Moreover, since $|T| = 4$, $S_T \simeq S_4$, so $G$ is isomorphic to a subgroup of $S_4$.
\end{proof}
Note: we can say more (see the solution of Exercise 1.2.9). The full group of isometries of a regular tetrahedron is isomorphic to $S_4$, and the subgroup of rotations (rigid motions) is a subgroup of cardinality $12$, which is isomorphic to the alternating group $ A_4$.

Exercise 1.7.20 (Group of rigid motions of a regular tetrahedron)

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Exercise 1.7.20 (Group of rigid motions of a regular tetrahedron)

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