The group of symmetries of the $d$ dimensional regular simplex is isomorphic to $S_{d+1}$
($S_n$ is the symmetric group of $n$ elements = the group of permutations)
I was able to prove the above proposition for $d=3$ the following way:
- Let $G$ be the group of the symmetries of the $3$-dimensional regular simplex (regular tetrahedron).
- There is an obvious homomorphism $\varphi \colon G \to S_4$, sending a symmetry to the corresponding permutation of the vertices.
- If a symmetry fixes all the vertices, it must be the identity symmetry, so $\ker \varphi = \{ \text{id} \} \Rightarrow \varphi $ is injective.
- Every transposition of neighbours is in the image $\text{Im} \varphi$.
- Since $\varphi$ is a group homomorphism $\text{Im}\varphi$ is a subgroup of $S_4$.
- $S_4$ is generated by these transpositions $\Rightarrow \text{Im}\varphi = S_4 \Rightarrow \varphi$ is surjective.
- So $\varphi$ is a bijective homomorphism $\Rightarrow \varphi$ is an isomorphism $ \Rightarrow G \cong S_4 $
But I don't know how to generalize this proof for dimension $d$. My main problem is that I think that not every transposition lies in Im$\varphi$. Am I right? If I'm wrong about it, can someone show me that all the transpositions are in the image of the homomorphism? If I'm right about this, can someone help me to prove the theorem in dimension $d$?
Thanks in advance!
Every transposition is still in the image of $\varphi$ in the general case. If you position the simplex in $\mathbb{R}^d$ and fix two points $P_i,P_j$ out of vertices $P_1,\ldots,P_{n+1}$ of the simplex, then the hyperplane determined by all $P_k$ for $k\neq i,j$ along with the midpoint of $\overline{P_iP_j}$ is of dimension $d-1$, and the reflection across this hyperplane exchanges $P_i$ and $P_j$ while fixing all other vertices (and preserving the shape of the simplex).
Everything else in your post continues to hold in the $d$-dimensional case, so the proof generalizes.