# Presentation of full symmetry group of tetrahedron

If we consider a regular $n$-gon in plane, then if $s_1,s_2$ denote the reflection symmetries of this polygon with lines of reflections very closely chosen, then we obtain the presentation of the dihedral group as $$\langle s_1,s_2: s_1^2=s_2^2=(s_1s_2)^n=1\rangle.$$ Now consider tetrahedron. It can be shown that the full symmetry group of this is $S_4$, being permutation group of vertices.

But, do anybody have a simple idea to draw pictorially the three planes of reflections $s_1,s_2,s_2$ which yield the following relations: $$\langle s_1,s_2,s_2 : s_1^2=s_2^2=s_3^2=(s_1s_3)^2=(s_1s_2)^3=(s_2s_3)^3=1\rangle$$

• The symmetry group of the tetrahedron is $A_4$, not $S_4$. Only even permutations are allowed because if you swap the positions of two vertices of a tetrahedron, you necessarily must swap the other pair as well. – Ethan Alwaise Oct 12 '16 at 4:03
• I am a bit confused by your question though because the dihedral group is the group of rigid symmetries of the $n$-gon, not just all permutations of the vertices. That's why you can visualize it with reflections. – Ethan Alwaise Oct 12 '16 at 4:07
• In short, the groups considered here are "reflection groups", i.e. (sub)groups of orthogonal group $O_n(R)$, which are generated by reflections, and these examples are the most basic examples of this subject. – p Groups Oct 12 '16 at 4:10
• @Ethan: it's correct that the group of isometries is $S_4$. The group of motions (determinant 1 isometries) is indeed $A_4$. A reflection can fix 2 vertices and swap the other 2. – YCor Oct 12 '16 at 4:20
• One way to visualize is to draw the tetrahedron on a sphere. It's made of 4 regular triangles with angles $2\pi/3$. If you draw medians in one of these triangles, you cut it into 6 triangles of angle $\pi/3$ (twice) and $\pi/2$. Reflections over the edges of these triangles are the generators of the isometry group with the presentation as you provide. – YCor Oct 12 '16 at 4:28

I think a three dimensional representation of $S_4$ will do the trick. The elements $s_1, s_2$ and $s_3$ are represented by the matrices: $$M_1 = \begin{pmatrix} 0 & -1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}, M_2 = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix}, M_3 = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$ One can verify that they satisfy the same relationships as in the question. To find the visual components one has to look at the eigenvectors with eigenvalue $-1$, being the normals to the reflecting planes, they are $$n_1= \begin{pmatrix} 1 \\ 1\\ 0 \end{pmatrix}, n_2= \begin{pmatrix} 1 \\ 0\\ -1 \end{pmatrix}, n_3= \begin{pmatrix} 1 \\ -1\\ 0 \end{pmatrix}$$