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 Post subject: Symetry group of Tetrahedron Posted: Sat Nov 03, 2018 6:16 pm

Joined: Wed Nov 15, 2017 12:37 pm
Posts: 20
Exercise :

Find the Symmetry Group of :

- The Tetrahedron
- The Cube
- The sphere with radius $r=1$ on $\mathbb R^3$

**Discussion :**

I know that the answer to the first question is $S_4$ through lectures but I do not know how to prove it and most questions online revolve around extended questions of these and not these. Sorry for not providing an attempt, but I am really trying to grasp on how to solve them and work on these questions. I would appreciate any help, hint or thorough solution/discussion to help me get how I work on these.

Top   Post subject: Re: Symetry group of Tetrahedron Posted: Tue Nov 06, 2018 4:48 pm

Joined: Sat Nov 14, 2015 6:32 am
Posts: 159
Location: Melbourne, Australia
1. The tetrahedron is a regular solid with $4$ vertices and $4$ triangular faces. The symmetry group is the alternating group $\mathcal{A}_4$.
2. The symmetry group of a cube is isomorphic to $\mathcal{S}_4$ , the permutation group on 4 elements. If we number the vertices of the cube from $1$ to $4$ and where opposite vertices are given the same number, the permutation corresponding to a symmetry can be read out from one of the faces. If we e.g. let the top face be numbered $1, 2, 3, 4$ clockwise, then perform a rotation around the $1$-diagonal, we see that the numbers of the top face has changed to $1, 3, 4, 2$, so this symmetry corresponds to the permutation $(234)$. Proceeding in this way we can classify the $24$ elements of the symmetry group by their order.
3. The symmetry group of the sphere, including reflections, is known as the orthogonal group $O(3)$. The symmetry group with rotations only is the special orthogonal group $SO(3)$. Both have infinite order, which follow from that rotations by any real angle and reflections about any plane are part of the symmetry group. The rotations form $SO(3)$, and the rotations and reflections form $O(3)$, without any missing symmetries.

Hope I helped!

_________________
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$

Top   Post subject: Re: Symetry group of Tetrahedron Posted: Fri Nov 16, 2018 11:33 am
 Team Member Joined: Mon Nov 09, 2015 1:36 am
Posts: 460
Location: Ioannina, Greece
We give a solution in the case of the tetrahedron:

Definition: A symmetry of a (regular) tetrahedron $S$ is a linear transformation $T:\mathbb{R}^3\longrightarrow\mathbb{R}^3$ with orthogonal matrix which also leaves tetrahedron $S$ unchanged(*), i.e. $T(S)=S$.

• Definition: An axis of symmetry of a (regular) tetrahedron is an axis which leaves the tetrahedron unchanged after a rotation of the tetrahedron around that axis, by a certain angle.
Attachment: Lets find the axes of symmetry: The axes $R_1$, $R_2$, $R_3$ and $R_4$ which passes through the vertices $1$, $2$, $3$ and $4$, respectively, and the centroids of the respective opposite faces of the tetrahedron, are axes of symmetry, since they leave it unchanged if it is rotated (clockwise) around one of them by an $\frac{2\pi}{3}$-angle, or by an $\frac{4\pi}{3}$-angle.
The rotation around the $R_1$-axis by an $\frac{2\pi}{3}$-angle can be represented (described) by the permutation $\rho_1=({2\,3\,4})$. The permutation $\rho_1^2=({2\,4\,3})$ represents the rotation around the $R_1$-axis by an $\frac{4\pi}{3}$-angle.
Similarly we have \begin{align*}
\rho_2=({1\,3\,4})\,, \quad \rho_2^2=({1\,4\,3})\,,\quad \rho_2^3=id\\
\rho_3=({1\,2\,4})\,, \quad \rho_3^2=({1\,4\,2})\,,\quad \rho_3^3=id\\
\rho_4=({1\,2\,3})\,, \quad \rho_4^2=({1\,3\,2})\,,\quad \rho_4^3=id\,.
\end{align*} The axes $T_1$, $T_2$ and $T_3$ which passes from the midpoints of the respective opposite edges of the tetrahedron, are axes of symmetry, since they leave it unchanged if it is rotated around one of them by an $\pi$-angle.
We have the permutations \begin{align*}
&\tau_1=({1\,2})\,({3\,4})\,,\quad \tau_2=({1\,4})\,({2\,3})\\
&\tau_3=({1\,3})\,({2\,4})\,,
\end{align*}which represent (describe) the rotation, by an $\pi$-angle, around the $T_1$-axis, the $T_2$-axis and the $T_3$-axis, respectively.
Also holds: \begin{align*}
\rho_1\,\rho_2=\tau_2\,, \quad \rho_1\,\rho_3=\rho_2\,, \quad \rho_1\,\rho_4=\tau_3\,,\\
\rho_2\,\rho_3=\tau_1\,, \quad \rho_2\,\rho_4=\rho_3\,, \quad \rho_3\,\rho_4=\tau_2\,,\\
\tau_1\,\tau_2=\tau_3\,, \quad \tau_2\,\tau_3=\tau_1\,, \quad \tau_1\,\tau_3=\tau_2\,.
\end{align*}The set ${\cal{T}}=\big\langle{\,\rho_{i}\,,\,\tau_{j} \; | \; i=1,2,3,4, \ j=1,2,3}\big\rangle$ equipped with composition of permutations is a group isomorphic to alternating group ${\cal{A}}_4$.
• Definition: A plane of symmetry of a (regular) tetrahedron is a plane which leaves the tetrahedron unchanged after a reflection of the tetrahedron regarding that plane.
Attachment: Similarly the reflections regarding a plane of symmetry can be represented as permutations.

In total all the above permutations form a group isomorphic to symmetric group ${\cal{S}}_4$.

(*) Leaves the tetrahedron unchanged, in the sence that we have the same picture of it, without distinguishing different faces, different edges or different vertices at first.

_________________
Grigorios Kostakos

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