Symmetry group of Tetrahedron Exercise :
Find the Symmetry Group of :


*

*The Tetrahedron

*The Cube

*The sphere with radius $r=1$ on $\mathbb R^3$


Discussion :
I am having a hard time understanding and solving exercises like this on my Algebra and Applications course. For example, 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/here 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. 
 A: A completely "rigorous" proof for these are not so trivial. First, you need to know that what is really meant by symmetry group of each of these objects is really the group of isometric isomorphisms  of the object. In the most general context, if you have a metric space $(X,d)$, then the group is given by the group of bijections
$$
\phi:X\to X
$$
with the property that $d(\phi(x),\phi(y))=d(x,y)$. 
For Euclidean space, you have the following nice fact
Fact:
Any isometric isomorphism of Euclidean space is affine, invertible and its associated linear map is orthogonal
For polytopes and spheres, this implies that any isometric isomorphism actually comes from a restriction of an orthogonal matrix. Given this, we can prove the following:
Prop:
Let ${\rm Iso}(T)$ denote the symmetry group of the tetrahedron $T$. We have an isomorphism
$${\rm Iso}(T)\cong S_4\;.$$
Proof:
The vertices of the tetrahedron are all equidistant from one another and this distance is the maximum distance between any two points in the tetrahedron. Since symmetries preserve distance, any symmetry must therefore permute the vertices (of which there are four). This assignment gives a group homomorphism $g:{\rm Iso}(T)\to S_4\;.$
Any three of the four vertices of $T$ also form a basis for the ambient Euclidean space. Any permutation of the vertices therefore determines an invertible linear map on the ambient space and (since these vectors are all equidistant) this map must be orthogonal. The restriction of this map to $T$ preserves $T$ and therefore determines a symmetry. This defines a group homomorphism $f:S_4\to {\rm Iso}(T)\;.$
Using the fact that linear maps are determined by their value on a basis, it's immediate from the assignments that $fg=gf={\rm id}$, so that this association defines an isomorphism of groups ${\rm Iso}(T)\cong S_4$. 
I'll leave the other two to you, but the idea for each of these is essentially the same. 
A: Here are a few hints.
As for the cube, see this answer I gave on a similar question. The work comes in proving what I say there is true. 
With the tetrahedron, you can do something similar. If you consider the point directly in the center of the tetrahedron and draw four lines, each connecting a vertex of the tetrahedron to this central point, look at how rotations or flips move these four lines around. 
For the sphere, think of points in the sphere as unit vectors in $\mathbb{R}^3$, and a symmetry of the sphere as matrices which move these vectors around. That is, given $v\in\mathbb{R}^3$ with $\|v\|=1$, for what matrices $A$ is it true that $\|Av\|=1$?
Edit: Perhaps I should mention that you should be explicit in whether or not you’re considering orientation-reversing maps. Different ideas for computing these groups may or may not count these maps, and so there can be confusion about why different groups pop up for the same object, e.g. $A_4$ vs. $S_4$ for the tetrahedron.
A: Since you've identified the course as "Algebra and Applications" perhaps an intuitive geometric argument will be rigorous enough.
To understand the symmetries of the regular tetrahedron, label the vertices $1,2,3,4$. Then a symmetry is known when you know how it permutes the vertices, so the group of symmetries is a subset (in fact a subgroup) of $S_4$. If you stare hard enough at the tetrahedron you can see a plane of reflection that swaps $1$ and $2$ while leaving $3$ and $4$ fixed. Clearly there's such a swapping reflection for each pair of vertices (an argument from informal symmetry!). Since the $2$-cycles generate $S_4$ the group of the tetrahedron must be all of $S_4$.
If you want only the proper symmetries (no reflections) think about the rotations through a third of a circle about the altitudes of the tetrahedron. See what they generate.
For the cube, number the vertices $1$ through $8$. If you manipulate an actual cube you should be able to find the $12$ permutations in $S_8$ that capture all the proper symmetries. If you want reflections too you should find another $12$.
For the sphere, you have to go back to linear algebra and think about linear transformations that preserve length. 
