In our group theory course, we are learning about stabilizers, applied to solid geometric bodies. I do not fully understand what stabilizers are for solid bodies, how you can find them and why the order of the elements is so important.
An example of a question was: Color a tetrahedron in such a way that its stabilizer is $V_{4}$ using 4 colors.
In the explanation, I saw someone speaking about the fact that all elements in $V_{4}$ have order 2 (Excluding $e$), and that this was an important fact. Can anyone explain to me what the connection is between the order of the elements in a group and constructing the stabilizer for a geometric solid?
Let $G$ be a group acting on a set $S$, and let $s$ be an element of this set. Then the stabilizer of an element $s$ is a subgroup of $G$ consisting of all elements $g \in G$ for which $g \cdot s = s$. This means that when the elements of the stabilizer act on $s$, it remains unchanged.
In a problem where a colored tetrahedron is considered, the stabilizer of a colored tetrahedron is a subset of all transformations (rotations and reflections) that leave the colored tetrahedron unchanged.
The order of an element $g$ in the group $G$ is the smallest positive integer $n$ such that $g^n = e$, where $e$ is the neutral element of the group. In the context of the tetrahedron transformation group, the element order corresponds to the number of transformations that must be performed to return to the original state.
The relationship between element orders and stabilizer is that all stabilizer elements have a specific order that reflects exactly how they affect the object. In the context of your tetrahedron example, it says that all elements in $V_4$ are of order $2$ (excluding $e$). This means that every transformation in $V_4$ (except the neutral one) changes the state of the tetrahedron, and you need to repeat it twice to return to the original state.