Linear algebra (Osnabrück 2024-2025)/Part II/Lecture 51

For two opposite semiaxes ${\displaystyle {}H}$ and ${\displaystyle {}-H}$, we have ${\displaystyle {}G_{H}=G_{-H}}$. For two semiaxes ${\displaystyle {}H_{1}}$ and ${\displaystyle {}H_{2}}$ that do not belong to the same axis (in particular, they are different) we have the relation ${\displaystyle {}G_{H_{1}}\cap G_{H_{2}}=\operatorname {Id} }$, because an isometry with two rotation axes is the identity. Since ${\displaystyle {}G}$ is the union of all ${\displaystyle {}G_{H}}$, ${\displaystyle {}H\in {\mathfrak {H}}(G)}$, we have a union

${\displaystyle {}G\setminus \{\operatorname {Id} \}=\bigcup _{H\in {\mathfrak {H}}(G)}{\left(G_{H}\setminus \{\operatorname {Id} \}\right)}\,,}$

where every group element ${\displaystyle {}g\neq \operatorname {Id} }$ appears twice on the right-hand side. Therefore,

${\displaystyle {}2(n-1)=\sum _{H\in {\mathfrak {H}}(G)}{\left(\operatorname {ord} \,(G_{H})-1\right)}\,.}$

The classes ${\displaystyle {}K_{i}}$ contain ${\displaystyle {}n/n_{i}}$ elements. Hence,

${\displaystyle {}2(n-1)=\sum _{H\in {\mathfrak {H}}(G)}{\left(\operatorname {ord} \,(G_{H})-1\right)}=\sum _{i=1}^{m}\sum _{H\in K_{i}}(n_{i}-1)=\sum _{i=1}^{m}{\frac {n}{n_{i}}}(n_{i}-1)\,.}$

Dividing by ${\displaystyle {}n}$ yields the claim.

For ${\displaystyle {}m=1}$, we muss have ${\displaystyle {}n_{1}={\frac {n}{-n+2}}}$ but this does not have a solution for ${\displaystyle {}n\geq 2}$.  For ${\displaystyle {}m=3}$, the condition becomes

${\displaystyle {}1+{\frac {2}{n}}={\frac {1}{n_{1}}}+{\frac {1}{n_{2}}}+{\frac {1}{n_{3}}}\,}$

with ${\displaystyle {}n_{1}\leq n_{2}\leq n_{3}}$. The left-hand side is ${\displaystyle {}>1}$. Because of ${\displaystyle {}n_{i}\geq 2}$, at least one them is ${\displaystyle {}n_{i}=2}$. So suppose ${\displaystyle {}n_{1}=2}$. For ${\displaystyle {}n_{2}=4}$, the right-hand side is again ${\displaystyle {}\leq 1}$, so that ${\displaystyle {}n_{2}=3}$ holds. The value ${\displaystyle {}n_{3}=3}$ leads to the solution ${\displaystyle {}n=12}$, the value ${\displaystyle {}n_{3}=4}$ leads to the solution ${\displaystyle {}n=24}$, and the value ${\displaystyle {}n_{3}=5}$ leads to the solution ${\displaystyle {}n=60}$. For ${\displaystyle {}n_{3}\geq 6}$, the right-hand side is again ${\displaystyle {}\leq 1}$, so that there is no further solution.
 For ${\displaystyle {}m\geq 4}$, the condition has the form

${\displaystyle {}m-2+{\frac {2}{n}}={\frac {1}{n_{1}}}+{\frac {1}{n_{2}}}+{\frac {1}{n_{3}}}+{\frac {1}{n_{4}}}+\cdots +{\frac {1}{n_{m}}}\,.}$

This has no solution, because the right-hand side is ${\displaystyle {}\leq m-2}$, since the first four summands yield at most ${\displaystyle {}2}$, and the further summands can be bounded by ${\displaystyle {}m-4}$.

Because of Lemma 51.1 and Lemma 51.2 , we must have ${\displaystyle {}n=n_{1}=n_{2}}$, and every class of semiaxes contains only one semiaxis. Therefore, there does only exist one rotation axis, and this symmetry group is, according to Lemma 34.1 , isomorphis to a symmetry group in the plane orthogonal to this axis. Due to Theorem 50.4 , this group is isomorphic to the cyclic group of order ${\displaystyle {}n}$.

There are three classes of semiaxes, say ${\displaystyle {}K_{1}}$, ${\displaystyle {}K_{2}}$, ${\displaystyle {}K_{3}}$. Two of them have order ${\displaystyle {}2}$ (each class with ${\displaystyle {}k}$ semiaxes), and one has order ${\displaystyle {}k}$ and two semiaxes (the numbers of the semiaxes follow with ${\displaystyle {}n_{1}=n_{2}=2}$ from Lemma 51.2 ). For ${\displaystyle {}k\geq 3}$, the two semiaxes from the third class form a line, because the opposite semiaxis has the same ordwe. For ${\displaystyle {}k=2}$, every semiaxis is equivalent to its opposite semiaxis. We denote the axis of ${\displaystyle {}K_{3}}$ by ${\displaystyle {}A_{3}}$. Every group element with another rotation axis must transform the two semiaxes from ${\displaystyle {}K_{3}}$ into each other, so that all other axes are orthogonal to ${\displaystyle {}A_{3}}$. Let ${\displaystyle {}g}$ denote a generating rotation around ${\displaystyle {}A_{3}}$. For a semiaxis ${\displaystyle {}H_{1}}$ from ${\displaystyle {}K_{1}}$, the

${\displaystyle g^{i}(H_{1}),i=0,\ldots ,k-1,}$

are exactly all semiaxes from ${\displaystyle {}K_{1}}$. These (or rather the points on these axes of norm ${\displaystyle {}1}$) form a regular ${\displaystyle {}k}$-gon in the plane orthogonal to ${\displaystyle {}A_{3}}$. The same holds for ${\displaystyle {}g^{i}(H_{2})}$ with ${\displaystyle {}H_{2}\in K_{2}}$. Every rotation about ${\displaystyle {}180}$ degree around an axis from ${\displaystyle {}K_{1}}$ permutes the semiaxes from ${\displaystyle {}K_{2}}$. Therefore, the semiaxes from ${\displaystyle {}K_{2}}$ yield a "bisection“ of the ${\displaystyle {}k}$-gon. Hence, the group is the (improper) symmetry group of a regular ${\displaystyle {}k}$-gon, or the proper symmetry group of a bipyramid over a regular ${\displaystyle {}k}$-gon, that is, it is a dihedral group ${\displaystyle {}D_{k}}$.

Due to the definition of a class of semiaxes, for ${\displaystyle {}H\in K}$ we also have ${\displaystyle {}g(H)\in K}$ for all ${\displaystyle {}g\in G}$. Therefore, the mapping ${\displaystyle {}\sigma }$ is well-defined. Let ${\displaystyle {}f,g\in G}$. Then we have

${\displaystyle {}\sigma _{g\circ f}(H)=(g\circ f)(H)=g(f(H))=g(\sigma _{f}(H))=\sigma _{g}(\sigma _{f}(H))=(\sigma _{g}\circ \sigma _{f})(H)\,.}$

Due to the condition, there are three classes of semiaxes of order ${\displaystyle {}2,3}$ and ${\displaystyle {}3}$; the number of elements in these classes is ${\displaystyle {}6,4,}$ and ${\displaystyle {}4}$. We consider a class of semiaxes ${\displaystyle {}K}$ of order ${\displaystyle {}3}$, with the four equivalent semiaxes, and the corresponding group homomorphism

${\displaystyle G\longrightarrow \operatorname {Perm} \,(K),g\longmapsto \sigma _{g}.}$

Let ${\displaystyle {}g\in G}$ be a rotation by ${\displaystyle {}120}$ degree around a semiaxis ${\displaystyle {}H\in K}$. It fixes ${\displaystyle {}H}$, and it gives a permutation of the three other semiaxes in the class. This permutation can not be the identity; otherwise, ${\displaystyle {}g}$ would fix two axes and then ${\displaystyle {}g}$ were the identity. Since ${\displaystyle {}g}$ has order ${\displaystyle {}3}$, this permutation is a ${\displaystyle {}3}$-cycle. In particular, the four semiaxes belong to different axes, and the rotation ${\displaystyle {}g^{2}}$ induces the other ${\displaystyle {}3}$-cycle. Since we can perform this consideration with every semiaxis from ${\displaystyle {}K}$, we can deduce that ${\displaystyle {}G}$ induces every ${\displaystyle {}3}$-cycle of the permutation group of the four semiaxes. This means that the image of the group homomorphism is exactly the alternating group ${\displaystyle {}A_{4}}$; hence, ${\displaystyle {}G\cong A_{4}}$. This group is isomorphic to the tetrahedral group, see Exercise 50.25 .

We consider the class of semiaxes ${\displaystyle {}K_{2}}$ of order ${\displaystyle {}3}$, which contains ${\displaystyle {}8}$ semiaxes equivalent to each other. For such a semiaxis ${\displaystyle {}H}$, the opposite semiaxis belongs also to a class of semiaxes with the same order. Therefore, ${\displaystyle {}-H}$ belongs to ${\displaystyle {}K_{2}}$, so that semiaxes of ${\displaystyle {}K_{2}}$ belong to four axes. We denote the set of these axes by ${\displaystyle {}{\mathfrak {A}}}$. We consider the group homomorphism

${\displaystyle G\longrightarrow \operatorname {Perm} ({\mathfrak {A}}),g\longmapsto {\left(\sigma _{g}:A\mapsto g(A)\right)}.}$

Here, we look at the action on the axes, not at the action on the semiaxes. We claim that not three of these four axes lie in a plane. For if ${\displaystyle {}A_{1},A_{2},A_{3}\subseteq E}$, then a rotation about ${\displaystyle {}120}$ degree around ${\displaystyle {}A_{1}}$, call it ${\displaystyle {}f}$, would produce the equivalent axes ${\displaystyle {}f(A_{2})}$ and ${\displaystyle {}f(A_{3})}$, but these can not lie in the plane ${\displaystyle {}E}$, and they can not both equal ${\displaystyle {}A_{4}}$. Suppose that the element ${\displaystyle {}g\in G}$ has the property, that ${\displaystyle {}\sigma _{g}}$ is the identity. This means that all lines ${\displaystyle {}A\in {\mathfrak {A}}}$ are mapped to themselves. Due to Exercise 51.15 , ${\displaystyle {}g}$ is the identity. Therefore, according to the kernel criterion, the group homomorphism is injective. Hence, we have an isomorphism. In particular, we can apply this reasoning for the cube group; this gives the isomorphism between the cube group and the permutation group ${\displaystyle {}S_{4}}$.