Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics I / 7 / b

For $X\in {\mathrm{SO}}_3(\mathbb{R})$, the defining identity $X^{T}X=I$ givesThus an orthogonal transformation preserves the Euclidean norm and sends the unit sphere to itself. Restricting the given group action to this invariant set preserves both action axioms.

If $XN=N$, the first column of $X$ is $N=e_1$. The other columns are unit vectors perpendicular to it, so the first row is also $(1,0,0)$. Hence $X=\mathrm{diag}(1,B)$ with $B^{T}B=I_2$ and $\det B=1$. Conversely, every such matrix lies in $G$ and fixes $N$. Since every element of the special orthogonal group in two dimensions has the planar rotation matrix form, the stabilizer subgroup isGeometrically these are exactly the rotations in three dimensions about the first coordinate axis.

Let $v\in S$. Choose a unit vector $u$ perpendicular to $v$, and put $w=v\times u$. The cross product identities show that $(v,u,w)$ is an orthonormal basis with positive orientation: $u\times (v\times u)=v$, soThus the matrix $A_v$ with these columns belongs to the special orthogonal group $G$ and satisfies $A_{v}N=v$. For any $v,z\in S$, the element $A_z{A}_v^{-1}$ sends $v$ to $z$. Therefore the group action on the unit sphere is transitive.

For any integer $n\ge 1$, let $R=R_{2\pi /n}$ in the stabilizer subgroup from i. Since $R_{\theta }{R}_{\varphi }=R_{\theta +\varphi }$ and $R_{\theta }=I$ exactly when $\theta \in 2\pi \mathbb{Z}$, this rotation matrix has order exactly $n$. Thereforeis a subgroup of $G$ of order $n$, including the trivial case $n=1$.

For infinitely many non-Abelian examples, let $n\ge 3$, retain $R=R_{2\pi /n}$, and take the half-turnDirect multiplication gives $J^2=I$ and $JRJ=R^{-1}$. Every word in $R,J$ therefore reduces to $R^k$ or $R^{k}J$, $0\le k<n$. The $R^k$ are distinct, as are the $R^{k}J$, and no element from the first family equals one from the second: they send $N$ to $N$ and $-N$, respectively. Hence $⟨R,J⟩$ has exactly $2n$ elements and is a dihedral group.

It is non-Abelian because $R\ne R^{-1}$ for $n\ge 3$, so $JR\ne RJ$. Different values of $n$ give different group orders, and therefore different subgroups. Thus $G$ has infinitely many non-Abelian subgroups, already among its finite subgroups.