 Solution

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.