= Solution
{c}
For any integer $n\geq1$, let $R=R_{2\pi/n}$ in the <stabilizer subgroup> from <7/b/i>. Since $R_\theta R_\phi=R_{\theta+\phi}$ and $R_\theta=I$ exactly when $\theta\in2\pi\mathbb Z$, this <rotation matrix> has order exactly $n$. Therefore
$$
\boxed{\langle R_{2\pi/n}\rangle\cong C_n}
$$
is a <subgroup> of $G$ of order $n$, including the trivial case $n=1$.
For infinitely many non-Abelian examples, let $n\geq3$, retain $R=R_{2\pi/n}$, and take the half-turn
$$
J=\operatorname{diag}(-1,1,-1)\in G.
$$
Direct multiplication gives $J^2=I$ and $JRJ=R^{-1}$. Every word in $R,J$ therefore reduces to $R^k$ or $R^kJ$, $0\leq k<n$. The $R^k$ are distinct, as are the $R^kJ$, and no element from the first family equals one from the second: they send $N$ to $N$ and $-N$, respectively. Hence $\langle R,J\rangle$ has exactly $2n$ elements and is a <dihedral group>.
It is non-Abelian because $R\ne R^{-1}$ for $n\geq3$, so $JR\ne RJ$. Different values of $n$ give different <group orders>, and therefore different <subgroups>. Thus \b[$G$ has infinitely many non-Abelian subgroups], already among its finite <subgroups>.
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