= Solution
Write $s=g_1$ and $r=g_2$. Direct <matrix multiplication> gives
$$
r^4=s^2=I_3,\qquad srs=r^{-1}.
$$
Every product reduces to $r^ks^\ell$, with $0\leq k<4$ and $0\leq\ell<2$. These eight <matrices> are distinct and are precisely the listed elements:
$$
g_3=r^2,\quad g_4=sr=r^{-1}s,\quad g_5=rs,\quad
g_6=srs=r^{-1},\quad g_7=r^2s.
$$
This is the <dihedral group> of <group order> eight. Its element $r^2$ commutes with both generators. Conjugation by $s$ interchanges $r$ and $r^{-1}$, while conjugation by $r$ sends $s$ to $r^2s$. It also interchanges $rs$ and $r^{-1}s$. The resulting <conjugacy classes> are
$$
\boxed{\{e\},\quad\{r^2\},\quad\{s,r^2s\},\quad
\{r,r^{-1}\},\quad\{sr,rs\}.}
$$
These are the stated $C_0,\ldots,C_4$ in order. To see that the last two reflection classes do not merge, conjugation changes the exponent in $r^ks$ by an even integer or changes its sign, preserving its parity. The five sets exhaust all eight elements.
Both generators are real <orthogonal matrices> with <determinant> $+1$. Hence every element obeys $g^\dagger g=I_3$ and $\det g=1$. The normal-form calculation establishes closure and inverses, so
$$
\boxed{G\leq SO(3)\leq SU(3).}
$$
In particular the embedding in the <special unitary group> uses the given real <matrices> viewed as complex ones.
Back to article page
