 Solution

Put $r=g_1$ and $s=g_2$. Direct matrix multiplication givesMoving every $s$ to the right shows that every word in these generators of a group has the form $r^a{s}^b$, with $a\in \{0,1,2\}$ and $b\in \{0,1\}$. The three $r^a$ are distinct diagonal matrices, and the three $r^{a}s$ are distinct off-diagonal matrices. HenceSince $srs=r^2\ne r$, this is a non-abelian group. Its group presentation identifies it with the dihedral group of an equilateral triangle, equivalently the symmetric group $S_3$.

The conjugacy classes areIndeed $s$ interchanges $r$ and $r^2$ by conjugation, while $r^{a}s{r}^{-a}=r^{2a}s$ runs through all three reflections. No reflection is conjugate to a rotation, since conjugation preserves order of a group element.

There are three inequivalent complex irreducible representations, one for each conjugacy class. The sum of squares of irreducible representation dimensions gives $d_1^2+d_2^2+d_3^2=6$. One representation is trivial, and the only possible positive integer dimensions are