By the dimensionality theorem, $$\sum_i d_i^2 = |G|,$$ where $d_i$ is the dimension of the $i$th irreducible representation, we can infer that the dihedral group $D_3$ has two one dimensional irreps and a single two dimensional one.
$D_3$ is the symmetry group of the triangle and a presentation of $D_3$ is $\left\{e, g, g^2, h, gh, g^2h \right\}$ where $g$ is a rotation by $\pi/3$ and $h$ is a reflection through a vertex of the triangle.
I have read in some notes that a second one dimensional irrep is given by declaring that $\rho(e) = \rho(g) = \rho(g^2) = 1$ and $\rho(h) = \rho(gh) = \rho(g^2 h) = -1$. Why is this a genuine representation of $D_3$?
See link here :http://mathworld.wolfram.com/DihedralGroupD3.html It argues based on the orthogonality of the rows on the table they provided there. They say given the six $1's$ of the trivial representation by orthogonality the other one dimensional irrep must have three $+1$s and three $-1$s. Why is this a plausible argument? What stops you from choosing other numbers?
Also, is there an error in their matrix representation for element $D$ there at the bottom?
Thanks.