I am having some problems reading character table. I will describe what I understand and thought to be right.

Consider the $$\pi$$ MOs formed from overlap of p-orbitals in benzene. These two degenerate $$\pi$$ MOs below have $$E_{1g}$$ symmetry: I don't understand why that is the case. Based on what I have read, if a symmetry operation $$g$$ maps an orbital to itself, e.g. then $$\chi'(g) = 1$$.

Elif $$g$$ maps the orbital to negative itself (invert the sign of orbital), then $$\chi'(g) = -1$$.

Else, $$\chi'(g) = 0$$.

For two degenerate orbitals as above, final $$\chi(g)$$ is the sum of $$\chi'(g)$$ for two orbitals, where $$\chi(g)$$ is character of a symmetry operation $$g$$, that is recorded in the character table.

Applying this, consider the $$C_6$$ of benzene, Not mapped to $$\pm$$ itself, so $$\chi'(g) = 0$$. Also not mapped to $$\pm$$ itself, so $$\chi'(g) = 0$$.

So $$\chi(g) = 0 + 0 = 0$$.

But if we look at the character table for $$D_{6h}$$, for the symmetry species $$E_{1g}$$, $$\chi(C_6) = 1 \neq 0$$.

Enlighten me, please, I must have understood something terribly wrong.