Let $G (\neq 1)$ be a finite group acting faithfully (preserving the orientation) on a compact Riemann surface $X$ of genus $g \geq 2$. This induces an action of $G$ on the space ${\mathcal{H}}^1(X)$ of holomorphic differentials. Let $\chi$ denote the character afforded by this action.
It follows from the Lefschetz fixed point formula :
$$ \chi(g) + {\overline{\chi(g)}} = 2 - |{\mathrm{Fix}}_X(g)| $$
(where ${\mathrm{Fix}}_X(g)$ denote the set of points of $X$ fixed by $g \in G, g \neq 1$) that $\chi$ must be faithful.
Now it is known that existence of irreducible faithful characters pose a severe restriction on a group. For example we must have $Z(G)$ cyclic.
Now start with a non-trivial finite group $G$ with $Z(G)$ not cyclic and let $G$ acts on $X$ (for some genus $g \geq 2$). Then the character $\chi$ obtained from this must be reducible. However, this also means that its irreducible components acts faithfully on ${\mathcal{H}}^1(X)$.
Isn't this a contradiction? Or am I missing something very trivial?
A certain overlook to the question posed above, my apologies:
Once you break the character into irreducible components of ${\mathcal{H}}^1(X)$, the submodules are not necessarily coming naturally coming from a Riemann surface.
So, there is no contradiction.