Divisibility of a divisor Let $X$ be a smooth complex projective curve and $f \colon X \to Y$ an étale Galois cover, whose Galois group $G$ is finite and of order $r$. For any $g \in G$, define $$\Delta_g = \{(x, \, g \cdot x) \; | \; x \in X \} \subset X \times X,$$ so that every $\Delta_g$ is a smooth divisor isomorphic to the diagonal $\Delta=\Delta_1$. 
The fact that $f$ is étale implies $\Delta_g \cap \Delta_h = \emptyset$ if $g \neq h$, so that the reducible divisor $$D=\sum_{g \in G} \Delta_g$$ is smooth (note that $D$ is the fibre product $X \times_Y X$).

Question. Is $D$ a $r$-divisible element in $\mathrm{Pic}(X \times X)$? 

 A: The answer is no. Before I explain why, let me reduce the problem to a problem in algebraic topology. Consider the natural map $Pic(X\times X)\stackrel{cycle}{\to} H_2(X\times X)\to End(H_1(X))$ where the second map sends a cycle $\sigma$ do the endomorphism $y\mapsto (\pi_2)_*(\pi_1^*y \cap \sigma)$. 
For the divisor $D$ as in the question, the associated endomorphism is the map $f^*f_*=\sum_{g\in G} g: H_1(X)\to H_1(X)$. Hence, it is enough to find an example of a curve $X$ with an algebraic action of a group $G$ such that the sum of the elements of $G$ acts on $H_1$ by an endomorphism which is not divisible by $|G|$. Since every finite cover of an algebraic curve is algebraizable, this reduces the problem to a purely topological question about orientable surfaces. 
Consider the example of a surface $M_3$ of genus 3. Let $C_2$ acts on it by rotation of $180^\circ$ around the middle hole (i.e. draw a torus with two handles and consider a rotation of the torus that interchange the handles). Then, as a representation of $C_2$, the first homology factors as $\mathbb{Z}^2 \oplus \mathbb{Z}[C_2]^2$, the first factors corresponds to the middle torus and the second to the two interchanged handles. Hence, it is enough to show that summation over the elements of $C_2$ on this module give rise to an endomorphism which is not divisible by $2$. But this is true already for $\mathbb{Z}[C_2]$, since the matrix 
\begin{pmatrix}
1 & 1 \\
1 & 1 
\end{pmatrix}
is not a multiple of $2$.
