Fix an algebraically closed field $k$ and a smooth projective (geometrically) integral genus $\geq 2$ curve $C$ over $k$. Denote by $J_C$ the Jacobian variety of $C$.
Chapter VI, Proposition 11 of Serre's Algebraic groups and class fields states that
Let $\pi:Y \rightarrow X$ be an abelian covering of the curve $X$. Then there exists a smallest modulus $\mathfrak{m}$ such that $Y$ is the pull-back of an isogeny $J' \rightarrow J_{\mathfrak{m}}$ and the support of this modulus is equal to the set of points $P \in X$ which ramify in $Y$.
Of course in our case $\mathfrak{m} = 0$ and so it follows that
Unramified abelian coverings of an algebraic curve are in one to one correspondence with isogenies of its Jacobian.
For a finite etale abelian group scheme $G$, the above corollary should imply that the embedding $C \rightarrow J_C$ induces the isomorphism of etale cohomology groups $H^1(J_C, G) \xrightarrow{\sim} H^1(C,G)$. This map is defined by sending a $J_C$-torsor $Y \rightarrow J_C$ to the pull-back $Y \times _{J_C} C \rightarrow C$.
Now, as seen here, the pull-back class $[Y \times _{J_C} C] \in H^1(C,G)$ is an unramified abelian covering, but I am having trouble showing that $\varphi:Y \rightarrow J_C$ is an isogeny in the first place. We have $\mathrm{Aut}(k(Y)/k(J_C)) \cong G$ since $\varphi$ is an abelian covering but I have no idea what the (finite) kernel of this map is. I guess this kernel would have something to do with $G$, any idea how should I think about this?
