Let $f: A \rightarrow B$ be a morphism of abelian varieties defined over a finite field $k$. Let $G$ be a finite group of $A$ and $\pi:A\rightarrow A/G$ the quotient morphism. 


Looking at just the group structure, it is enough to have that $G\subset \ker f$ to ensure the existence of a group morphism $g$ such that $f=g\circ \pi$.


When is this $g$ in fact a morphism of varieties?



morphism of abelian variety
This is Theorem 4 on page 73 of Mumford's *Abelian Varieties*, where you will also find a proof. Here's the statement: Let $X$ be an abeian variety. There is a 1-1 correspondence between the two sets of objects:


(a) finite subgroups $K\subset X$


(b) separable isogenies $f:X\to Y$, where two isogenies $f\_1:X\to Y\_1$
and $f\_2:X\to Y\_2$ are considered equal if there is an isomorphism $h:Y\_1\to Y\_2$ such that $f\_2=h\circ f\_1$, which is set up by $K=\text{ker}(f)$, and $Y=X/K$.


