Do isogenies between AVs over finite fields separate finite subgroups? Waterhouse in his thesis (Abelian varieties over finite fields, Ann. scient. \'Ec. Norm. Sup., t. 2, 1969, p 521-560) seems to use without comments the following fact:
Let $k$ be a finite field, and let $A$, $B$ be two abelian varieties over $k$ that are $k$-isogenous. Consider the set $I(A,B)$ of all the $k$-isogenies from $A$ to $B$. Then for any finite, non-trivial, subgroup $H$ of $A$, there is $\varphi\in I(A,B)$ that does not vanish identically on $H$.
This fact is used implicitly in lines 8-9 page 533, right after the definition of kernel ideal.
Does anyone have an argument to see it? Also, I do not know what role the assumption that the base field $k$ is finite should play. Thanks.
[EDIT: Actually what Waterhouse uses is that, under the assumption of the second paragraph  above, there is a ${\it morphism}$ $\varphi:A\rightarrow B$ that does not vanish identically on $H$]
[EDIT 2: I report here Waterhouse's statement. Let $A$ be an abelian variety over a finite field $k$, and let $R$ be its $k$-endomorphism ring. Let $I$ be a left ideal of $R$ that contains an isogeny of $A$. Define $H(I)$ to be the finite subgroup of $A$ given by the intersection of all ker($\varphi$), as $\varphi$ ranges in $I$.
By definition, $I$ is a kernel ideal if $I=$ { $r\in R: r\cdot H(I)=0$ }.
Here comes the line I can't verify:
"Every $I$ is contained in a kernel ideal $J$ with $H(J)=H(I)$, namely $J=$ { $r\in R: r\cdot H(I)=0$ }."
The question is "how do we know that $J$, as just defined, is a kernel ideal?" I think this question is just a reformulation of the main question I asked above.]
 A: Let me describe a natural straightforward generalization of Chris Wutrich's counterexample.
Let $B$ be a $g$-dimensional  abelian variety over a  field $k$ and assume that $End_k(B)$ is a principal ideal domain. Let $A$ be another abelian variety over $k$ that is not $k$-isomorphic to $B$ but $k$-isogenous to it. Then the group $Hom_k(A,B)$ becomes a free $End_k(B)$-module of rank 1. This means that there exists a (generator) isogeny $\lambda:A \to B$ such that every $k$-homomorphism $v: A\to B$ is a composition $u\lambda$ of $\lambda$ and a certain $u \in End_k(B)$. In particular, $ker(v)$ always contains $\ker(\lambda)$. Since $A$ and $B$ are not isomorphic over $k$, the isogeny $\lambda$ is not an isomorphism and therefore $H:=ker(\lambda)\subset A$ is nontrivial but is killed by every $k$-homomorphism from $A$ to $B$.
In order to construct explicit examples (over finite fields) pick any imaginary quadratic field $K$ of discriminant 1 amd let $O$ be the ring of integers in $K$, which is PID.  Then for a ``half" of the primes $p$ there exist a finite field $k$ of characteristic $p$ and an ordinary elliptic curve $B$ over $k$ with
$End_k(B)=End(B)=O$. Enlarging (if necessary) $k$, we may find an elliptic curve $A$ over $K$ that is not $k$-isomorphic to $A$ but $k$-isogenous to it. For example, if a prime $\ell$ is different from $p$ and  inert in $O$ then we may pick a cyclic order $\ell$ subgroup $C$ in $B(k)$ (enlarging $k$ if necessary) and put $A=B/C$. Then the cyclic order $\ell$ subgroup $H=B_{\ell}/C\subset A$ is killed by every homomorphism $A \to B$.
A: As the comments and the answer above explain, the answer to my original question is NO. For completeness, I include the (very easy) proof of the statement in the second edit of my question (whose notation will be adopted) which is a weaker statement than that I was asking in the first place.
The group of homomorphisms Hom$_k(A/H(I),A)$ has the property that the intersection of all the kernels of its elements is the trivial subgroup of $A/H(I)$. To see this, one can just observe that the ideal $I$ can be identified with a subset of Hom$_k(A/H(I),A)$, and this subset already has the property that the intersection of all the kernels of its elements is trivial.
It now immediately follows that $H(I)=H(J)$.
I would like to thank the authors of the comments and answer above for their interest.
