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.]