I am currently working on elliptic curves (over finite fields), and I asked myself the following question.
Let $\phi:E \longrightarrow E'$ be an isogeny and $\hat \phi:E' \longrightarrow E$ the corresponding dual isogeny. Is it true that $\ker(\phi)\cong\ker(\hat\phi)$ ?
I don't see any obvious reason why this should be true, and in fact it feels like it's not but I can't prove it. I just need a yes/no answer, but a proof is also welcome.
Furthermore we may suppose that we're working over a finite field $\mathbb{F}_{p^2}$, and the order of the isogeny is a prime power $l^a$ coprime to the characteristic of the field. Also the kernel of $\phi$ is cyclic and $E(\mathbb{F}_{p^2})\cong E'(\mathbb{F}_{p^2})$, but both $E(\mathbb{F}_{p^2})$ and $E(\mathbb{F}_{p^2})$ contain the $l^a$ torsion so it doesn't exclude the possibility that $\ker(\hat\phi)\cong C_{l^{a/2}}\times C_{l^{a/2}}$ or something else non-cyclic.
Let $E$ be an ordinary elliptic curve over $\Bbb F_p$. Let $E'=E$ and $\phi$ be the Frobenius map. Then $\ker\phi$ is trivial. But $\hat\phi\circ\phi=[p]$ which has kernel of order $p$ (since $E$ is ordinary) so $\ker\hat\phi$ has order $p$.
For isogenies of degree coprime to the characteristic, the kernels are isomorphic.