 Solution

The Hasse theorem for elliptic curves states that for $E/{\mathbb{F}}_q$,Let $\pi$ be the Frobenius isogeny of an elliptic curve. Its fixed points are exactly $E({\mathbb{F}}_q)$, and $1-\pi$ is separable because its differential is the identity. ThereforeFor every endomorphism $\alpha$, the dual isogeny gives $\alpha \hat{\alpha }=[\mathrm{deg}\alpha ]$, and the degree satisfies the parallelogram lawConsequently, if $a=1+q-\mathrm{deg}(1-\pi )$ is the trace of $\pi$, thenThis is nonnegative for every pair of integers $m,n$. By approximating the minimizing real ratio $m/n=a/2$, the quadratic polynomial can be nonnegative for all rational ratios only if its discriminant is nonpositive. Thus $a^2\le 4q$, and the displayed point-count identity proves Hasse's theorem.

An explicit example with nonisomorphic groups occurs over ${\mathbb{F}}_{43}$. PutOn $E_1$, the point $(30,19)$ has order seven and its nonzero cyclic subgroup has $x$-coordinates $30,31,40$. Vélu formulas give the quotient coefficientsso the degree-seven quotient iswhich is $E_2$. Direct quadratic-residue counts give $\#E_1({\mathbb{F}}_{43})=\#E_2({\mathbb{F}}_{43})=49$. The points killed by seven on $E_1$ are only the displayed kernel and $O$, while $(1,8)$ and $(3,2)$ are independent seven-torsion points on $E_2$. Hence

Finally, let $p<43$ and suppose two curves are linked by an isogeny of degree seven. The isogeny and its dual induce inverse isomorphisms on every prime-to-seven primary subgroup, so nonisomorphic rational point groups would require $49$ to divide their common order. Since this order is less than $57$ by Hasse, one group would then be cyclic of order $49$ and the other would contain all of $E[7]$. The Weil pairing on rational seven-torsion forces ${\mu }_7\subset {\mathbb{F}}_p$, hence $p\equiv 1(\mathrm{mod}7)$ unless $p=7$. The only prime below $43$ congruent to one modulo seven is $29$, but Hasse gives $\#E({\mathbb{F}}_{29})<49$; for $p=7$ the same inequality is immediate. No such pair exists below $43$.