I'm trying to solve the following problem, but I feel like I'm missing some key fact.
Problem: Let $p\equiv 2\pmod{3}$ be an odd prime and consider the elliptic curve $E(\mathbb{F}_p)$ defined by $$y^2=x^3+B$$ where $B\not\equiv 0\pmod{p}$. Prove that if $E[2]\not\subseteq E(\mathbb{F}_p)$, then $E(\mathbb{F_p})$ is a cyclic group.
It is well known that if $p\equiv 2\pmod{3}$ and $B\in \mathbb{F}_p^{\times}$, then the given curve is supersingular, thus $E(\mathbb{F}_p)$ has order $p+1$. Since the group of points of an elliptic curve on a finite field is either cyclic or the direct sum of two cyclic groups, let's write $$E(\mathbb{F}_p)\simeq \mathbb{Z}_{n_1}\oplus \mathbb{Z}_{n_2}$$ where $n_1\mid n_2$. Then it suffices to prove that $n_1=1$.
Since $p$ is odd, we know that $E[2]\simeq \mathbb{Z_2}\oplus\mathbb{Z_2}$, thus the fact that $E[2]\not\subseteq E(\mathbb{F}_p)$ implies that $n_1$ is odd. Also, since $$n_1n_2=|E(\mathbb{F}_p)|=p+1\equiv 0\pmod{3}$$ we have that $3\mid n_1n_2$. Notice that if $E[3]\subseteq E(\mathbb{F}_p)$, then by a well known fact we would have that $K$ contains a primitive third root of unity, which is not true as $3\nmid p-1$. Thus, since $E[3]\simeq\mathbb{Z_3}\oplus\mathbb{Z_3}$ and $E(\mathbb{F}_p)$ has an element of order $3$, this should show that $3\mid n_2$ but $3\nmid n_1$. If we repeat the same argument for any odd prime divisor $q$ of $n_1n_2$, we should get that $q\mid n_2$ but $q\nmid n_1$, thus $n_1=1$.
Is this reasoning correct, and if so, is there a better way to solve this? Also, I feel like the hypothesis that $E[2]\not\subseteq E(\mathbb{F}_p)$ is necessary, since otherwise we could have something like $$E(\mathbb{F}_p)\simeq \mathbb{Z}_2\oplus \mathbb{Z}_{(p+1)/2}$$ but I haven't been able to construct any such counterexample.