How to prove that $\text{End}_{\mathbb{F}_p}(E)$ is commutative for a given elliptic curve E? Given a prime $p$ and considering the finite field $\mathbb{F}_p$, I need to see that $\text{End}_{\mathbb{F}_p}$(E) is commutative using orders. It is known that $\text{End}_{\mathbb{F}_p} \subseteq \text{End}(E)$, and I have seen how $\text{End}(E)$ is one of the following:
$\mathbb{Z}$
An order in an imaginary quadratic field
An order in a quaternion algebra
The key is, if I show that $\text{End}_{\mathbb{F}_p}(E)$ is exactly an imaginary quadratic order, then it would be necessarily a commutative ring, but how to show that $\text{End}_{\mathbb{F}_p}(E)$ is indeed an imaginary quadratic order? This fact should be true regardless the curve is either ordinary or supersingular, or, in other words, regardless the inclusion $\text{End}_{\mathbb{F}_p}\subseteq \text{End}(E)$ is strict or not.
Any help on how to approach this will be appreciated, thanks.

$\text{End}_{\Bbb F_p}(E)$ consists of the endomorphisms of $E$ that commute
with the Frobenius automorphism $F$. As $p$ is prime, $F$ has norm $p$, and
so $\Bbb Z[F]$ is a quadratic imaginary order.
If $E$ is ordinary, $\text{End}(E)$ is a quadratic imaginary order
and so $\text{End}_{\Bbb F_p}(E)=\text{End}(E)$.
If $E$ is supersingular, then $\text{End}(E)$ is a non-commutative
quaternion order, and the elements that commute with $F$ are just those in
$\Bbb Q(F)\cap\text{End}(E)$, which form an order in the quadratic field
$\Bbb Q(F)$.
In both cases, $\text{End}_{\Bbb F_p}(E)$ may be an order strictly containing
$\Bbb Z [F]$.
