supersingular if and only if $E[p]\cong 0$?

Question

Using the definitions that an elliptic curve $E$ over a finite field $K$ of characteristics $p$ is supersingular if $E[p^r]＝0$ for all $r≧1$, how can I show that $E$ is supersingular if and only if $E[p]＝0$ ?

$E[p^r]$ for all $r≧1$⇨ $E[p]$ is obvious, but the inverse direction is unclear..

Answer 1

Well, if $E[p]=0$ then multiplication by $p$ is injective. So multiplication by $p^r$ is an $r$-fold composition of injective maps, so also injective.

Another way to see it: if $p^rx = 0$ then $p(p^{r-1}x)=0$, so $E[p]=0$ implies $p^{r-1}x = 0$. Repeat until you get $x = 0$.

Answer 2

$E[p]$ is larger than $\{O\}$ iff $E[p^r]$ is larger than $\{O\}$.

$E[p]\subset E[p^r]$ and if $E[p^r]$ containins a non-trivial element $Q$ of order $p^m$ then $p^{m-1}Q$ is a non-trivial element of $E[p]$.