Question about the location of the r-torsion in the quotient group used in the Tate pairing I'm working through Pairings for Beginners by Craig Costello, and am trying to understand the Tate pairing.

He defines $rE = \{r*P | P \in E(\mathbb{F}_{q^k})\}$ and then forms the quotient group $E(\mathbb{F}_{q^k})/rE$. He then claims:


  if $r^2 || \#E(\mathbb{F}_{q^k})$, then $E[r] \cap rE = \{\mathcal{O}\}$.


where $\cdot||\cdot$ means divides, but just once, and $E[r]$ is the r-torsion. 

My question

Why is this true? I definitely agree that $\mathcal{O}$ is in this intersection, since $r[P] = \mathcal{O}$ for any $P$ in $E[r]$, but I don't see why this should be the only element in the intersection. Moreover, if there were some point $Q \neq \mathcal{O}$ in $E[r] \cap rE$, then $Q$ would have order $r^2$, which by Lagrange's Theorem, would be possible under the assumption that $r^2 || \#E(\mathbb{F}_{q^k})$.
The group $E[r]$ is isomorphic to $\mathbb{Z}/r \mathbb{Z} \times \mathbb{Z}/r \mathbb{Z}$, which has $r^2$ elements, all of order $r$. Since your element $Q$ has order $r^2$, $E(\mathbb{F_{q^k}})$ has also a cyclic subgroup of order $r^2$. 
This group has $r (r-1)$ elements of order $r^2$ which cannot be elements from $E[r]$. This implies that $\#E(\mathbb{F_{q^k}})$ must be at least divisible by $r^3$.