An elliptic curve has a point of order $n$ iff $E[n]\cong\mathbb{Z}/n\times\mu_n$? Let $E$ be an elliptic curve over some field $k$, and let $n$ be coprime to the characteristic of $k$. Then I claim that $E(k)$ has a point of order $n$ if and only if $E[n]\cong\mathbb{Z}/n\times\mu_n$ as group schemes over $k$.
Is this correct?
I see this through the Weil Pairing, though it's possible I've made a mistake.
The intuition is - elliptic curves are famously self-dual. Then $E[n]$ should also be self-dual. If $E[n]$ admits a point of order $n$, then it has a subgroup which is isomorphic to $\mathbb{Z}/n$, and since the dual of $\mathbb{Z}/n$ is $\mu_n$, it should also have a subgroup isomorphic to $\mu_n$, so it must be the direct product?
Is this right? Are there any subtleties I should be careful of?
Okay, I messed up somewhere....thinking about it now.
 A: It's slightly more subtle than you present it here:
Lemma. If $E$ has a $k$-rational $n$-torsion point $P$, then we have a short exact sequence
$$0 \to \mathbb Z/n \to E[n] \to \mu_n \to 0.$$
(You can think of this either as a short exact sequence of finite group schemes, or of $\Gamma_k$-modules, where $\Gamma_k = \operatorname{Gal}(\bar k/k)$.)
Proof. The $n$-torsion point $P$ gives an injection $\mathbb Z/n \to E[n]$, so we get a short exact sequence
$$0 \to \mathbb Z/n \to E[n] \to G \to 0,$$
for some group $G$ of rank $n$. Cartier duality and the Weil pairing then turn this into a sequence
$$0 \to G^\vee \to E[n] \to \mu_n \to 0,$$
realising $\mu_n$ as a quotient of $E[n]$. Since the Weil pairing is alternating, the orthogonal complement of $\mathbb Z/n$ is $\mathbb Z/n$, so $G^\vee = \mathbb Z/n$ (as subgroups of $E[n]$). $\square$
However, the sequence need not be split:
Example. Consider the elliptic curve $E$ over $\mathbb Q$ given by $y^2 = x^3 + x$. Because the polynomial $x^3 + x = x(x^2+1)$ has exactly one rational (even real) root, we have
$$E[2](\mathbb Q) = \{O, (0,0)\}.$$
If the sequence splits, then $E[2] \cong \mathbb Z/2 \times \mu_2 = \mathbb Z/2 \times \mathbb Z/2$, since $\mathbb Q$ has a second root of unity. This would force $\# E[2](\mathbb Q) = 4$, which is false. So let's take a closer look:
The other $2$-torsion points are defined over $\mathbb Q(i)$. If we let $P = (0,0)$ and $Q = (i,0)$, then the action of $G = \operatorname{Gal}(\mathbb Q(i)/\mathbb Q)$ on $E[2]$ is given by
$$\sigma P = P, \hspace{2cm} \sigma Q = Q + P.$$
Indeed, $\sigma Q$ has to be the other non-rational $2$-torsion point, which is $Q + P$.
Now on the quotient $E[2]/\langle P\rangle$, the Galois action has become trivial: $\sigma Q = Q + P$, which means that $Q$ is fixed mod $P$. This is also what the lemma says, since $\mu_2 = \mathbb Z/2$.
