I'd like to verify an answer I have for the following question:
Given that for an elliptic curve over a finite field $E/\mathbb{F}_{q}$ we have an isomorphism $$ E(\mathbb{F}_{q}) \cong \mathbb{Z}/m\mathbb{Z} \oplus \mathbb{Z}/mn\mathbb{Z} $$ where gcd$(q, m) = 1$, show that $q \equiv 1 \ \text{mod} \ m$.
My solution is as follows:
Let $S \in E(\mathbb{F}_{q})$ be a point contained in the first invariant factor of the decomposition above, so that $S \in E[m]$, and let $T \in E[m]$. We note that since $E[m] \cong \mathbb{Z}/m\mathbb{Z} \oplus \mathbb{Z}/m\mathbb{Z}$, that $E[m] \subset E(\mathbb{F}_{q})$. This implies (see Silverman AEC Chapter 3, Corollary 8.1.1) that the the image $\mu_{m}$ of the Weil $e_{m}$-pairing is contained in $\mathbb{F}_{q}^{*}$. In particular we have \begin{align} e_{m}([q]S, T) &= e_{m}(S, T)^{q} \\ &= e_{m}(S, T) \end{align} so that by the non-degeneracy of the Weil pairing we have $$ [q]S = S $$ for any $S \in \mathbb{Z}/m\mathbb{Z}$, so multiplcation by $q$ acts as the identity mod $m$ so $q \equiv 1 \ \text{mod} \ m$.
Any corrections or ways I could streamline this proof would be appreciated!
As you say, $\mu_m\subseteq\Bbb F_q^*$. By Lagrange's theorem, $m\mid q-1$.