The group $E(\mathbb{F}_q)/NE(\mathbb{F}_q)$ Let $E$ be an elliptic curve defined over the finite field $\mathbb{F}_q$ and let $N\geq 1$. We also make the following assumptions:
$T \in E(\mathbb{F}_q)[N]$ is a point of exact order $N$,  $E(\mathbb{F}_q)[N^2]=\mathbb{Z}T$ and  $N$ divides $q-1$.
Why is the group $E(\mathbb{F}_q)/NE(\mathbb{F}_q)$    cyclic of order $N$ ? 
 A: The $\mathbb{F}_q$-rational points of an elliptic curve $E$ defined over
$\mathbb{F}_q$ form a group that is isomorphic to
$$
E(\mathbb{F}_q)\simeq C_{n_1}\oplus C_{n_2},
$$
where $n_2\mid\gcd(n_1,q-1)$. This is not trivial. Because $E(\overline{\mathbb{F}_q})[N]$ is generated by at most two elements for all $N$, the same holds for $E(\mathbb{F}_q)$. Thus the group is a direct sum of two cyclic groups of orders $n_2\mid n_1$ (think invariant factors of a finite abelian group). The fact that $n_2$ must also be a factor $q-1$ is more difficult to see. The only argument that I can recall at this time relies on the properties of the Weil pairing. Because all of $E[n_2]\simeq C_{n_2}\oplus C_{n_2}$ is rational over $\mathbb{F}_q$, the Weil pairing must also take values in $\mathbb{F}_q$, and thus $\mu_{n_2}$ must be a subgroup of $\mathbb{F}_q^*,$ so $n_2\mid q-1$.
In your question we are given that there is a point $T$ of order $N$. Therefore $N\mid n_1$.
Furthermore, we know that 
$$E(\mathbb{F}_q)[N^2]\simeq C_{\gcd(N^2,n_1)}\oplus C_{\gcd(N^2,n_2)}$$
is cyclic of order $N$ (as it is generated by $T$). If $N$ has any common factors with $n_2$, then the $N$-torsion (a fortiori $N^2$-torsion) cannot be cyclic, so $N$ and $n_2$ must be coprime. This already implies that $[N]$ is an isomorphism from the subgroup $C_{n_2}$ to itself. Let us look at the other summand $C_{n_1}$. Because $N\mid n_1$
and $\gcd(N^2,n_1)=N$, we can write $n_1=N\cdot n'$ with $N$ and $n'$ coprime. Consequently
$C_{n_1}\simeq C_N\oplus C_{n'}$, and under this isomorphism $[N]C_{n_1}=1\oplus C_{n'}$.
The claim follows from these observations.
