Let $G$ be a finite group with precisely one non-trivial proper normal subgroup $N$. Suppose $\#N = 5$. If $f: G \rightarrow \mathbb{Z} / 6 \mathbb{Z}$ is a surjective homomorphism, what are the possible orders for the group $G$?
So far I have the following: Since $f$ is surjective, we can use Homomorphism Theorem telling us $$G/\operatorname{Ker}(f) \cong \mathbb{Z} / 6 \mathbb{Z} $$ Where $\operatorname{Ker}(f) $ is a normal subgroup of $G$. Then, since N is the only proper normal subgroup we get (not sure about this step) $$ G/N \cong \mathbb{Z} / 6 \mathbb{Z} $$ and hence $$ \#(G/N) = \frac{\#G}{\#N} = \frac{\#G}{5} = \# \mathbb{Z} / 6 \mathbb{Z} = 6$$ So $\#G = 5 \times 6 = 30$? Is this the correct approach?