This is my attempt. I am not sure as for its plausibility.
$Attempt$: Let $G$ be a group of order $185$. Then $G=185=5\cdot 37$. The $Sylow-p$ subgroups are unique and normal and therefore $G$ is nilpotent and so $G=P_1\times P_2$ where $P_1$ is the $Sylow-5$ normal subgroup while $P_2$ is the other. Since $5$ and $37$ are prime then $P_1$ and $P_2$ are cyclic. Is a product of cyclic groups a cyclic group? By what I know so far, $G$ is either isomorphic to $\Bbb{Z}_5\times \Bbb{Z}_{37}$ or to $\Bbb{Z}_{185}$ but $\Bbb{Z}_5\times \Bbb{Z}_{37}$ is isomorphic to $\Bbb{Z}_{185}$ because $(5,37)=1$ then what does it mean?? I am so confused...
As you said the $37$ group is normal, hence $$G=\mathbb{Z}_{37}\rtimes \mathbb{Z}_5.$$ Now since $|Aut(\mathbb{Z}_{37})|=36$, there are no elements of order $5$ there hence the action is trivial and $$G=\mathbb{Z}_{37}\times \mathbb{Z}_5.$$ Which is cyclic.