In Gallian's abstract algebra, there are two problems of the same form. Prove that any group of order $33$ is cyclic, and prove that any group of order $35$ is cyclic.
In the first case, I'd assume there are no elements of order $33$, then there must be elements of orders $3$ and $11$ since these elements are prime. Say $a,b \in G$, $|a|=3,|b|=11$. Then after some steps the conclusion is that $|ab|=33$ and the group is cyclic.
However to prove a group $G$ is order $35$. Assuming there is no element of order $35$, all nonidentity elements have orders $7$ or $5$. However, there must be elements of both orders. Consider the subgroup of order $7$. Then this is the only subgroup of order $7$, therefore it is normal. Call this subgroup $H$. Then $H$ is abelian so $H \leq C(H)$, $N(H)=G$. If $C(H)=H$, then $|N(H)/C(H)|=5$ however by a theorem, this must be isomorphic to a subgroup of ${\rm Aut}(H)$, but $|{\rm Aut}(H)|=6$ a contradiction. Thus $C(H)=G$ and the product of an element in $H$ and element of order $5$ in $G$ has order $35$, So $g$ is cyclic. Where $C(H)$ is the centralizer of $H$ in $G$, $N(H)$ is the normalizer and ${\rm Aut}(H)$ is the group of automorphisms of $H$.
To me, the first method for proving the similar statements is much easier. My question is, is there any reason you cannot use the first method of proof, to prove the second statement?
