Gallian's "Contemporary Abstract Algebra", chapter 7 problem 54
Suppose that $G$ is a group of order $105$ with the property that $G$ has exactly one subgroup for each divisor of $105$. Prove that $G$ is cyclic
Sylow theorems, normal subgroups, and factor groups have not been covered yet at this point in the book, so I suspect they are not needed.
Pick $\alpha \in G $ such that $\alpha$ is not in any of the subgroups of order $1, 3, 5, \dots 35$. Then it would have to have order 105 by Lagrange. We know such an $\alpha$ exists because $|H_1| + |H_3| + \dots + |H_{35}| \leq 1+3+5+7+15+21+35 < 105 = |G|$, where $H_i$ denotes the subgroup of order $i$.
Is this correct?