At most one subgroup of every order dividing $\lvert G\rvert$ implies $G$ cyclic [closed]

Suppose we have a finite group $G$ of finite order $n$. For every $d\mid n$, $G$ has at most one subgroup of order $d$. Show that $G$ is cyclic.

Solution 1:

Let $a_d$ denote the number of elements of order $d$ and $u_d$ the number of cyclic subgroups of order $d$. Then $a_d = \varphi(d) u_d$, where $\varphi$ is the Euler phi function. Reason: A cyclic group of order $d$ contains exactly $\varphi(d)$ elements of order $d$, and each element $g$ of order $d$ is contained in exactly one cyclic subgroup or order $d$, namely $\langle g\rangle$.

The assumption on $G$ implies $u_d \leq 1$ for all $d\mid n$. We get $$ n = \lvert G\rvert = \sum_{d\mid n} a_d = \sum_{d\mid n} \varphi(d) u_d \leq \sum_{d\mid n} \varphi(d) \cdot 1 = n\text{.} $$ So the "$\leq$" is in fact an equality, implying $u_d = 1$ for every divisor $d$ of $n$. In particular $u_n = 1$, and hence $G$ is cyclic.