A cyclic group $G$ has only one proper subgroup and it is of order $p$ (a prime). What is the order of $G$? The best I could come up with :
$G$ consists of :


*

*The identity $e$,

*Generators of $G$,

*Generators of the subgroup (say, $H$) which is of order $p$.
If order of $G$ is $n$, then there are $φ(n)$ generators of $G$. Since $H$ is a proper subgroup, all the $p-1$ generators of $H$ are different from those of $G$. And then there's $e$. So order of $G$ is $φ(n)+p$. 
 A: We know that $G$ is a cyclic group, and it contains an order $p$ element. It follows from Lagrange's Theorem that
$$G = \mathbb Z_{n} \quad \text{ for some } \quad n : p | n$$
We also know that the subgroup is proper, implying that $n>p$. It's also the only proper subgroup. If $G = Z_{ap}$ for $a \neq p$, then there would be a proper subgroup of some different order $a$-- specifically the group $\langle p\rangle$ consisting of $\{p,2p,3p, \dots ap=0\}$.
We have that $n > p$, and also that $n \neq a p$ where $a \neq p$. The conclusion is that $n = p^2$.
Note that this proof did not make use of the fact that $p$ is prime. If $p$ were not prime, then you would necessarily have more than one proper subgroup.
A: I think the order of group should be $p^2$  as it is cyclic for every divisor we have a unique subgroup and $p^2$  has only three divisors  hence unique subgroup of order p.
Hope it works
A: 
  
*
  
*If $G=n$, then $G$ has a unique subgroup of order $d$ for each divisor $d$ of $n$ such that $\langle a^{\frac{n}{d}}\rangle$ and $\mid a^{\frac{n}{d}}\rangle\mid=d$
  

So, $n=p^2$. Otherwise you can obtain more than one subgroup.

  
*
  
*If $G$ is infinite cylic then we know that G is isomorphic to $\Bbb Z\rightarrow G$ with $n\rightarrow g^n$. 
  

We know that subgroups of $\Bbb Z$ has the form $n\Bbb Z=\{na|  n\in \Bbb N, a\in \Bbb Z \}$. If it has prime order subgroup we can find $n\in \Bbb N$ such that $pn=1$. This is a contradiction. Actually, we know that any subgroup of $(\Bbb Z, +)$ is infinite.
A: Your approach is actually good. Say $n=|G|$. Then an element is in one of the following classes


*

*the identity;

*the generators of the unique subgroup of order $p$;

*the generators of $G$.


In particular, $G$ cannot be infinite. Also, $n=|G|$ cannot have prime divisors different from $p$ (Cauchy), so $n=p^k$ for some $k$ and $\varphi(n)=p^k-p^{k-1}$. Therefore we have
$$
p^k=1+(p-1)+\varphi(p^k)=p+p^{k}-p^{k-1}
$$
which implies $k=2$.

A different proof, without counting. Let's drop the assumption that $G$ is cyclic, which is not necessary.
If $G$ is infinite, then it cannot have an element of infinite order; but an infinite cyclic group has infinitely many subgroups: contradiction.
Hence all elements of $G$ must have finite order. Take an element $x$ of order $m$; if a prime $q$ different from $p$ divides $m$, then $G$ has a subgroup of order $q$, namely $\langle x^{m/q}\rangle$.
Therefore the order of each element is of the form $p^k$.
Suppose that all non identity elements have order $p$. Take an element $a\in G$, $a\ne 1$. If $b\in G$, $b\ne1$, we must have $\langle b\rangle=\langle a\rangle$ or we'd get two distinct subgroups of order $p$. Hence $G=\langle a\rangle$ is cyclic of order $p$: a contradiction.
Hence there is an element $y$ having order $p^k$ with $k\ge2$. Then $\langle y^{p^{k-2}}\rangle$ is a subgroup of order $p^2$: by the assumption on $G$, this subgroup cannot be proper, hence $G$ is cyclic of order $p^2$.
