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Consider the following statement:

Let $G$ be a finite group, and $H\lneq G$ a proper subgroup that is maximum, meaning that every proper subgroup $H'\lneq G$ is contained in $H$. If $[G:H]=p$ with $p$ prime, then $G$ is cyclic of order $p^n$.

If every proper subgroup is contained in $H$, then we can take $\sigma \in G\setminus H$ (that is nonempty), and so $\langle \sigma \rangle \nsubseteq H$, then $\langle \sigma \rangle$ cannot be a proper subgroup, thus $\langle \sigma \rangle =G$. But why is the cardinality of $G$ a power of $p$?

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Suppose $G$ is not a $p$-group, then any Sylow-$p$ subgroup of $G$ is a non-trivial (since $[G:H] = p$) proper subgroup of $G$. By assumption, $H$ contains this Sylow-$p$ subgroup, but this contradicts the fact that $[G:H] = p$.

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You know $G$ is cyclic. It cannot be infinite cyclic, since the infinite cyclic group does not have the given property (it has distinct maximal subgroups). So $G$ is cyclic of finite order $n$. Moreover, if $H$ is of order $k$, then $n=kp$. Write $k=p^am$ with $\gcd(p,m)=1$.

A cyclic group of order $n$ has unique proper subgroups of order $d$ for every proper divisor $d$ of $n$. All those subgroups are subgroups of $H$, hence their order must divide $k=p^am$. So if $d$ is a proper divisor of $n=p^{a+1}m$, then it is a divisor of $p^am$. In particular, $p^{a+1}$ is not a proper divisor of $n$ (since it is not a divisor of $p^am$); but it plainly is a divisor of $n$. So $p^{a+1}=n$ and $n$ is a power of $p$, as desired.

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