# If $H$ is a maximum subgroup of a finite group $G$ of index $p$ then $G$ is cyclic of order $p^n$

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 $$p$$?

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 $$[G:H] = p$$.
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.