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**Drexel28** Yes. That is in fact what I was refering to. The fundamental theorem of cylic groups. So basically my proof was based on two things

1. $\displaystyle p$ is prime, therefore by LT any element $\displaystyle g$ of $\displaystyle G$ must have order $\displaystyle p^m\quad 1\le m \le n$.

2. Using that fact we know that there are three possiblities for any $\displaystyle g\in G$. Namely: $\displaystyle |g|=p$ or $\displaystyle |g|=p^\ell \quad 1<\ell<n$ , $\displaystyle |g|=n$ where $\displaystyle 1<\ell<n$. If the first case is true then we are done. If the second is true then $\displaystyle \langle g\rangle$ is a cyclic subgroup of $\displaystyle G$ of order $\displaystyle p^\ell$ and using the fundamental theorem of cyclic groups we can conclude there is a subgroup of $\displaystyle \langle g\rangle$ of order $\displaystyle p$, which of course means that the gnerator of that group is of order $\displaystyle p$. If the third case is true then $\displaystyle G$ itself is cyclic, in which case we may apply the fundamental theorem again.

Is there something wrong with that argument?