Let $p$ be a prime. If a group has more than $p − 1$ elements of order $p$, why can’t the group be cyclic?

If $G=\langle g\rangle $ is cyclic of order $n$ and $n=ap$, $p$ prime, then precisely the elements $g^a, g^{2a}, \ldots g^{(p-1)a}$ have order $p$.

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Or to put it differently: Let $H$ be the kernel of $x\mapsto x^p$. Then $H$ consists of all elements having order a divisor of $p$, while the image consists of all elements $g^{kp}$. As the order of the image is thus $\frac np$, the order of the kernel $H$ must be $p$. But as $H$ also contains one element of order $\ne p$ (namely the neutral element), we conclude that there are at most $p-1$ elements of order $p$.

Note that this second method does not make use of the fact that $p$ is prime. It is enough to assume that $p>1$ and $p|n$.

Hints:

1) If $\,G\,$ is a cyclic group of order $\,n\,$ , then for any divisor $\,d\,$ of $\,n\,$ there exists exactly one unique subgroup of $\,G\,$ of order $\,d\,$ , which is also cyclic .

2) A cyclic group of order $\,m\,$ has $\,\phi(m)\,$ different generators , with $\,\phi=$ the totient function of Euler.

3) For any prime $\,p\;,\;\;\phi(p)=p-1\,$ ...

If $G$ is infinite cyclic group ,it can't have the element of prime order . So suppose $G$ is finite cyclic group which contains some elements of order $p$, $p$ must divide $|G|$ by Lagrange's Theorem. Then by the Fundamental Theorem of Finite Cyclic Groups, for the divisor $p$ of $|G|$, there is exactly one subgroup of order $p$. In that subgroup of order $p$, there are $\phi(p)=p-1$ elements of order $p$. Hence, there are only $p-1$ elements of order $p$ in $G$.

If there are more than $p-1$ elements of order $p$, then there must be one more subgroups of order $p$, which contradicts with the Fundamental Theorem of Finite Cyclic Groups. Hence $G$ is not cyclic.