Given a prime number why cant a cyclic group have more than $p-1$ elements of order $p$ The other day I stumbled across this question:

Let $p$ be a prime number. If a group has more than $p-1$ elements of order $p$. Can it be cyclic?

I've tried to solve this by trying to prove that given a cyclic group $G=\langle g\rangle$ and $a_i=g^{k_i}$, $a_j=g^{k_j}$ both order $p$ where $k_i = k_j \pmod{p}$ then $a_i=a_j$ thus proving that there can only be $p-1$ different numbers of order $p".
To do this, I've done the following:
I've first picked a number $a$ such that $O(a)=p$. We know that $a=g^k$ therefore, $O(a)=O(g^k)=\frac{O(g)}{\gcd(O(g), k)} \Rightarrow \gcd(O(g), k)=\frac{O(g)}{O(a)}= \frac{O(g)}{p}=l$
As $\gcd(O(g), k)=l$ then $l|k \Rightarrow k=l\cdot m$
Which means that $l=\gcd(O(g), k)=\gcd(l\cdot p, l\cdot m) \Rightarrow p\nmid m$
With this in mind, we can rewrite $m$ as $m_n=n\cdot p + r$ and $k$ as $k_n=l\cdot m_n = l\cdot n\cdot p + l\cdot r$
Therefore $a_i=g^{k_i}=g^{l\cdot i\cdot p + l\cdot r}=g^{l\cdot i\cdot p}\cdot g^{l\cdot r}=(g^{O(g)})^i\cdot g^{k_0}=1_G\cdot g^{k_0} = (g^{O(g)})^j\cdot g^{k_0}=g^{k_j}$ where $k_i = k_j \pmod{p}$
The problem with this proof is that I'm not using the condition of p being prime or at least I'm not sure where if I'm using it unknowingly.
I would apreciate any advise you could give me as I'm new to group theory and it's still a confusing field to me.
Thanks in advance.
 A: 
Theorem: If $d\mid n$ and $d>0$, then the number of elements of order $d$ in a cyclic group of order $n$ is $\varphi (d)$, where $\varphi$ is Euler's totient function.

Source: Theorem 4.4 of Gallian's, "Contemporary Abstract Algebra (Eighth Edition)".
A: The assumption that $p$ ia a prime number isn't needed, all you need to assume is that $p\gt1$. Here is an easy proof using the facts that a cyclic group is
Abelian and a subgroup of a cyclic group is cyclic.
Let $G$ be a cyclic group and let $H=\{g\in G:g^p=e\}$. Since $G$ is Abelian, $H$ is a subgroup of $G$, so $H$ is a cyclic group. A generator of $H$ has order at most $p$, so $|H|\le p$ and $|H\setminus\{e\}|\le p-1$. Since $p\gt1$, every element of order $p$ in $G$ belongs to $H\setminus\{e\}$, Q.E.D.
A: If $G$ is cyclic of order $n$ then $G$ is group-isomorphic to the additive group on $\mathbb Z/n\mathbb Z$. So we just need to look at that case. If an element $g$ of the group has order $p$ then $p|n$. Trivially this gives us $p-1$ elements of the same order by taking $g^k$ for $k=1,\ldots,p-1$.
Now suppose $q^p = e = g^p$. This means in terms of $\mathbb Z/n\mathbb Z$ that $pq = 0 = pg$ (or take the discrete logarithm).
This means that both $q$ and $g$ are multiple of $n/p$. So if we take $g=n/p$ this implies $q = kg$ for some $k$, which implies that $q$ has to be one of the $p-1$ solutions we’ve already found.
EDIT: By the way we also see that there are either $0$ or $p-1$ such elements.
