Let p be a prime. If a group has more than $p-1$ elements of order $p$, then prove that the group can't be cyclic. Result: Let G be a group which has more than $p-1$ elements of order $p$. I need to prove that such a group can't be cylic. $p$ is a prime number.  
 Let's consider the case when G is finite. I want to prove the result using contradiction.
 Proof:  On the contrary, assume that G is cyclic. Hence, there exists $a\in G$ such that $G=<a>$. Let $|G|=n$  
Case 1: $p$ divides $n$ 
In this case, no. of elements of order $p =\phi (p)$, where $\phi$ is Euler Phi function.
Since it is known that, $\phi (p)\le p-1$, we have a contradiction.  
Case 2: $p$ doesn't divide $n$ 
gcd $(p,n)=1$ 
Let $|a^p|=t$. This implies that $a^{pt}=e$, where $e$ is identity of $G$. 
$\implies$ $n$ divides $pt \implies \exists$ integer $k$ such that $ kn=pt$. How do I arrive at a contradiction from here? 
PS: I don't know Lagrange's theorem, Cosets, Sylow's theorem, Isomorphisms yet. 
 A: Here is a proof without using Lagrange's theorem. Suppose $p$ does not divide $|G|=n$. Since $G=\langle a\rangle $ we must have $|a|=n$. Now, by our assumption there is an element $x\in G$ or order $p$. Since it is an element in the group there must be some $0\leq t\leq n-1$ such that $x=a^t$. Then:
$a^{pt}=x^p=e$
$n$ is the order of $a$, so this implies $n|pt$. But by our assumption $\gcd(p,n)=1$, so we conclude that $n|t$. But since $0\leq t\leq n-1$ this implies $t=0$. So $x=a^0=e$. It is a contradiction because $e$ has order $1$, not $p$. 
A: To complete your missing case it is enough to prove that if $G$ is a non-trivial, cyclic group of order $n$ and $\gcd(p, n)=1$ then there is no non-trivial element $a$ in $G$ such that $a^p$ is trivial. (This is simply a special case of Lagrange's theorem for cyclic groups, but it's proof is elementary.)
To see this, suppose $a^p=e$ the identity. As $\gcd(p, n)$ are coprime, there are integers $k, l\in\mathbb{Z}$ such that $kp+ln=1$. Therefore, as $a^{ln}=(a^n)^l=e^l=e$, we have:
$$
\begin{align*}
a^{kp+ln}&=a^1\\
a^{kp}&=a\\
(a^{p})^k&=a\\
e^k&=a
\end{align*}
$$
This is a contradiction, as $e^k=e$ (and $a\neq e$ as $G$ is non-trivial).
A: If $G$ is infinite, then $G$ is $\mathbb{Z}$. But $\mathbb{Z}$ has no element
with finite order except the identity. So assume $ G$ is finite and cyclic. Let
$a ∈ G$ and $∣a∣ = p$. Then for any $e ≠ b ∈ ⟨a⟩$,  order of $b$ divides $p$ and $∣b∣ = p$.
Hence, $⟨a⟩$ contains $p−1$ elements of order $p$. By the hypothesis, there exists $c ∉ ⟨a⟩$
such that $∣c∣ = p$. Then $⟨c⟩$ is another subgroup of $G$ with order $p$. But by the
Fundamental Theorem of Cyclic Groups, $G$ can have only one subgroup of order $p$,
we have a contradiction.
A: Suppose that $G$ is either an infinite cyclic group or a cyclic group of finite order $n$, with generator $g$. Then, there are three cases:

Case 1: $G$ is finite, and $p$ divides $n$, let's say $n=kp$.

In this case, the elements of order $p$ are the non-identity elements of $G$ whose $p$th power is the identity. If $(g^m)^p=e$, then $n=kp \vert mp$, so $k \vert m$. The number of positive multiples of $k$ less than $n=kp$ is exactly equal to $p-1$, so $G$ has exactly $p-1$ elements of order $p$.

Case 2: $G$ is finite, and $p$ does not divide $n$.

In this case, suppose that $(g^m)^p=e$. Then, $n \vert mp$, and since $n$ and $p$ are coprime, $n \vert m$. Hence, $g^m=e$, so $G$ has no elements of order $p$.

Case 3: $G$ is infinite.

In this case, the group $G$ is torsion-free, so it cannot have any elements of order $p$.
The above proves that if $G$ is cyclic, then it has either $0$ or $p-1$ elements of order $p$. By contraposition, if $G$ has more than $p-1$ elements of order $p$, then it cannot be cyclic.
