$p$ is an odd prime, then prove that there is no group with exactly $p$ elements of order $p$.

Question

Question

If $p$ is an odd prime, then prove that there is no group with exactly $p$ elements of order $p$.

Attempt

Assume that such a group $G$ exists.

If $x_1 \in G$ and if $ |x_1|=p$ then $ |{x_1}^{-1}|=p$

hence such elements occur in pairs,

$p$ being an odd prime implies $x=x^{-1}$ for some element x whose order is $p$. Otherwise there would be even number of elements of order $p$

$x=x^{-1}$

$\implies x^2=e$

$\implies |x|=2$

$\implies 2$ is an odd prime.

A big contradiction.

Answer 1

Other approach, pick an element of given ones of order $p$. But then all non trivial powers of this element give rise to elements of order $p$, hence there are at least $p-1$ of them. If $G$ has precisely $p$ elements of order $p$, then there must be precisely one other, giving also rise to $p-1$ elements of order $p$. Hence $2(p-1)=p$, so $p=2$ a contradiction. By the way: there are no groups (finite or infinite) with exactly $2$ elements of order $2$.