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.