$G$ contains at least $r(p-1)$ elements of order $p$

Question

Suppose a group $G$ has $r$ distinct subgroups of prime order $p$. Show that $G$ contains at least $r(p-1)$ elements of order $p$.

Aside: I know how to use this to prove that a group of order $56$ has a proper normal subgroup.

Answer 1

First of all, note that the subgroups of prime order $p$ must be cyclic (why?). Which means that every element in them apart from $e$ is of order $p$. And for another fact, notice that two distinct subgroups of order $p$ can't have anything apart from $e$ in common, otherwise they wouldn't be distinct. Which means we have $r(p-1)$ such elements of order $p$ ($p-1$ because we're excluding $e$).

Answer 2

IF $\;P,Q\;$ are two different subgroups of order $\;p\;$ , then clearly $\;P\cap Q=\{1\}\;$ (why?), and from here the claim, as there are $\;p-1\;$ elements of order $\;p\;$ in each such subgroup.