Let $G$ be a finite group and let $p$ a prime number that divides $|G|$. Let $h$ denote the number of subgroups of $G$ of order $p$. Let $G$ be a finite group and let $p$ a prime number that divides $|G|$. Let $h$ denote the number of subgroups of $G$ of order $p$. 
Prove that:
i)  There are $h(p-1)$ elements of order $p$ in $G$
ii) $h\equiv 1 \pmod p$

I'm not yet introduced to normal subgroups. This is an exercise from the chapter of group actions. But I don't see how I can use my knowlegde about group actions to solve this exercise. 
 A: The first part is straightforward. An element $a$ of order $p$ generates a subgroups $\langle a \rangle$ of order $p$, which consists of the identity $e$, and of $p-1$ elements of order $p$. Two (distinct) such subgroups $A$ intersect in the identity. Therefore the set of elements of order $p$ is the disjoint union of all these $A \setminus \{ e \}$.
For the second part, let $A$ be one the subgroups of order $p$, and consider the action of $A$ by conjugacy on the set $\mathcal{S}$ of subgroups of order $p$.
The orbits of the action have length $1$ or $p$. Clearly the orbit of $A$ has length $1$. If the orbit of $B \ne A$ has length $1$ (that is, $B$ is fixed by $A$ in this action), then $\langle A, B \rangle$ is an elementary abelian group of order $p^2$ which has $p+1$ subgroups of order $p$, one of which is $A$, so this yields $p$ subgroups of order $p$ (all fixed by $A$ in this action) other than $A$.
Then there are the orbits of length $p$. Summing all up, we see that the number $h$ of subgroups in $\mathcal{S}$ is congruent to $1$ modulo $p$.
