Cauchy's Theorem $\#$ of elements order $p$ In the stronger statement of Cauchy's Theorem it states the the number of elements of order $p$ is a multiple of $p$.
http://en.wikipedia.org/wiki/Cauchy%27s_theorem_(group_theory)#Proof_2
I noticed that they didn't count the actual number. In general is this hard to count? 
I want to count the number of elements of order $p$ in a $p$-group $G$ with $|G|=p^n$ such that all elements of order $p$ form a subgroup of $G$(including $e$) 
Let $ X = \{\,(x_1,\cdots,x_p) \in G^p : x_1x_2...x_p = e\, \}  $
Let the cyclic group $C_p$ of order $p$ act on $X$
Then $|X^c|$ gives me the number I want.
I tried counting with the class equation 
$$|X| = |X^c| + \sum[C_p:Stab(x)]$$ 
$(p^n)^{p-1} = |X^c| + p + p+p+p+p......+p  =|X^c| +p(1+1+1+1+.....+1+1+1)$
.... I can't come out with a number
I'm sure someone has proven what that number is, it's been 100+ years. If any, can i get an reference, preferably elementary but if its hard...ill try to understand it, on what type of groups  do we know about the number of elements order $p$ there are (such that $p| \ |G|$) 
Thanks ;D
 A: As other people have mentioned, it is impossible to say anything exact without more information.
But here is one example of an improvement.
Let $G$ be a finite group with order divisible by the prime $p$. Denote the number of elements of order $p$ in $G$ by $o_p(G)$. Let $P$ be a Sylow $p$-subgroup of $G$, of order $p^n$. Call $P$ exceptional if $P$ is cyclic, generalized quaternion, dihedral or semidihedral.

*

*$n = 1$: In this case $o_p(G) \equiv rp - 1 \mod{p^2}$ for some $0 \leq r \leq p-1$, and all values occur for some $G$.


*$n \geq 2$: If $P$ is exceptional, then $o_p(G) \equiv p-1 \mod{p^2}$.


*$n \geq 2$: If $P$ is not exceptional, then $o_p(G) \equiv -1 \mod{p^2}$.
This is proven in the following article by Herzog.

Marcel Herzog, Counting group elements of order $p$ modulo $p^2$, Proc. Amer. Math. Soc. 66 (1977), 247-250. DOI

With more information about the structure of $P$ you can tell more about $o_p(G)$. For example, if $P$ is cyclic, then $o_p(G) \equiv p-1 \mod{p^n}$, which is better than the above result when $n \geq 3$.

More examples of results like this can probably be found by applying theorems from the paper "On a Theorem of Frobenius" by Philip Hall. For example, the following is corollary 4.31 (pg. 497).

Theorem: The number of subgroups of order $p$ in $G$ is congruent to
$$1 + p + p^2 + \cdots + p^{m-1} \mod{p^m}$$
where $m = \omega(P)$ or $m = p-1$ according as the $p$-Sylow subgroup $P$ of $G$ is regular or not.

Here $\omega(P)$ is the integer such that $[P : \mho_1(P)] = p^{\omega(P)}$, where $\mho_1(P)$ is the subgroup generated by $p$-th powers of $P$.
By the above result, it follows that $o_p(G) \equiv -1 \mod{p^m}$.
A: There is no exact formula. The only thing you can say is that the number of elements of order $p$ is $\equiv -1$ mod $p$. It follows from the proof of James McKay you are referring to: note that in this Proof $2$ you mention the identity element is counted. 
