Cauchy's Theorem $\#$ of elements order $p$

Question

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

Answer 1

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 $n \geq 3$.

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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 $o_p(G) \equiv -1 \mod{p^m}$.

Answer 2

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.