# Clarifications on proof that the fixed points of order $p$, $i_p(G)\equiv -1\pmod{p}$

I'm reading this paper by Marcel Herzog on jstor: http://www.jstor.org/stable/2040939?seq=1

I want to follow up on a few things about the short proof of Theorem 1, found on the bottom of page 1 of the pdf.

The proof takes $E$ to be a elementary abelian $p$-group of maximal order, how can we be sure one exists?

I understand that each orbit divides the order of $E$, by the orbit-stabilizer theorem, so why does $i_p(G)\equiv f\pmod{p}$?

Here $i_p(G)$ is the number of elements of order $p$ in $G$, and $f$ is the number of fixed points on the set of elements of order $p$ when acted on by conjugation by $E$.

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By calling it an "aper", are you accusing Herzog of plagiarism? :-) –  Henning Makholm Sep 2 '11 at 23:43
typo corrected. –  Gerry Myerson Sep 3 '11 at 0:33
Thanks for mentioning the paper. I had been struggling to take the p-group case and prove it for general finite groups. I had Kulakoff and Berkovich and Isaacs for the p-group case, and Berkovich seemed to just assume we knew it worked for all finite groups too. –  Jack Schmidt Sep 3 '11 at 1:27