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I'm reading this paper by Marcel Herzog on jstor:

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
up vote 2 down vote accepted

I'm not sure I understand your first question. Every finite group contains elementary abelian p-groups of maximum order since the identity subgroup is an elementary abelian p-group, and so Zorn's lemma for finite sets applies.

For your second question, this is just like the class equation for proving a non-identity p-group has a non-identity center. Every orbit has order a power of p, so there are two kinds of orbits: those where the power is 0 (the fixed points, orbits of size 1), and those where the power is positive (those orbits whose order is divisible by p). Mod p, all orbits of the second type disappear, and so the total number of points, ip(G), is equivalent mod p to the number of orbits of the first type, f.

The paper in question is:

  • Herzog, Marcel. Counting group elements of order p modulo p2. Proc. Amer. Math. Soc. 66 (1977), no. 2, 247–250. MR466316 DOI:10.2307/2040939
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Thanks Jack Schmidt. – Gotye Sep 3 '11 at 21:56

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