# Where is maximality used in this proof that $p$ divides the number of elements of order $p$ plus $1$.

If $G$ is a finite group, then if $n$ is the number of elements of order $p$, $p\mid n+1$ when $n\neq 0$.

Suppose $n\neq 0$. Since there is an element of order $p$, $p$ divides the order of the group, so we can take a maximal abelian $p$-group $H$. Let $P$ denote the set of all elements of order $P$, and let $H$ act on $P$ by conjugation. The orbits must divide the order of $H$, and partition $P$. So modulo $p$, $n\equiv f\pmod{p}$ where $f$ is the number of fixed points.

Of course, $H$ fixes all its elements of order $p$, (which form a subgroup when the identity is adjointed), so it has $p^m-1$ elements in $P$ for some $m$.

By maximality, it follows somehow that $H$ fixes only those $p^m-1$ elements in $P$. Why can't it fix more? If it did fix more, does that contradict maximality?

After that it follows easily that $n\equiv f=p^m-1\equiv-1\pmod{p}$. Thank you for your time.

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The statement of your first sentence is not true. There are more than $2+1$ elements of order $2$ in $\mathbb Z_2^3$, so their number cannot divide $2+1$. –  Mariano Suárez-Alvarez Dec 12 '11 at 15:19
Why do you think this is true? In the group $G={(\mathbb Z/p\mathbb Z)}^k$, all $p^k-1$ non-zero elements of $G$ are of order $p$, but $p^k-1$ does not divide $p+1$. –  Thomas Andrews Dec 12 '11 at 15:22
Ah, I see from the subject line that you really want to show $p|n+1$. –  Thomas Andrews Dec 12 '11 at 15:26
@all Sorry, I made an unfortunate mix up. I really should mind my $p$s and $n$s. –  Waldott Dec 12 '11 at 15:28

In general, if $H$ is an abelian subgroup of $G$ and $g\in G\setminus H$ such that $hgh^{-1}= g$ for all $h$ in $H$, then the set $H'=\{g^kh:k\in\mathbb Z, h\in H\}$ is also an abelian subgroup of $G$.

If $H$ is finite and $g$ of order $q$, the $|H'|=q|H|$.

In your case, $H$ is a maximal abelian $p$-group, so if it fixes some element $g\in G\setminus H$ of order $p$, then $H'$ is a bigger abelian $p$-group contained on $G$.

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I see now, thanks for the quick response. –  Waldott Dec 12 '11 at 15:47

If there were some element $g$ outside $H$ of order $p$ and fixed under conjugation by $H$, then it would commute with $H$ and so $H$ and $g$ together generate an Abelian $p$-subgroup strictly containing $H$, contradicting its maximality as Abelian $p$-subgroup.

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I see, thanks. And this generated group is precisely that mentioned in Thomas Andrews' answer. –  Waldott Dec 12 '11 at 15:48