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If I have nontrivial $p$-groups $G, H$, where $p$ is a prime number, and $H$ acts on $G$ by automorphisms, how can I show that the set of fixed points of the action (the set $\{ x \in G : h*x = x,~\forall h \in H \}$) is nontrivial (has more than $1$ element)?

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Number of fixed pts =number of elements in set mod p. –  user641 Sep 23 '12 at 19:08
@SteveD Can you elaborate? I know that the number of fixed points of each automorphism $h \in H$ must divide $|G|$, since it's a subgroup. Which set are you talking about when you say "set mod $p$"? –  david Sep 23 '12 at 19:12
He means that the number of fixed points is congruent modulo $p$ to the number of the elements in the set he described. (I don't know if he's right but I just read his comment.) –  Patrick Da Silva Sep 23 '12 at 19:21

1 Answer 1

up vote 4 down vote accepted

The general theorem that you need here, and which Steve aludes, is:

Theorem: Let $\,P\,$ be a finite $\,p-\,$ group , $\,X\,$ a finite set, and suppose $\,P\,$ acts on $\,X\,$ . Define $\,X_P:=\{x\in X\;:\;ax=x\,\,\forall\,a\in P\}\,$ . Then, $\,|X_P|=|X|\pmod p\,$

Proof: We know that $\,|X|=\sum|\mathcal Orb(x)|\,$ , where the sum runs over different (and thus disjoint) orbits.

Now, for any $\,x\in X\,\,,\,\,|\mathcal Orb(x)|\,$ is a power of $\,p\,$ , so if there are $\,n\,$ orbits with one single element, then

$$|X_P|=n\Longrightarrow |X|=n+\,(\text{powers of }\,p)\,\,\,\;\;\;\;\;\square$$

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