Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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)?

share|improve this question
    
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
1  
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
add comment

1 Answer

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

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.