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$G$ is a $p$-group, which means $|G|=p^n$ for $n\in \mathbb{Z^+}$.

Now,if $p$ does not divide $|S|$, for S is a set that G acts upon, how do I show that there exists $a\in S$ such that $G_a=G$

So how do I do this, I tried to find a relation with the order of $S$ but I really don't know how.

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Sorry, you know, when I think too long on something, I tend to have the illusion that the rest of the world will have to know how the prob goes and how to do it already, haha. – Akaichan Mar 19 '13 at 2:09
This is not true unless you are more specific about $a\in S$. For instance, let $C_p$ act on itself by addition, and act trivially on $\{\bullet\}$, and define $S=C_p\cup\{\bullet\}$. Then $p\nmid\#S=p+1$ but if $a\in C_p$ then ${\rm Stab}_G(a)=0\ne C_p$. Is it possible you want $p$ to not divide the orbit of $a$? – anon Mar 19 '13 at 2:14
Your definition of $\,G_a\,$ is...weird. It doesn't need that $\,\forall g\in G\,$ inside those brackets. – DonAntonio Mar 19 '13 at 2:17
I'm re-commenting due to a foolish typo I did in my last comment, which I then erased. Thanks to Anon.......The claim is almost sure false. What is true is $\,|S^G|=|S|\pmod p\,$ , with $\,S^G:=\{s\in S\;;\;gs=s\,\;\,\forall\,g\in G\}\,$ – DonAntonio Mar 19 '13 at 2:30
Actually, all is it asked is to show that $G_a=G$, and I thought that showing the $\forall g$ part is the way to go, it is my own claim, not the claim that I have to prove – Akaichan Mar 19 '13 at 5:00
up vote 2 down vote accepted

If you check my last comment under your question, from that lemma it follows that there must exist $\,a\in S\,$ s.t. $\,ga=a\;,\;\;\forall\,g\in G\,$ (why?).

For this very particular $\,a\in S\,$ it is certainly true that $\,G_a=G\,$ ...:)

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