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$G$ is a p-group and $S$ is a set that $G$ acts on. p does not divide $|S|$. Why is there at least one element $a\in S$ such that $|O(a)|=1$, or in other words, $G_a=G$?

I tried to ask this question yesterday but did not word it right, so the one who helped me claim that this part is clearly true and I really do not understand why, and I couldn't get a straight answer for the past day.

Any of you are welcome to vote to delete this question (after some discussion with me about the problem I hope), but I just really want to know why it is "clearly true, because I still couldn't reason it right.

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"The one who helped you" it's me, and there were more trying to make sense of your messy question. Anyway, I wrote there that there's this claim $\,|S^G|=|S|\pmod p\,$ ,and from this it is clear, in your case, that htere's at least one fixed element for the whole group. – DonAntonio Mar 20 '13 at 3:49
up vote 4 down vote accepted

The order of $G$ is $p^n$ for some $n$. The set $S$ can be partitioned into orbits and the orbit stabilizer relation tells you that the size of an orbit divides the order of the group, so the orbits are all of size $p^m$ for some $m \leq n$.

If each orbit had size $p^m$ with $m > 0$ then $p$ would divide the size of each orbit and hence divide $|S|$.

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Since $|S|$ does not divide $p$, $\exists a\in S$ such that $|O(a)|$ does not divides $p$.

Orbit-Stablizer Theorem says $\forall a\in S,|O(a)||G_a|=|G|$, which means $|O(a)|$ must be a power of $p$ as $|G|$ is a $p$-group. Therefore, for that $a$, $|O(a)|$ must equal to $1$.

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How about this? If $G$ acts on $S$, then $S=\bigcup_{a\in S}O(a)$ (the union of all orbits of a).

So $|S|=\sum_{a\in S}|O(a)|=\sum_{a\in S}{|G|\over |G_a|}$.

Now, p does not divide $|S|$, which means p does not divide $\sum O(a)$ or there exists $a\in S$ that p does not divide $|G|\over |G_a|$.

Since $G$ is a p-group, if p does not divide $|G|\over |G_a|$, this means ${|G|\over |G_a|}=p^0=1$

Or $|G|=|G_a|$, so $G_a=\{g\in G: g.a=g \ \forall \ g\in G\}$.

Thus $G=G_a$

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Looks perfectly OK, and is essentially what the other answers are saying. – Marc van Leeuwen Mar 21 '13 at 5:33

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