Group $G$ acting on $\Omega$ such that each $\alpha \in \Omega$ has unique $p$-element fixing $\alpha$.

Let $G$ be a group acting on a set $\Omega$ and let $p$ be a prime. Suppose that for each $\alpha \in\Omega$ there is a $p$-element $x \in G$ such that $\alpha$ is the only point fixed by $x$. If $\Omega$ is finite, show that $G$ is transitive on $\Omega$; and if $\Omega$ is infinite, show that $G$ has no finite orbit on $\Omega$.

Any hints how to solve this problem?

EDIT: My initial thoughts were all wrong as I am assumed uniqueness of the $p$-element there.

• The problem does not guarantee the uniqueness of $x_\alpha$, nor could it. This invalidates most of what you've written here. – Alex G. Aug 15 '15 at 12:43
• Yes. Thank you for pointing out, somehow I thought the $p$-element is unique, but of course this is not given by the exercise. – StefanH Aug 15 '15 at 12:51

First suppose that $\Omega$ is finite, and so $G/K$ is finite, where $K$ is the kernel of the action, so we can assume that $K=1$ and $G$ is finite.
For any $\alpha \in \Omega$, there is a nontrivial $p$-subgroup $Q(\alpha)$ of $G$ with the unique fixed point $\alpha$. Let $P(\alpha)$ be a Sylow $p$-subgroup of $G$ containing $Q(\alpha)$. Since the orbits of $P$ are unions of orbits of $Q$ and have length a power of $p$, $P(\alpha)$ must fix $\alpha$, which is its unique fixed point.
Now for $\alpha,\beta \in \Omega$, an element $g \in G$ conjugating $P(\alpha)$ to $P(\beta)$ must map $\alpha$ to $\beta$, so $G$ is transitive.
Now suppose that $\Omega$ is infinite and has a finite orbit $\Delta$. Then $|\Delta| \equiv 1 \bmod p$. Let $\alpha \in \Omega \setminus \Delta$. Any $p$-element fixing $\alpha$ must also fix some point in $\Delta$, contrary to assumption.
• I understand the finite case, but your last line I do not get, why do we have $|\Delta| \equiv 1 \pmod{p}$ and why must any $p$-element fix some point in $\Delta$? – StefanH Aug 15 '15 at 15:36
• Okay, I guess I got the first one. If $\Delta$ is a finite orbit, choose some $\alpha \in \Delta$. Then $\Delta^{Q(\alpha)} = \Delta$. So by the $p$-group fixed point theorem we have $|\mbox{fix}(Q(\alpha))| \equiv |\Delta| \pmod{p}$, and with $\mbox{fix}(Q(\alpha)) = \{\alpha\}$ the first congruence follows. For the other assertion: Any $p$-element $x$ such that $\Delta^x = \Delta$ with $|\Delta| \equiv 1 \pmod{p}$ has to fix some element from $\Delta$ I have a feeling that it must be right, but I cannot formulate a formal argument for it... – StefanH Aug 15 '15 at 17:12
• ... ohh its just the $p$-group fixed point theorem again. Set $X = \langle x \rangle$, which is a $p$-group and $\Delta^X = \Delta$, hence $|\mbox{fix}(X)| \equiv |\Delta| \pmod{p}$, so that $|\mbox{fix}(X)| \equiv 1 \pmod{p}$. So we have points in $\Delta$ fixed by all elements from $X$, in particular there exists some point fixed by $x$. – StefanH Aug 15 '15 at 18:28