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.

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. 
