Let $p$ be a prime and let $X$ be a finite set whose cardinality is divisible by $p$. Assume a group $G$ of order $p^n$ for $n>1$ acts on $X$. Show that if $G$ fixes an element in $X$ then it fixes more than one element.
No idea with this problem. Maybe there is some useful key facts about such kind of group. May I please ask to pointing them out or prove it directly? Many thanks!
On
If $x \in X$, then the orbit stabilizer theorem tells us that $|O(x)| = (G:G_x)$, where $|O(x)|$ is the order of the orbit containing $x$ and $G_x$ is the stabilizer of $x$. If no other element is fixed, then there exists $y \in X$ such $|O(y)| = (G:G_y) \neq 1$ and hence $O(y) = p^m$ but $X$ has only $p$ elements so $|O(y)| = p$. Thus $O(y) = X$ and we no longer have $O(x) \cap O(y) = \emptyset$
Use the fact that $X$ can be decomposed into disjoint union of orbits. Assume that there is exactly one element fixed in $X$ by $G$. Now, the other orbits must have length dividing $p^{n}$ and are strictly bigger than $1$, and hence must be divisible by $p$. Hence $X$ is a disjoint union of sets, one of length one, and the others of length divisible by $p$, so $|X|=1+kp$ for some $k >0$. Contradiction. We must have another fixed point.