# $p$-Group Acting on a Finite Set and the Number of Fixed Points ## Problem 359

Let $P$ be a $p$-group acting on a finite set $X$.
Let
$X^P=\{ x \in X \mid g\cdot x=x \text{ for all } g\in P \}.$

The prove that
$|X^P|\equiv |X| \pmod{p}.$ Add to solve later

## Proof.

Let $\calO(x)$ denote the orbit of $x\in X$ under the action of the group $P$.

Let $X^P=\{x_1, x_2, \dots, x_m\}$.
The orbits of an element in $X^p$ under the action of $P$ is the element itself, that is, $\calO(x_i)=\{x_i\}$ for $i=1,\dots, m$. Let $x_{m+1}, x_{m+2},\dots, x_n$ be representatives of other orbits of $X$.

Then we have the decomposition of the set $X$ into a disjoint union of orbits
$X=\calO(x_1)\sqcup \cdots \sqcup \calO(x_m)\sqcup \calO(x_{m+1})\sqcup \cdots \sqcup \calO(x_n).$

For $j=m+1, \dots, n$, the orbit-stabilizer theorem gives
$|\calO(x_j)|=[P:\Stab_P(x_j)]=p^{\alpha_j}$ for some positive integer $\alpha_j$. Here $\alpha_j \neq 0$ otherwise $x_j \in X^P$.

Therefore we have
\begin{align*}
|X|&=\sum_{i=1}^m|\calO(x_i)|+\sum_{j=m+1}^n|\calO(x_j)|\\
&=\sum_{i=1}^m 1 +\sum_{j=m+1}^n p^{\alpha_j}\\
&=|X^P|+\sum_{j=m+1}^n p^{\alpha_j}\\
&\equiv |X^P| \pmod{p}.
\end{align*}
This completes the proof. Add to solve later

### More from my site

#### You may also like...

This site uses Akismet to reduce spam. Learn how your comment data is processed.

###### More in Group Theory ##### Order of Product of Two Elements in a Group

Let $G$ be a group. Let $a$ and $b$ be elements of $G$. If the order of $a, b$ are...

Close