If G acts on X, show that there must be a fixed point for this action. Please help.

Question

Suppose that $G$ is a group of order $p^k$, where $p$ is prime and $k$ is a positive integer. Suppose that $X$ is a finite set and assume that $p$ does not divide the $|X|$. If $G$ acts on $X$, show that there must be a fixed point for this action. i.e show that there is some element $x\in X$ whose stabilizer $G_x=\{g.x;g\in G\}$ is all of $G$.

Answer 1

This is a standard application of the following fixed-point theorem. Suppose $G$ is a $p$-group (which is the case here). Let $X^G$ be the set of fixed points of $G$. Then, $$|X| \equiv |X^G| \pmod{p}.$$

Since $p \not \mid |X|$, it follows that $|X^G| \not \equiv 0 \pmod{p}$. In particular, $|X^G| \neq 0$, so there exists some fixed point.

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Here's a proof of the above equality: The orbits of $G$ partition $X$, so that $|X| = |X^G| + \sum |B_i|$, where $B_i$ are the orbits. But, since $G$ is a $p$-group, each $|B_i|$ is divided by $p$ (do you see why?)

Added Here's why $|B_i| \equiv 0 \pmod{p}$: Since $B_i$ is an orbit, $G$ acts on $B_i$, possibly with fixed points. Now, let $G^{B_i}$ be the set of $g \in G$ fixing some fixed $x \in B_i$. Then $G^{B_i}$ is a subgroup of $G$, hence of order $p^k$ for some $k$. The quotient group $G/G^{B_i}$ is then also a $p$-group, and I claim that it acts bijectively on $B_i$. First, let $y \in B_i$. Then $y=g \cdot x$ for some $g \in G$ (because $y$ and $x$ are in the same orbit). Now, assume $y$ is also equal to $g^\prime \cdot x$. But then $g^\prime g^{-1}$ fixes $x$, hence lies in $G^{B_i}$, and thus $g$ and $g^\prime$ represent the same element in the quotient $G/G^{B_i}$. Thus, all the failure of injectivity lies in $G^{B_i}$, so after modding out by it, we see that $|G/G^{B_i}|=|B_i|$, hence $|B_i| \equiv 0 \pmod{p}$.

Answer 2

The orbit-stabilizer theorem (easy to prove!) says that whenever you have a group acting on a set $X$, for any $x\in X$, the size of the orbit $Gx$ is $|G|/|\text{Stab}_G(x)|$. This means that since $|G| = p^k$, that every orbit will have size a power of p (possibly $p^0$!). Since the orbits partition $X$, and since $p\nmid|X|$, this means that it's impossible to partition $|X|$ into orbits each with size a positive power of $p$. Hence, there must exist orbits with size $p^0 = 1$. These orbits are just points of $X$ fixed by all of $G$.

Answer 3

Put together facts you should know:

• A $G$-set $X$ is the disjoint union of its orbits, so $|X|$ is the sum of the sizes of the orbits.
• The size of an orbit divides $|G|$ by orbit-stabilizer.
• Fixed points correspond to orbits of size $1$.