Show that the following group action has a non-zero singleton orbit.

Question

Let F be a finite field of characteristics prime p.Let G be a group of order $p^r$ for some r.Let G acting on $F^n$ for n>1.Then show that there exist a non-zero vector in $F^n$ whose orbit will be singleton.

Answer 1

We have the follwing nice theorem:

Th.: Let $\;G\;$ be a finite $\;p$- group acting on a finite set $\;X\;$ , and put

$$X^g:=\{x\in X\;;\;x^g=x\;\forall\;g\in G\}$$

Then,

$$|X^G|=|X|\pmod p$$

Proof: We know

$$X=\bigsqcup\,\mathcal Orb(x)$$

Now,

$$|\mathcal Orb(x)|=[G:G_x]\;,\;\;G_x:=\{g\in G\;;\;x^g=x\}$$

and thus all the orbits are of size a power if $\;p\;$(including $\;p^0=1$) , and

$$x\in X^g\iff |\mathcal Orb(x)|=1$$

so that in fact

$$|X|=\sum_{x\in X^G}1+(\text{powers of}\;p)=|X^G|\pmod p$$

Your problem is now solved by noting that zero is always a fixed point in $\;\Bbb F^n\;$ , and thus there must be at least another fixed point since $\;|\Bbb F^n|=p^m\;,\;\;m\in\Bbb N\;$