Stabilizer of a Group Action

I am trying to solve the following problem.

Let $p$ me a prime and $L$ a linear map from $\mathbb{F}_p^n$ to $\mathbb{F}_p^n$. Suppose that there exist $k\geqslant 0$ such that $L^{p^k}=1$ where $1$ means the identity map from $\mathbb{F}_p^n$ to $\mathbb{F}_p^n$. Prove that there exist a non-zero vector $v$ such that $L(v)=v$.

I am trying to solve this in the context of group actions. Since $L^{p^k}=1$ the group generated by $L$ is cyclic of order $p^k,$ let's call it $G$. We define $\cdot: G \times \mathbb{F}_p^n \to \mathbb{F}_p^n$ by setting $L^m\cdot v = L^m(v).$ It is not that hard to verify that $\cdot$ is indeed an action.

Okay, now to conclude what I want I need to show that $L$ is an element of $Stab(v)$ for some non-zero $v$. It is equivalent to show that $Stab(v)=G$ for some $v$ or even that the $orb(v)=\{v\}$ for some non-zero $v$.

I have tried to do this directly but it does not seem easy, then I tried by contradiction: suppose that for all $v \in \mathbb{F}_p^n$, $L(v)\neq v$ then $L \notin Stab(v)$ for all non-zero $v$. Then since $Stab(b)\leqslant G$, its order has to be a power of $p$, but not $p^k$. Then we have two options, either the order is $1$ which would imply that the action is transitive or the order is a non-zero power of $p$. But I don't know how to derive a contradiction from here.

Another thing that I also though of was that this actions induces the homomorphism $\phi:G\to S_{\mathbb{F}_p^n}$ by setting $\phi(L^m)(v)=L^m(v)$. What I could prove was that $\phi$ is injective since $\phi(L^m)=1$ implies $\phi(L^m)(v)=v$ for all $v$, then $L^m(v)=v$ for all $v$ and then $L^m=1.$ From this we can see that $G$ is isomorphic to the group $\{L^m; m\in \mathbb{N}\}$.

I don't know anything else to think in the context of group actions, so I think I have already touched the key to solve it but I cannot see it by myself. Could someone take a look at this and show me WHERE I could give more attention to solve it?

Please, I would like to solve it by myself because I have an Algebra qualyfing exam next wednesday and I really need to learn, so I would really appreciate good hints just for me to give some steps more... I would not like to have a full answer.

Thank you very much!

Hint: In the ring $\mathrm{Hom}_{\mathbb F_p}(\mathbb F_p^n,\mathbb F_p^n)$, we have $L^{p^k}-1=(L-1)^{p^k}$.

• thanks! Maybe I can use it... – bttmbrcelo Aug 21 '17 at 23:00
• Wouldn't it imply that L(v)=v for all v? If this is true than the action is trivial isn't it? – bttmbrcelo Aug 21 '17 at 23:20
• @bttmbrcelo No, eg consider $L=\begin{bmatrix}1&1\\0&1\end{bmatrix}$. btw I think it's easier to approach this question using generalized eigenspaces rather than a group action. – stewbasic Aug 21 '17 at 23:23
• okay, thank you. A doubt... in general in a field $\mathbb{F}_p$ if I have $u^p=0$ can I say $u=0$? – bttmbrcelo Aug 21 '17 at 23:30
• @bttmbrcelo Yes, but $\mathrm{End}_{\mathbb F_p}(\mathbb F_p^n,\mathbb F_p^n)$ is not a field, just an $\mathbb F_p$-algebra. Even over the real numbers you can't multiple two nonzero numbers to get zero, but you can with real matrices. – stewbasic Aug 22 '17 at 0:03

Here's another approach, based on the

Lemma: Let $G$ be a finite $p$-group, acting on the finite set $S$. If $F$ is the set of fixed points of this action, then $$|F|\equiv|S|\pmod{p}$$

The proof is easy, and is really just the fact that an orbit of size bigger than $1$ has to have size dividing $G$, and hence is $0\pmod{p}$.

Now let $G=\langle L\rangle$, and $S=\mathbb{F}_p^n$, so that $F=\{v\in\mathbb{F}_p^n\mid L(v)=v\}$. By the above lemma, $$|F|\equiv|S|\equiv0\pmod{p}$$

Since $0\in F$, we see that $|F|>0$, so that $|F|\ge p$. Thus, there is a nonzero vector $v\in F$.

• Thank you! Very nice answer! – bttmbrcelo Aug 22 '17 at 5:04