# How does the multiplicative group of a finite field, considered as a vector space, act on subspaces?

Given that a finite vector space $V = \operatorname{GF}(p)^n$ corresponds to the finite field $F = \operatorname{GF}(p^n)$, I'm wondering about how the multiplicative subgroup of $F$, $F^*$, acts on subspaces of $V$. Since $F$ is a field, we know that any element of $F^*$ is a linear automorphism of V, and is in $\operatorname{GL}(V)$. It therefore maps $k$-dimensional subspaces to $k$-dimensional subspaces. Further, we know that $F^*$ is the cyclic group of order $p^n - 1$. However, there are a couple things I'm wondering:

1) When will the action, for a given $p^n$ and $k$, be semi-regular: i.e. when will there be at most one $x \in F^*$ that maps between two $k$-dimensional subspaces $U_1$ and $U_2$.

2) What will the orbits under this action look like? Particularly, when will several subspaces in the orbit of a given $t$-dimensional space lie in a $t+1$-dimensional subspace, for example when will several 3-space in the same orbit lie in a single 4-space. I know that in the highly specific case where $p=2$ and $gcd(n,6)=1$, there are a number of 3-spaces (projective planes) containing three 2-spaces (projective lines) in the same orbit equal to the number of 2-spaces (projective lines) in the vector space. But do we know that, in general, an orbit will intersect like this?

I especially care about the case where $p=2$: if anyone knows anything about the action of $\operatorname{GF}(2^n)^*$ on its underlying vector space, or any books / articles that talk about this stuff, it would be tremendously appreciated.

Thank you!

(The question ultimately relates to a project where I am looking for designs over a finite field, and hence am trying to find collections of subspaces with certain regularities. Orbits under the multiplicative group seem very useful. Some related ideas come from: Thomas, Simon. "Designs over finite fields." Geometriae Dedicata 24.2 (1987): 237-242.)

• The action of the multiplicative group of the field is actually rather boring, as it fixes every subspace. It is just scalar multiplication. – Matt Samuel Jun 26 '15 at 22:56
• While it's true that $(\mathbb{F}_p)^n\cong\mathbb{F}_{p^n}$ as $\mathbb{F}_p$-vector spaces, I'd argue it's incorrect to state that one "corresponds" to the other. There is no canonical isomorphism, you have to choose one explicitly. – Zev Chonoles Jun 26 '15 at 22:57
• Matt - yes, but I am considering the multiplicative group of the entire field, rather than just the prime field. Zev - that's a good point. However the global properties of the orbits should not depend on which vector you choose to generate the field. – Kavi D. Jun 26 '15 at 23:27
• It is probably better to just not say that $F_{p^n}$ corresponds to anything. Just ask what is the action of the multiplicative group of $F_{p^n}$ on the set of the $F_p$-subspaces of $F_{p^n}$, and voilà. No ugly "corresponds". – Mariano Suárez-Álvarez Jun 26 '15 at 23:38
• Good point! I'm new to this - should I edit the question? – Kavi D. Jun 26 '15 at 23:45

The correspondence through the use of the Field Reduction map will give a canonical way to view $\mathbb{F}_{q^n}$ as $\mathbb{F}_{q}^{n}$.
1) $\mathbb{F}_{q^n}^{*}$ will not act semiregularly on the $k$-spaces of $PG(n-1,q)$ for odd $q$, because the elements in the base field $\mathbb{F}_{q}$ will act as the identity. Although for $q=2$ this is okay. In general, you can consider a subgroup $\langle \mu \rangle$, where $\mu \in \mathbb{F}_{q^{n}}^*$ and ask if this group acts semi-regularly.
For q=2, you are in a very special case; $\mathbb{F}_{q^n}^{*}$ acts as a Singer cycle on the space, a cyclic group acting sharply transitive on the points. The action on other subspaces has been studied in
• Field Reduction map - what's that? Any references where it is discussed? – Zev Chonoles Jul 5 '15 at 20:23