The relationship between subfields and subgroups of a finite field.

Question

I am trying to get my head around the structure $GF(p^n)$ when viewed as a vector space of dimension $n$ over $GF(p)$ (mainly the relationship between the additive and multiplicative structures). I'm unsure about a few things:

1) If $m\mid n$, and $k$ is a multiple of $m$ not dividing $n$, how many subspaces of dimension $k$ (amongst $\prod_{i=0}^{k-1}\frac{(p^n-p^i)}{(p^k-p^i)}$ of them) contain the unique subfield $GF(p^m)$? Moreover, are these "special" subspaces fixed under multiplication from $GF(p^m)$? I'm specifically interested in the size of the stabiliser of such a subspace under the action of the full multiplicative group.

2) Is it true that if gcd(m,n)=1, then no subspace of dimension $m$ can contain a subfield? (excluding the case m=1, of course)

3) If a subspace does not contain a subfield, must it have a trivial stabiliser under the action of the multiplicative group? i.e. are the subspaces partitioned into orbits of maximal size? $\bf{EDIT}$: not trivial stabiliser (obviously false), but stabilised by nothing more than $GF(p)^*$?

Any words of wisdom on techniques to deal with these kinds of questions would be greatly appreciated. If it makes any difference, I'm mainly interested in the case $p=2$.

Answer 1

Note that 1) can be reformulated in the following way: Given a finite field $F$ and a finite dimensional vector space $V$ with a subspace $V_0 \subseteq V$ of dimension $m$, how many subspaces of $V$ of dimension $k$ contain $V_0$? We are thus looking for the subspaces of $V/V_0$ of dimension $k-m$. Given that $V$ has dimension $n$, the question reduces to the problem of computing the number of $k-m$ dimensional subspaces of an $n-m$ dimensional $F$-vector space (which you already seem to know). Note that your assumption on $k$ is not necessary to do this.

To be stable under multiplication with $GF(p^m)$ theses subspaces would have to be $GF(p^m)$-subspaces of $GF(p^n)$ and here your assumption on $k$ to be a multiple of $m$ is clearly necessary for this to happen (I don't see why you would want $k$ not to divide $n$). Indeed any $GF(p^m)$-subspace of $GF(p^n)$ containing $GF(p^m)$ with dimension $\frac{k}{m}$ defines a $GF(p)$-subspace containing $GF(p^m)$ with dimension $k$. The converse however does not hold.

For 2) there are easy counterexamples. You can take any three dimensional $GF(p)$-subspace containing $GF(p^2)$ in $GF(p^4)$.

For 3) note that any subspace is stabilized by the multiplicative group of $GF(p)$ so there will be no trivial stabilizers.

Edit:

Asking for the stabilizer of a $GF(p)$-subspace of $GF(p^n)$ under the action of $GF(p^n)^*$ has not much to do with containing certain subfields. In fact if $m$ divides $n$ then we find a direct sum decomposition $GF(p^n) = GF(p^m) \oplus V_0$ for some $GF(p^m)$-subspace $V_0$ of $GF(p^n)$. As $1$ is not an element of $V_0$, it contains no subfield of $GF(p^n)$. However the stabilizer of $V_0$ will contain $GF(p^m)^*$.

In fact, the subfield of $GF(p^n)$ generated by the stabilizer of some subspace $V_0$ will be the largest subfield of $GF(p^n)$ such that $V_0$ is a vector space over this field and the stabilizer is then this field's multiplicative group.

Answer 2

Take $\mathbb{F}_{p^{n}}$, and suppose that $m | n$, say $n = mc$. Then we can view $\mathbb{F}_{p^{n}}$ as a $c$-dimensional vector space over $\mathbb{F}_{p^{m}}$. If you think of $k$ being a multiple of $m$ as well, then you can put $k = m d$, so $d \leq c$, and you can ask how many $d$-dimensional subspaces you have of $\mathbb{F}_{p^{m}}^{c}$; this gives the number of subspaces of dimension $k$ in $\mathbb{F}_{p}^{n}$ which are fixed under multiplication by elements of $\mathbb{F}_{p^{m}}$. If we want the number containing $\mathbb{F}_{p^m}$, (embedded as a one-dimensional subspace of $\mathbb{F}_{p^{m}}^{c}$), this is equal to the number of $d-1$ dimensional subspaces of $\mathbb{F}_{p^{m}}^{c-1}$.