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Let $n>2$ and $q$ be a prime power. I want to find all the embeddings of $\mathrm{GL}(n-1,q)$ into $\mathrm{GL}(n,q)$. An embedding I first thought of is as following. Let $V$ be an $n$-dimensional vector space over $\mathrm{GF}(q)$ and $H$ is a subgroup of $\mathrm{GL}(V)$ fixing a nonzero vector in $V$. Then $H$ is isomorphic to $\mathrm{GL}(n-1,q)$.

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  • $\begingroup$ Judging from your last sentence, I suppose you should correct your second sentence : your question really is “Are all embeddings of $GL(n-1,q)$ into $GL(n,q)$ natural?” not “Can $GL(n−1,q)$ only be embedded into $GL(n,q)$ in the natual way?” $\endgroup$ May 13 '13 at 10:08
  • $\begingroup$ @EwanDelanoy: I agree that the formulation you suggest is clearer than the one in the question, but is the latter really wrong? The only thing I can really find wrong with it is the use of "the": there are of course many natural ways to embed $GL(n-1,q)$ into $GL(n,q)$. The last sentence can be taken to define the notion of "natural" more precisely. $\endgroup$ May 13 '13 at 10:14
  • $\begingroup$ @MarcvanLeeuwen I agree with you, but I believe I’ll not be the only one confused at first by the unclear phrasing. $\endgroup$ May 13 '13 at 11:10
  • $\begingroup$ @EwanDelanoy@MarcvanLeeuwen Sorry for my quite unclear phrasing, I should have raised the qustion as Derek's. $\endgroup$ May 13 '13 at 13:02
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So the question should be: if $H$ is a subgroup of ${\rm GL}(n,q)$ isomorphic to ${\rm GL}(n-1,q)$, then does $H$ necessarily fix two complementary subspaces of $V$ of dimensions $1$ and $n-1$?

This is a nontrivial question. Let $K$ be the field of order $q$. Then the only simple $KH$-modules of dimension less than $n$ have dimensions 1 and $n-1$, so the constituents of $V$ as $KH$-module must have dimensions $1$ and $n-1$, but then the question is whether this module is completely reducible, and that depends on the 1-cohomology of $H$ on the $(n-1)$-dimensional constituent.

There are some instances in which there are embeddings that are not completely reducible, so the answer to the question is no. These include $n=3$, $q=2^k$ for $k \ge 2$, and also $n=4$, $q=2$. I think the answer is yes in all other cases, but I would not bet on it!

Added later: here is a rough description of the non-natural examples when $n=3$ and $q=2^k$. Consider first the subgroup of matrices of the form

$$\left( \begin{array}{ccc}a&b&0\\c&d&0\\x&y&1\end{array} \right).$$

in which $ab-cd=1$. This is a semidirect product of $K^2$ with ${\rm SL}(2,q)$ with the natural action. It turns out that $H^1({\rm SL}(2,2^k),K^2)$ is nonzero, and has dimension 1 over $K$, when $k \ge 2$.

Let $C \cong {\rm SL}(2,q)$ be the a complement of $K^2$ corresponding to a nonzero element of $H^1({\rm SL}(2,q),K^2)$. Then $C$ does not act completely reducibly on $V$, and the subgroup $H$ gerenrated by $C$ and scalar matrices is isomorphic to ${\rm GL}(2,q)$. Then (depending on whether your matrices act on the left or right) $H$ fixes a 1-dimensional or a 2-dimensional subspace of $V$, but not both, and the transpose of $H$ does the opposite. So there are just two conjugacy classes of non-natural ${\rm GL}(2,q)$s in ${\rm GL}(3,q)$.

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  • $\begingroup$ I just edited my question to get rid of the unclear phrasing "natrual" and then see your answer. I should have formulated my question exactly as you state here. However, now that I have generalised the quesion as finding all the embeddings, can we know all the emddings for the case $(n,q)=(3,2^k)$? $\endgroup$ May 13 '13 at 12:56
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If your last sentence is you definition of "natural", then the answer is that other embeddings are possible. Since the ones below follow a simple pattern, you will probably want to extend "natural" to include them as well.

You can always embed $\mathbf{GL}(n-1,F)$ into $\mathbf{GL}(n,F)$ by sending $A$ to the block matrix $$ \begin{pmatrix} A&0\\0&\det A\end{pmatrix}. $$ Unless $F=\mathbf{GF}(2)$, the image of this embedding does not have any fixed vectors.

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