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(EDIT) Just finished this problem (answered). I would appreciate a quick check on my work!

Here is the problem.

Let $F$ be a field and let $G$ be the subgroup of $GL_n(F)$ that stabilizes the standard basis vector $\begin{bmatrix} 1\\0\\...\\0 \end{bmatrix}$

  1. Show that $G$ has a subgroup $H$ isomorphic to $GL_{n-1}(F)$
  2. Show that $G$ has a normal subgroup $N$ isomorphic to the additive group $F^{n-1}$
  3. Show that $G$ is the semidirect product $N$ ⋊ $H$.
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What does "stabilizes" mean here? That $\,Gv= v\,$ , with $\,v\,$ that standard vector, or that the vector is an eigenvector? – DonAntonio Nov 28 '12 at 5:13
$G = stab(e_1) \le GL_n(F)$ – KUSH Nov 28 '12 at 5:18
As in the orbit/stabilizer theorem – KUSH Nov 28 '12 at 5:18
OK, thanks..... – DonAntonio Nov 28 '12 at 5:20
up vote 0 down vote accepted

Solution to 1

Let $H=\left\{ \begin{bmatrix} 1 & 0 & 0 & ...\\0 \\... & & M\\0 \end{bmatrix} : M \in GL_{n-1}(F) \right\}$

Take $a,b \in H$, say $M=A$ for $a$ and $M=B$ for $b$. Now $ab^{-1}= \begin{bmatrix} 1 & 0 & 0 & ...\\0 \\... & & AB^{-1}\\0 \end{bmatrix} \in H$

So by abbreviated subgroup test, $H \le G$. Now let $\phi:H \rightarrow GL_{n-1}(F)$ where phi takes $a\mapsto A$. This is clearly a bijection since the extra entries are just 0 and 1 and are always the same.

$\phi(ab) = \phi \bigg( \begin{bmatrix} 1 & 0 & 0 & ...\\0 \\... & & AB\\0 \end{bmatrix} \bigg) = \begin{bmatrix} AB \end{bmatrix} =\begin{bmatrix} A \end{bmatrix}\begin{bmatrix} B \end{bmatrix} = \phi(a)\phi(b)$. So yes $\phi$ is an isomorphism.

So $H \cong GL_{n-1}(F)$. $\Box$

Solution to 2

Let $N=\left\{ \begin{bmatrix} 1 & x_1 & x_2 & ...\\0 \\... & & I_{n—1}\\0 \end{bmatrix} : x_i \in F \forall i \right\}$

Note: $X = \begin{bmatrix} x_1 &x_2 &... \end{bmatrix} \Rightarrow X \in F^{n—1}$

I figured it out. Showing subgroup straightforward. Normalcy is also not bad.

Take $\psi:N\rightarrow F^{n-1}$ where $n\in N\mapsto X$. Bijection. $\psi(ab)=[A+B]=[A]+[B]=\psi(a)+\psi(b)$. $\Box$

Solution to 3

First of all, $H \bigcap N = I_n$ is pretty obvious by the construction.

Next, $NH = \begin{bmatrix} 1 & x_1 & x_2 & ...\\0 \\... & & M \\0 \end{bmatrix}$ is the entirety of $G$.

Clearly $NH \subset G$ and $NH \supset G$ $\Rightarrow G=NH$

Then, since $H \bigcap N = I_n$ and $G = NH$, we have 3 done by my professor's construction of semi-direct product.

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