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For my abstract algebra class I have to do some exercises concerning group actions of $GL_n(K)$ on the set of flags $$F_n = \{0 \subseteq V_1 \subseteq V_2 \subseteq \ldots \subseteq V_r = K^n \,\vert\, \text{dim}_K V_i = n_1 + \ldots + n_i\}$$

where $n_1 + \ldots + n_r = n$. Although I got to know the basic things such as orbit and stabilizer, I don't have any practice on how to apply these concepts and, concretely, for these exercise I don't know how to proceed.

The question is threefold and as follows. $\textit{First}$, show that the parabolic subgroup $$P_n = \Bigg\{ \begin{pmatrix} A_1 & B_{12} \ldots & B_{1r} \\ 0 & \ddots & \ddots & \vdots \\ \vdots & \ddots & \ddots & B_{r-1r} \\ 0 & \ldots & 0 & A_r \end{pmatrix}\Bigg\}$$

where $$A_i \in GL_{n_i}, \, 1\leq i \leq r, B_{ij}\in \text{Mat}(n_{i} \times n_{j}, K) \, 1\leq i < j \leq r $$

is a subgroup of $GL_{n}$ by using the group action of $GL_n(K)$ on $F_n$. $\textit{Secondly}$, with the help of this result, show that the corresponding Levi subgroup

$$M_n = \Bigg\{ \begin{pmatrix} A_1 & 0 \ldots & 0 \\ 0 & \ddots & \ddots & \vdots \\ \vdots & \ddots & \ddots & 0 \\ 0 & \ldots & 0 & A_r \end{pmatrix}\Bigg\}$$

where $$A_i \in GL_{n_i}$$ is in turn a subgroup of $P_n$. And $\textit{finally}$, show that the unipotent radical

$$U_n = \Bigg\{ \begin{pmatrix} I_1 & B_{12} \ldots & B_{1r} \\ 0 & \ddots & \ddots & \vdots \\ \vdots & \ddots & \ddots & B_{r-1r} \\ 0 & \ldots & 0 & I_r \end{pmatrix}\Bigg\}$$

where $$I_i = Id_{GL_{n}}, \, 1\leq i \leq r, B_{ij}\in \text{Mat}(n_{i} \times n_{j}, K) \, 1\leq i < j \leq r $$ is a normal subgroup in $P_n$. With that, also prove that $P_n/U_n \cong M_n$

What I know is that the stabilizer for a group G and a set M is the set $G_x = \{g \in G\vert g\cdot x = x\}$ with $x \in M$ and a subgroup of $G$. I would merely guess that in the first or second part I would have to show that along these lines. In short, I do not have a clear conceptual picture and, most of all, I wouldn't know how to write it down properly even in case I had it.

EDIT: Due to Derek Holt's comment, I tried to figure out how to show that $P_n$ is the stabilizer of the flags in which $V_i$ is the subspace spanned by the first $n_1 + \ldots + n_i$ basis vectors. Although I can somehow imagine it with simple block matrices of the kind of

$$\begin{pmatrix} A & B \\ C & D \end{pmatrix} $$

I do not know how to put it on paper. What I tried is to rewrite each vector $x \in K^n$ for all $V_i$ where $i \in \{1, \ldots, n\}$ to

$$ x = \begin{pmatrix} x_1 \\ \vdots \\ \vdots \\ x_n\end{pmatrix} = \begin{pmatrix} x_{V_i} \\ x'_{V_i} \end{pmatrix}$$

where $$ x \in V_i \Leftrightarrow x'_{V_i} = 0$$ Put informally, I think I 'see' that if I left-multiply some vector $x \in V_i \subset F_n$ with some matrix $p \in P_n$, then I should still get some vector $x^* \in V_i \subset F_n$. But how to write that down in a correct way? Also, for illustration purposes I tried to left-multiply the vector $$ x = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ 0 \\ \vdots \\ 0 \end{pmatrix}$$ where $x \in \mathbb{R}^n$ with the given matrix

$$ \begin{pmatrix} A_1 & B_{12} \ldots & B_{1r} \\ 0 & \ddots & \ddots & \vdots \\ \vdots & \ddots & \ddots & B_{r-1r} \\ 0 & \ldots & 0 & A_r \end{pmatrix}$$

in order to show that this really affects the resulting vector $x^*$ only in the first 3 coordinates, i.e. that it still lies in the subspace $V_i \in F_n$ it was before. But I got in some trouble regarding how to actually multiply it with the given block matrices. For example, $A_1$ would be in $GL_{n_1}(K)$, but how would I know what $n_1$ is?

With regard to the third part, I know what a normal subgroup is and how to check it, i.e. $xVx^{-1} \subset V \, \forall x \in P_n$. I already did that with triangular matrices and the triangular matrices where $1$s are on the diagonal. However, here are entire matrices within other matrices, so I assume this complicates things. Hence, what's the idea?

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    $\begingroup$ To start things off, $P_n$ is precisely the stabilizer of the flag in $F_n$ in which $V_i$ is the subspace spanned by the first $n_1 + \cdots + n_i$ vectors in the natural/standard basis of $K^n$. That explains why $P_n$ is a subgroup. $\endgroup$ – Derek Holt May 12 '17 at 20:35
  • $\begingroup$ I tried to work out your hint and edited my question above. As I wrote above, although it seems somehow right, I struggle with the proper notation and how to write it down. $\endgroup$ – Taufi May 13 '17 at 9:57
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    $\begingroup$ The group $U_n$ is the subgroup of $P_n$ consiting of those $g$ that induce the identoty map on each of the quotient spaces $V_{i+1}/V_i$. That explains conceptiually why it is a normal subgroup of $U_n$, althoguh you could of coruse verify that directly. $\endgroup$ – Derek Holt May 13 '17 at 11:21
  • $\begingroup$ Thanks for your further comment. Though before I come to last part, could you clarify once more how to formally prove that $P_n$ is a subgroup? The thoughts I made after your hint are rudimentary at best. Also, how to show that $M_n \leq P_n$? To me, this seems like a group of some kind of neutral elements that map each vector to itself. $\endgroup$ – Taufi May 13 '17 at 11:40
  • $\begingroup$ The elements in $P_n$ are precisely the invertible matrices that map each basis vector in $V_i$ to an element in $V_i$ - because they have $0$ coefficients for all basis elements not in $V_i$. I am not sure how to make that any clearer! The fact that it is a subgroup follows from the fact that it is the stabilizer of a flag - the stabilizer of a flag is a subgroup. $M_n$ is the diagonal direct sum of the groups ${\rm GL}(n_i,K)$ and so is a group. It is clearly a subset of $P_n$. $\endgroup$ – Derek Holt May 13 '17 at 13:23

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