Question on GL(n,F) representation Let A be the group of all invertible n x n matrices over F, A+/- the subgroups of all upper/lower matrices. 


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*F^n as an A-module is irreducible? Is this because F^n has only one orbit under A?

*Why is F^n indecomposable but not irreducible as an A+ or A- module? Is this statement equivalent to some theorem in linear algebra?
Thanks.
 A: This is for part $2$. 
First, we have the following $A^+$-submodules of $F^n$
$$V_k = \textrm{span}\{ e_i \ | \ 1\le i \le k\}$$
for all $0\le k \le n$, with $V_0 = (0)$, $V_n = F^n$ and $\dim V_k = k$ for all $k$. We will prove there are no other $A^+$ submodules.  Consider the following element of $A^+$, $g= I +n$, where $n(e_i) = e_{i-1}$, $n( e_1) = 0$. In fact, we can show that any subspace $W$ of $F^n$ invariant under $g$ equals one of the $V_k$. Indeed, let $k$  be the smallest index so that $V_k \supset W$. Then $W$ contains an element of the form $v = a_k e_k + a_{k-1} e_{k-1} + \ldots + a_1 e_1$, with $a_k \ne 0$. Now every subspace invariant under $g = I +n$ is also invariant under $n$. Therefore $v$,$n(v)$, $n^2(v)$, $\ldots$, $n^{k-1}(v)$ are in $W$. But these $k$ vectors form a basis of $V_k$. Therefore $W\supset V_k$ and so $W=V_k$.
We have determined all the $A^{+}$-submodules of $F^n$. We see that they are totally ordered by inclusion and so they cannot decompose $F^n$ in a nontrivial way. 
