A question about the proof of a theorem in Representation theory of groups

Question

My Question is about one part of the proof of theorem in the book "A Course in the Theory of Groups" by Derek J.S. Robinson. I highlight the part that my question is about.

We know that if $G$ is a subgroup of $GL(n, F)$, then of course the inclusion $G\hookrightarrow GL(n, F)$ is a matrix representation of $G$ over $F$, and $G$ may be called reducible, irreducible etc. according as this representation has the property stated.

Theorem 8.1.10: Let $G$ be a subgroup of $GL(n, F)$ where F is any field. If every element of $G$ is unipotent, then $G$ is conjugate to a subgroup of $U(n, F)$, the group of all upper unitriangular matrices.

Proof: Let G act on a vector space $V$ of dimension $n$. It is enough to prove that there is a series of FG-submodules $0=V_1 \le V_2 \le ... \le V_k=V$ such that G operates trivialy on $\frac{V_{i+1}}{V_i}$. For then, on choosing suitable bases for the $V_i$, we can represent the elements of $G$ by unitriangular matrices.

Suppose first that F is algebraically closed. We may assume that $G$ is irreducible, otherwise induction on $n$ yields the result.((I don't know how and what can be the start of induction??)) By hypothesis the trace of every element of $G$ equals $n$, so 8.1.9 may be applied to give $|G| = 1$.

Now suppose that F is not necessarily algebraically closed and write $\overline{F}$ for its algebraic closure. Let $\overline{V}=\overline{F}\otimes_F V$ and view this as an $\overline{F}G$-module. By the last paragraph there is a series of $\overline{F}G$-modules $0=\overline{V}_0 \le \overline{V}_1 \le ... \le \overline{V}_k=\overline{V}$ with $\frac{\overline{V}_{i+1}}{\overline{V}_i}$ a trivial module. We can identify $a$ in $V$ with $1\otimes a$ in $\overline{V}$, so that $V \le \overline{V}$, and define $V_i=V\cap \overline{V}_i$. Then the $V_i$ form a series of the required type.

Answer 1

The idea is as follows. If $n = 1$, there is not much to prove. Let us assume the claim holds for all subgroups of $GL(n',F)$ for $n' < n$ (induction hypothesis). If $G \subseteq GL(n,F)$ is now reducible then it is conjugate to a subgroup of $\left\{\left(\matrix{ A & C \\ 0 & B }\right): A \in GL(a,F), B \in GL(b,F), C \in F^{a \times b}\right\}$ for some $a,b > 0 $ such that $a + b = n$. Now we can use our induction hypothesis on the images of $G$ (or more precisely, of its conjugate) in $GL(a,F)$ and $GL(b,F)$ (note that these images are subgroups consisting only of unipotent elements) and get that these are conjugate to subgroups of the unitriangular matrices in $GL(a,F)$ and $GL(b,F)$, respectively. This makes $G$ conjugate to a subgroup of the unitriangular matrices of $GL(n,F)$.

Answer 2

Maybe the theorem should be restated as follows to make the induction clearer:

Theorem: Let $G$ be a group and $\rho: G\to \text{GL}_F(V)$ be a representation of $G$ on a finite-dimensional $F$-vector space $V$. Assume further that for any $g\in G$, its action $\rho(g)\in\text{GL}_F(V)$ is unipotent. Then there exists a $G$-stable flag $0=V_0\subset V_1\subset ...\subset V_k=V$ in $V$ such that $G$ acts trivially on $V_{i+1}/V_{i}$ for all $0\leq i<k$.

Once we know this, we may in particular apply it to faithful unipotent representations, i.e. elementwise unipotent subgroups $G\hookrightarrow\text{GL}(V)$, and get the desired base change.

Proof by induction on $n$. If $n=0$, the claim is trivial. If $n>0$, we may assume $\rho$ is irreducible, as otherwise we may consider the composition factors of the $G$-representation $V$ independently, and each of them has dimension smaller $n$. Now the rest of the proof stays the same. For the application of Theorem 8.1.9., you have to replace $G$ by $\rho(G)$, which is an irreducible subgroup of $\text{GL}_F(V)$ by the very definition of irreducible subgroups.

(After all, the induction is not necessary: What you prove is that the only irreducible, unipotent representation is the trivial one, and hence any composition series for your original representation will have the desired property)

Answer 3

If the given representation of G is reducible, we can conjugate (i.e. transform) it into upper block triangular form where the diagonal blocks are irreducible. Then, if we can prove that every irreducible representation is conjugate to an upper unitriangular representation, we can apply this fact to every diagonal block in the upper block triangular form of the original representation and conjugate it into upper unitriangular form. But an upper block triangular form where each diagonal block is upper unitriangular is itself upper unitriangular. So we have that the original representation, too, can be conjugated into upper unitriangular form. Does this help?