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If G is a linear group, then G has the largest normal subgroup consisting of unipotent matrices. This is called the unipotent radical of G. If G is brought by conjugation in Mn(K) to a block diagonal form with irreducible diagonal blocks, then the unipotent radical of G is the kernel of the projection onto the diagonal blocks. How to understand "If G is brought by conjugation in Mn(K) to a block diagonal form with irreducible diagonal blocks, then the unipotent radical of G is the kernel of the projection onto the diagonal blocks"? What does projection mean? Thanks.

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You can replace "brought by conjugation to" with "conjugate to" : Up to conjugacy, you can write $G$ as a group of block upper triangular, that is of the form

$ \left( \begin{matrix} A_1 & * & \ldots \\ & \ddots &*\\ 0& & A_r \end{matrix} \right) $

such that the blocks $A_i$ on the diagonal are irreducible (meaning you cannot get smaller blocks if you conjugate). You get such a decomposition by looking at the action of $G \subset \textrm{GL}_n$ on $K^n$ (you need a maximal sequence $\{0\} \subsetneq V_1 \subsetneq \ldots \subsetneq V_r = K^n$ of $G$-stable subspaces).

The unipotent radical is given by matrices in the group where all blocks are identity matrices.

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@Cohen, thank you very much. But it seems that some unipotent matrices are not upper triangular. Some matrices which are unipotent (so they are in the unipotent radical) but they are not the matrices given in your answer. –  LJR May 11 '11 at 1:22
    
The unipotent radical does NOT contain all unipotent matrices. It's the biggest NORMAL subgroup of $G$ consisting only of unipotents elements. Take for example, $G = \textrm{GL}_n$ or $\textrm{SL}_n$, then it's irreducible (it's not conjugate to a smaller block upper triangular matrices subgroup). So the decomposition reduces to just one block ($G$ itself). The result just states that's there is no (non trivial) normal subgroup of unipotent matrices although there are plenty of unipotent matrices (we say $G$ is reductive). In the general case, the blocks that appear are also reductive. –  Joel Cohen May 11 '11 at 2:59
    
@Cohen, thanks. –  LJR May 11 '11 at 3:01

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