Block Levi decomposition of standard parabolics

$L_{\alpha }$ is a subgroup of $P_{\alpha }$: it contains $I_n$; if $l,l^{\mathrm{\prime }}\in L_{\alpha }$ then $l{l}^{\mathrm{\prime }}\in L_{\alpha }$ and $l^{-1}\in L_{\alpha }$, because block diagonal matrices of the given block sizes multiply and invert blockwise by [L3], the inverse of an invertible block diagonal matrix having the inverted diagonal blocks.

The diagonal block map $\phi :P_{\alpha }\to L_{\alpha }$, $\phi (p):=\mathrm{diag}(A_{11},\dots ,A_{rr})$ formed from the diagonal blocks $A_{ii}$ of $p$, is a well-defined group homomorphism. It is well defined because $p$ is invertible and each diagonal block is invertible: since $p^{-1}\in P_{\alpha }$ by [L1], write its diagonal blocks as $B_{ii}$. The block product rule [L3] applied to $p{p}^{-1}=p^{-1}p=I_n$ gives $A_{ii}{B}_{ii}=B_{ii}{A}_{ii}=I_{a_i}$, so each $A_{ii}$ is invertible and belongs to ${\mathrm{GL}}_{a_i}({\mathbb{F}}_q)$ by [L4], so each diagonal block lies in the group ${\mathrm{GL}}_{a_i}({\mathbb{F}}_q)$ of [L5]. It is a homomorphism because the diagonal blocks of a product of upper block triangular matrices are the products of the diagonal blocks, by [L3].

The kernel of $\phi$ is $U_{\alpha }$: a matrix $p\in P_{\alpha }$ satisfies $\phi (p)=I_n$ exactly when $p_{kk}=1$ for every $k$ and $p_{kl}=0$ for $k\mathrm{\ne }l$ with $\mathrm{blk}(k)\ge \mathrm{blk}(l)$, which is the definition of $U_{\alpha }$; since $\phi$ is a homomorphism by step 1.2, this makes $U_{\alpha }$ a subgroup of $P_{\alpha }$ and $U_{\alpha }⊴P_{\alpha }$.

The map $\phi :P_{\alpha }\to L_{\alpha }$ is surjective, since $\phi (l)=l$ for every $l\in L_{\alpha }$; thus inclusion $L_{\alpha }\hookrightarrow P_{\alpha }$ is a section of $\phi$. Separately, the block-extraction map $\psi :L_{\alpha }\to {\prod }_{i=1}^r{\mathrm{GL}}_{a_i}({\mathbb{F}}_q)$, $\psi (l)=(A_{11},\dots ,A_{rr})$, is a group homomorphism by [L3]. Its inverse is the assembly map $(g_1,\dots ,g_r)\mapsto \mathrm{diag}(g_1,\dots ,g_r)$: both composites are identities by inspection of the blocks. This proves the asserted isomorphism for $L_{\alpha }$, while $\ker \phi =U_{\alpha }$ remains as in step 2.1.

Every $p\in P_{\alpha }$ factors as $p=\phi (p)\cdot \phi (p{)}^{-1}p$ with $\phi (p)\in L_{\alpha }$ by step 3.1 and $\phi (p{)}^{-1}p\in U_{\alpha }$ by step 2.1, so $P_{\alpha }=L_{\alpha }{U}_{\alpha }$; if $p=lu=l^{\mathrm{\prime }}{u}^{\mathrm{\prime }}$ with $l,l^{\mathrm{\prime }}\in L_{\alpha }$ and $u,u^{\mathrm{\prime }}\in U_{\alpha }$, then $l^{-1}{l}^{\mathrm{\prime }}=u(u^{\mathrm{\prime }}{)}^{-1}$ lies in $L_{\alpha }\cap U_{\alpha }$, and a block diagonal matrix in $U_{\alpha }$ has $u_{kk}=1$ and $u_{kl}=0$ for $k\mathrm{\ne }l$, hence equals $I_n$, so $l=l^{\mathrm{\prime }}$ and $u=u^{\mathrm{\prime }}$. Therefore the multiplication map $L_{\alpha }\times U_{\alpha }\to P_{\alpha }$ is a bijection and, with $L_{\alpha }\cap U_{\alpha }=\{I_n\}$ and $U_{\alpha }⊴P_{\alpha }$ from step 2.1, the group $P_{\alpha }$ is the internal semidirect product $P_{\alpha }=L_{\alpha }⋉U_{\alpha }$.

The extreme cases: at $\alpha =(n)$ there is one block, so $P_{(n)}=G$ by [L1] and every matrix is block diagonal of type $(n)$, whence $L_{(n)}=G$ and $U_{(n)}=\{I_n\}$; at $\alpha =(1^n)$ every block is a singleton, so $L_{(1^n)}$ is the set of diagonal matrices in $G$ and $U_{(1^n)}$ the set of unitriangular matrices, that is $L_{(1^n)}=T$ and $U_{(1^n)}=U$ by [L2], with $P_{(1^n)}=B$ and $B=T⋉U$. ∎