Problem 5.5.22 (Group of upper triangular matrices as semidirect product)

Proof. Let $G$ be the group of upper triangular matrices in ${\mathrm{GL}}_n(F)$.

(a)

If $$\begin{array}{llll}\hfill & P=\left(\begin{array}{ccccc}\hfill d_1\hfill & \hfill \ast \hfill & \hfill \hfill & \hfill \cdots \hfill & \hfill \ast \hfill \\ \hfill 0\hfill & \hfill \ddots \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill ⋮\hfill & \hfill \ddots \hfill & \hfill d_i\hfill & \hfill \ddots \hfill & \hfill ⋮\hfill \\ \hfill ⋮\hfill & \hfill \hfill & \hfill \hfill & \hfill \ddots \hfill & \hfill \ast \hfill \\ \hfill 0\hfill & \hfill \cdots \hfill & \hfill \hfill & \hfill 0\hfill & \hfill d_n\hfill \end{array}\right)\in G,A=\left(\begin{array}{ccccc}\hfill 1\hfill & \hfill \ast \hfill & \hfill \hfill & \hfill \cdots \hfill & \hfill \ast \hfill \\ \hfill 0\hfill & \hfill \ddots \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill ⋮\hfill & \hfill \ddots \hfill & \hfill 1\hfill & \hfill \ddots \hfill & \hfill ⋮\hfill \\ \hfill ⋮\hfill & \hfill \hfill & \hfill \hfill & \hfill \ddots \hfill & \hfill \ast \hfill \\ \hfill 0\hfill & \hfill \cdots \hfill & \hfill \hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right)\in U,& \hfill & \\ \hfill & & \hfill \end{array}$$

where * is for any element in $F$, then

$$\begin{array}{llll}\hfill \mathit{𝑃𝐴}{P}^{-1}& =\left(\begin{array}{ccccc}\hfill d_1\hfill & \hfill \ast \hfill & \hfill \hfill & \hfill \cdots \hfill & \hfill \ast \hfill \\ \hfill 0\hfill & \hfill \ddots \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill ⋮\hfill & \hfill \ddots \hfill & \hfill d_i\hfill & \hfill \ddots \hfill & \hfill ⋮\hfill \\ \hfill ⋮\hfill & \hfill \hfill & \hfill \hfill & \hfill \ddots \hfill & \hfill \ast \hfill \\ \hfill 0\hfill & \hfill \cdots \hfill & \hfill \hfill & \hfill 0\hfill & \hfill d_n\hfill \end{array}\right)\left(\begin{array}{ccccc}\hfill 1\hfill & \hfill \ast \hfill & \hfill \hfill & \hfill \cdots \hfill & \hfill \ast \hfill \\ \hfill 0\hfill & \hfill \ddots \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill ⋮\hfill & \hfill \ddots \hfill & \hfill 1\hfill & \hfill \ddots \hfill & \hfill ⋮\hfill \\ \hfill ⋮\hfill & \hfill \hfill & \hfill \hfill & \hfill \ddots \hfill & \hfill \ast \hfill \\ \hfill 0\hfill & \hfill \cdots \hfill & \hfill \hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right)\left(\begin{array}{ccccc}\hfill d_1^{-1}\hfill & \hfill \ast \hfill & \hfill \hfill & \hfill \cdots \hfill & \hfill \ast \hfill \\ \hfill 0\hfill & \hfill \ddots \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill ⋮\hfill & \hfill \ddots \hfill & \hfill d_i^{-1}\hfill & \hfill \ddots \hfill & \hfill ⋮\hfill \\ \hfill ⋮\hfill & \hfill \hfill & \hfill \hfill & \hfill \ddots \hfill & \hfill \ast \hfill \\ \hfill 0\hfill & \hfill \cdots \hfill & \hfill \hfill & \hfill 0\hfill & \hfill d_n^{-1}\hfill \end{array}\right)& \hfill & \\ \hfill & =\left(\begin{array}{ccccc}\hfill 1\hfill & \hfill \ast \hfill & \hfill \hfill & \hfill \cdots \hfill & \hfill \ast \hfill \\ \hfill 0\hfill & \hfill \ddots \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill ⋮\hfill & \hfill \ddots \hfill & \hfill 1\hfill & \hfill \ddots \hfill & \hfill ⋮\hfill \\ \hfill ⋮\hfill & \hfill \hfill & \hfill \hfill & \hfill \ddots \hfill & \hfill \ast \hfill \\ \hfill 0\hfill & \hfill \cdots \hfill & \hfill \hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right)\in U,& \hfill & \end{array}$$

so $U⊴G$.

By definition of $U$ and $D$,

$$U\cap D=\{I\}.$$

If

$$\begin{array}{lll}\hfill M=\left(\begin{array}{ccccc}\hfill a_{1,1}\hfill & \hfill \cdots \hfill & \hfill a_{1,i}\hfill & \hfill \cdots \hfill & \hfill a_{1,n}\hfill \\ \hfill 0\hfill & \hfill \ddots \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill ⋮\hfill & \hfill \ddots \hfill & \hfill a_{i,i}\hfill & \hfill \cdots \hfill & \hfill a_{i,n}\hfill \\ \hfill ⋮\hfill & \hfill \hfill & \hfill \hfill & \hfill \ddots \hfill & \hfill \hfill \\ \hfill 0\hfill & \hfill \cdots \hfill & \hfill \hfill & \hfill 0\hfill & \hfill a_{n,n}\hfill \end{array}\right)& & \hfill \\ \hfill & & \hfill \end{array}$$

is any matrix in $G$, then $M=(M{D}^{-1})D$, where $D=\mathrm{diag}(a_{1,1},\dots ,a_{n,n})\in D$ and $M{D}^{-1}\in U$. Explicitly,

$$\begin{array}{llll}\hfill M& =\left(\begin{array}{cccccc}\hfill 1\hfill & \hfill a_{12}{a}_{22}^{-1}\hfill & \hfill \cdots \hfill & \hfill a_{1,i}{a}_{i,i}^{-1}\hfill & \hfill \cdots \hfill & \hfill a_{1,n}{a}_{n,n}^{-1}\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill \ddots \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill ⋮\hfill & \hfill \hfill & \hfill \ddots \hfill & \hfill 1\hfill & \hfill \cdots \hfill & \hfill a_{i,n}{a}_{n,n}^{-1}\hfill \\ \hfill ⋮\hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \ddots \hfill & \hfill \hfill \\ \hfill 0\hfill & \hfill \hfill & \hfill \cdots \hfill & \hfill \hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right)\left(\begin{array}{ccccc}\hfill a_{1,1}\hfill & \hfill 0\hfill & \hfill \cdots \hfill & \hfill \cdots \hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \ddots \hfill & \hfill \hfill & \hfill \hfill & \hfill ⋮\hfill & \hfill \hfill \\ \hfill ⋮\hfill & \hfill \ddots \hfill & \hfill a_{i,i}\hfill & \hfill \ddots \hfill & \hfill \hfill \\ \hfill ⋮\hfill & \hfill \hfill & \hfill \hfill & \hfill \ddots \hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \cdots \hfill & \hfill \hfill & \hfill 0\hfill & \hfill a_{n,n}\hfill \end{array}\right)\in \mathit{𝑈𝐷},& \hfill & \\ \hfill & & \hfill \end{array}$$

so $G=\mathit{𝑈𝐷}$.

Since

• $U⊴G$,
• $U\cap D=\{I\}$,
• $G=\mathit{𝑈𝐷}$,

by Theorem 9,

$$G\simeq U{⋊}_{\phi }D,$$

where $\phi :D\to \mathrm{Aut}(U)$ is the homomorphism associated to the action by conjugation.

(b)

Let $n=2$. By Exercise 3.1.11, $$\begin{array}{llll}\hfill & \sigma \{\begin{array}{ccc}\hfill F\hfill & \hfill \to \hfill & U\hfill \\ \hfill \alpha \hfill & \hfill \mapsto \hfill & \left(\begin{array}{cc}\hfill 1\hfill & \hfill \alpha \hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right)\hfill \end{array}\tau \{\begin{array}{ccc}\hfill F^{\times }\times F^{\times }\hfill & \hfill \to \hfill & D\hfill \\ \hfill (\gamma ,\delta )\hfill & \hfill \mapsto \hfill & \left(\begin{array}{cc}\hfill \gamma \hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \delta \hfill \end{array}\right)\hfill \end{array}& \hfill & \end{array}$$

are isomorphisms, so that $F^{\times }\times F^{\times }\simeq D$ and $F\simeq U$.

By part (a), $G\simeq U{⋊}_{\phi }D$, where

$$\phi \{\begin{array}{ccc}\hfill D\hfill & \hfill \to \hfill & \mathrm{Aut}(U)\hfill \\ \hfill \left(\begin{array}{cc}\hfill \gamma \hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \delta \hfill \end{array}\right)\hfill & \hfill \mapsto \hfill & {\phi }_{\gamma ,\delta }\{\begin{array}{ccc}\hfill U\hfill & \hfill \to \hfill & U\hfill \\ \hfill \left(\begin{array}{cc}\hfill 1\hfill & \hfill \alpha \hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right)\hfill & \hfill \mapsto \hfill & \left(\begin{array}{cc}\hfill \gamma \hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \delta \hfill \end{array}\right)\left(\begin{array}{cc}\hfill 1\hfill & \hfill \alpha \hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right){\left(\begin{array}{cc}\hfill \gamma \hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \delta \hfill \end{array}\right)}^{-1}\hfill \end{array}\hfill \end{array}$$

If $u_{\alpha }=\left(\begin{array}{cc}\hfill 1\hfill & \hfill \alpha \hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right)$, then $\phi (\mathrm{diag}(\gamma ,\delta ))={\phi }_{\gamma ,\delta }$, and

$$\begin{array}{llll}\hfill {\phi }_{\gamma ,\delta }(u_{\alpha })& =\left(\begin{array}{cc}\hfill \gamma \hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \delta \hfill \end{array}\right)\left(\begin{array}{cc}\hfill 1\hfill & \hfill \alpha \hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right){\left(\begin{array}{cc}\hfill \gamma \hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \delta \hfill \end{array}\right)}^{-1}& \hfill & \\ \hfill & =\left(\begin{array}{cc}\hfill 1\hfill & \hfill \mathit{𝛼𝛾}{\delta }^{-1}\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right)& \hfill & \\ \hfill & =u_{\mathit{𝛼𝛾}{\delta }^{-1}}.& \hfill & \end{array}$$

Consider

$${\phi }^{\prime }\{\begin{array}{ccc}\hfill F^{\times }\times F^{\times }\hfill & \hfill \to \hfill & \mathrm{Aut}(F)\simeq F^{\times }\hfill \\ \hfill (\alpha ,\gamma )\hfill & \hfill \to \hfill & {\phi }_{\gamma ,\delta }^{\prime }\{\begin{array}{ccc}\hfill F\hfill & \hfill \to \hfill & F\hfill \\ \hfill \alpha \hfill & \hfill \mapsto \hfill & \mathit{𝛼𝛾}{\delta }^{-1}\hfill \end{array}\hfill \end{array}$$

By the proof of Theorem B in the Appendix, where $H=U,K=D,H^{\prime }=F,K^{\prime }=F^{\times }\times F^{\times },f={\sigma }^{-1},g={\tau }^{-1}$,

$$G\simeq U{⋊}_{\phi }F\simeq F{⋊}_{{\phi }^{\prime }}(F^{\times }\times F^{\times }).$$

Indeed, consider ${\chi }_f\in \mathrm{Aut}(U)$ defined by ${\chi }_f(\psi )=f\circ \psi \circ f^{-1}$, and $\lambda :U{⋊}_{\phi }D\to F{⋊}_{{\phi }^{\prime }}(F^{\times }\times F^{\times })$ defined for $u=\left(\begin{array}{cc}\hfill 1\hfill & \hfill \alpha \hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right)\in U$ and $d=\mathrm{diag}(\gamma ,\delta )=\left(\begin{array}{cc}\hfill \gamma \hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \delta \hfill \end{array}\right)\in D$ by

$$\begin{array}{llll}\hfill \lambda (u,d)& =(f(u),g(d))& \hfill & \\ \hfill & =({\sigma }^{-1}(u),{\tau }^{-1}(d))& \hfill & \\ \hfill & =(\alpha ,(\gamma ,\delta )).& \hfill & \end{array}$$

Then for all $(\gamma ,\delta )\in K^{\prime }=F^{\times }\times F^{\times }$, and for all $\alpha \in U^{\prime }=F$,

$$\begin{array}{llll}\hfill [({\chi }_f\circ \phi \circ g^{-1})(\gamma ,\delta )](\alpha )& =\left\{{\chi }_f[\phi (\mathrm{diag}(\gamma ,\delta ))]\right\}(\alpha )& \hfill & \\ \hfill & =[{\chi }_f({\phi }_{\gamma ,\delta })](\alpha )& \hfill & \\ \hfill & =(f\circ {\phi }_{\gamma ,\delta }\circ f^{-1})(\alpha )& \hfill & \\ \hfill & =f({\phi }_{\gamma ,\delta }(u_{\alpha }))& \hfill & \\ \hfill & =f(u_{\mathit{𝛼𝛾}{\delta }^{-1}})& \hfill & \\ \hfill & =\mathit{𝛼𝛾}{\delta }^{-1}& \hfill & \\ \hfill & =[{\phi }^{\prime }(\gamma ,\delta )](\alpha ).& \hfill & \end{array}$$

Therefore

$${\phi }^{\prime }={\chi }_f\circ \phi \circ g^{-1}.$$

By the proof of Theorem B of the Appendix, $\lambda$ is an isomorphism, and so

$$G=U{⋊}_{\phi }D\simeq F{⋊}_{{\phi }^{\prime }}(F^{\times }\times F^{\times }),$$

where ${\phi }^{\prime }={\chi }_f\circ \phi \circ g^{-1}$ is the homomorphism

$${\phi }^{\prime }\{\begin{array}{ccc}\hfill F^{\times }\times F^{\times }\hfill & \hfill \to \hfill & \mathrm{Aut}(F)\simeq F^{\times }\hfill \\ \hfill (\alpha ,\gamma )\hfill & \hfill \to \hfill & {\phi }_{\gamma ,\delta }^{\prime }\{\begin{array}{ccc}\hfill F\hfill & \hfill \to \hfill & F\hfill \\ \hfill \alpha \hfill & \hfill \mapsto \hfill & \mathit{𝛼𝛾}{\delta }^{-1}\hfill \end{array}\hfill \end{array}$$

Proof. To each $\psi \in \mathrm{Aut}(H)$, we can associate ${\psi }^{\prime }\in \mathrm{Aut}(H^{\prime })$ making the following diagram commutative:

namely ${\psi }^{\prime }=f\circ \psi \circ f^{-1}$.

This allows us to define

If ${\xi }_{f^{-1}}:\mathrm{Aut}(H^{\prime })\to \mathrm{Aut}(H)$ is the map defined by ${\xi }_{f^{-1}}({\psi }^{\prime })=f^{-1}\circ {\psi }^{\prime }\circ f$, then, for all $\psi \in \mathrm{Aut}(H)$ and all ${\psi }^{\prime }\in \mathrm{Aut}(H^{\prime })$,

therefore ${\xi }_{f^{-1}}\circ {\chi }_f=1_{\mathrm{Aut}(H)},{\chi }_f\circ {\xi }_{f^{-1}}=1_{\mathrm{Aut}(H^{\prime })}$, which shows that ${\chi }_f$ is bijective.

Furthermore, if ${\psi }_1,{\psi }_2\in \mathrm{Aut}(H)$

therefore ${\chi }_f$ is an isomorphism of $\mathrm{Aut}(H)$ on $\mathrm{Aut}(H^{\prime })$.

We can now use ${\chi }_f$ to match each homomorphism $\phi :K\to \mathrm{Aut}(H)$ to a homomorphism ${\phi }^{\prime }:K^{\prime }\to \mathrm{Aut}(H^{\prime })$ making the following diagram commutative:

or ${\phi }^{\prime }={\chi }_f\circ \phi \circ g^{-1}$, which is indeed a homomorphism, because it is composed of three homomorphisms.

Let us now construct the isomorphism $H{⋊}_{\phi }K\to H^{\prime }{⋊}_{{\phi }^{\prime }}{K}^{\prime }.$ Consider

Then $\lambda$ is bijective, of reciprocal $(h^{\prime },k^{\prime })\mapsto (f^{-1}(h),g^{-1}(k))$. Let’s check that $\lambda$ is indeed a homomorphism. If $(a,b),(c,d)\in H{⋊}_{\phi }K$, then

And

It remains to prove $f({\phi }_b(c))={\phi }_{g(b)}^{\prime }(f(c))$, for all $b\in K$ and all $c\in H$, which amounts to showing that $f\circ {\phi }_b={\phi }_{g(b)}^{\prime }\circ f$, i.e.

By definition of ${\phi }^{\prime }$ and ${\chi }_f$, using ${\phi }_b=\phi (b)$,

This completes the proof that $\lambda$ is an isomorphism. □