Exercise 2.1.17 (Subgroups of $\mathrm{GL}_n(F)$)

Proof. Put

• $H\subset {\mathrm{GL}}_n(F)$ by definition, and $I_n\in H$, so $H\ne \varnothing$.

• Let $A={(a_{\mathit{𝑖𝑗}})}_{(i,j)\in {[[1,n]]}^2},B={(b_{\mathit{𝑖𝑗}})}_{(i,j)\in {[[1,n]]}^2}$ be elements of $G$.

  Then $a_{\mathit{𝑖𝑗}}=b_{\mathit{𝑖𝑗}}=0$ for all $i>j$, and $a_{\mathit{𝑖𝑖}}=b_{\mathit{𝑖𝑖}}=1$ for all $i$.

  Put $C=\mathit{𝐴𝐵}\in {\mathrm{GL}}_n(F)$, where $C={(c_{\mathit{𝑖𝑗}})}_{(i,j)\in {[[1,n]]}^2}$. By Exercice 16, $C$ is an upper triangular matrix, so for all $(i,j)\in {[[1,n]]}^2,$

  $$i>j\Rightarrow c_{\mathit{𝑖𝑗}}=0.$$

  Suppose now that $i=j$. Then

  $$c_{\mathit{𝑖𝑖}}=\underset{j=1}{\overset{n}{\sum }}{a}_{\mathit{𝑖𝑗}}{b}_{\mathit{𝑗𝑖}}.$$

  Since $a_{i}j=0$ if $i>j$ and $b_{\mathit{𝑗𝑖}}=0$ if $j>i$, then $a_{\mathit{𝑖𝑗}}{b}_{\mathit{𝑗𝑖}}$ for all $(i,j)\in {[[1,n]]}^2$ such that $i\ne j$. Therefore

  $$c_{\mathit{𝑖𝑖}}=a_{\mathit{𝑖𝑖}}{b}_{\mathit{𝑖𝑖}}=1.$$

  This shows that $C=\mathit{𝐴𝐵}\in H$.

• Let $A={(a_{\mathit{𝑖𝑗}})}_{(i,j)\in {[[1,n]]}^2}\in H\subset {\mathrm{GL}}_n(F)$, and $A^{-1}={(d_{\mathit{𝑖𝑗}})}_{(i,j)\in {[[1,n]]}^2}.$

  By Exercise 16, $A^{-1}$ is an upper triangular matrix, so for all $(i,j)\in {[[1,n]]}^2,$

  $$i>j\Rightarrow d_{\mathit{𝑖𝑗}}=0.$$

  Since $a_{\mathit{𝑖𝑖}}=1$ for all $i$, $\det <!--\; FUNCTION\; APPLICATION\; -->(A)=\underset{i=1}{\overset{n}{\prod <!--\; FUNCTION\; APPLICATION\; -->}}{a}_{\mathit{𝑖𝑖}}=1$.

  For every $i\in [[1,n]]$, the cofactor $c_{\mathit{𝑖𝑖}}$ is given by $c_{\mathit{𝑖𝑖}}={(-1)}^{i+i}=m_{\mathit{𝑖𝑖}}$, where $m_{\mathit{𝑖𝑖}}$ is the corresponding minor, and $d_{\mathit{𝑖𝑖}}=(1∕\det <!--\; FUNCTION\; APPLICATION\; -->(A))c_{\mathit{𝑖𝑖}}=c_{\mathit{𝑖𝑖}}$, so

  $$\begin{array}{llll}\hfill d_{\mathit{𝑖𝑖}}=c_{\mathit{𝑖𝑖}}=m_{\mathit{𝑖𝑖}}& =\left|\begin{array}{cccccc}\hfill a_{1,1}\hfill & \hfill \cdots \hfill & \hfill a_{1,i-1}\hfill & \hfill a_{1,i+1}\hfill & \hfill \cdots \hfill & \hfill a_{1,n}\hfill \\ \hfill 0\hfill & \hfill \ddots \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill & \hfill \hfill \\ \hfill ⋮\hfill & \hfill \ddots \hfill & \hfill a_{i-1,i-1}\hfill & \hfill a_{i-1,i+1}\hfill & \hfill \cdots \hfill & \hfill a_{i-1,n}\hfill \\ \hfill 0\hfill & \hfill \cdots \hfill & \hfill 0\hfill & \hfill a_{i+1,i+1}\hfill & \hfill \cdots \hfill & \hfill a_{i+1,n}\hfill \\ \hfill ⋮\hfill & \hfill \hfill & \hfill \hfill & \hfill \ddots \hfill & \hfill \ddots \hfill & \hfill \hfill \\ \hfill 0\hfill & \hfill \cdots \hfill & \hfill \hfill & \hfill \hfill & \hfill 0\hfill & \hfill a_{n,n}\hfill \end{array}\right|& \hfill & \\ \hfill & =a_{1,1}\cdots a_{i-1,i-1}{a}_{i+1,i+1}\cdots a_{n,n}& \hfill & \\ \hfill & =\underset{k\in [[1,n]],k\ne i}{\prod }{a}_{\mathit{𝑘𝑘}}& \hfill & \\ \hfill & =1.& \hfill & \end{array}$$

  Therefore $A^{-1}\in H$.

In conclusion, $H=\left\{{(a_{\mathit{𝑖𝑗}})}_{(i,j)\in {[[1,n]]}^2}\in {\mathrm{GL}}_n(F)\mid \forall <!--\; FUNCTION\; APPLICATION\; -->(i,j)\in {[[1,n]]}^2,a_{\mathit{𝑖𝑗}}=0\text{ and }\forall <!--\; FUNCTION\; APPLICATION\; -->i\in [[1,n]],a_{\mathit{𝑖𝑖}}=1\right\}$ is a subgroup of ${\mathrm{GL}}_n(F)$. □