Verification exercise in Lang's Algebra In the section on the simplicity of $PSL_n(F)$. The first line in page 544 (Lang Algebra, Third edition) there is remark that "This is easily seen..." I am afraid, I fail to see how. Can someone tell me how to go about that verification? In detail, the problem is as follows.
Theorem- Let G be a $SL_n(F)$-invariant subgroup of $GL_n(F)$ such that G is not contained in the center of $GL_n(F)$, then $SL_n(F)$ is contained in $ G$.
Proof: Let $A\in G$ s.t there exists $u\neq0$ such that $Au$ is not a scalar multiple of u, say $Au= v$. (Such an A is available because G is not in the center)
Let $u,v$ be contained in a hyperplane $H= kerf$ , where f is a linear functional on $F^n$.  Define T by $T=T_u$, where $T_u (x)= x+f(x)u$ for all $x\in F^n$.
Let $B=ATA^{-1}T^{-1}$.
To show that $B\neq I$, where I is the identity in $GL_n(F)$.
It says that it is easy to verify that $ATA^{-1} \neq T$ by checking RHS and LHS on arbitrary $x\in F^n$. I have tried this but am getting nowhere.
 A: This is missing some context: e.g., I don't understand what an "$\operatorname{SL}_n\left(F\right)$-invariant subgroup of $\operatorname{GL}_n\left(F\right)$" is, and why a hyperplane $H$ containing $u$ and $v$ exists (is $n \geq 3$ a requirement?).
Anyway, I guess I can still answer the concrete question: We shall check that $AT \neq TA$. Indeed, pick $x \in F^n$ such that $f\left(x\right) \neq 0$. (Such $x$ exists, since $\ker F$ is a hyperplane.) Thus, $f\left(x\right) v$ is a nonzero multiple of $v$, and hence not a multiple of $u$ (since $v$ is not a multiple of $u$). In particular, $f\left(x\right) v \neq f\left(Ax\right) u$.
Now,
$\left(AT-TA\right)x = A\underbrace{Tx}_{=x + f\left(x\right) u} - \underbrace{TAx}_{=Ax + f\left(Ax\right) u} = A\left(x + f\left(x\right) u\right) - \left(Ax + f\left(Ax\right) u\right)$
$= Ax + f\left(x\right)Au - Ax - f\left(Ax\right) u = f\left(x\right)\underbrace{Au}_{=v} - f\left(Ax\right) u = f\left(x\right) v - f\left(Ax\right) u \neq 0$
(because $f\left(x\right) v \neq f\left(Ax\right) u$), so that $AT-TA \neq 0$ and thus $AT \neq TA$.
