Lie algebra of dimension 2

Lie algebra of dimension 2

From Humphreys' Introduction to Lie Algebras and Representation Theory:
We can determine (up to isomorphism) all Lie algebra of dimension $2$.
Start with a basis $x,y$ of $L$. Clearly, all products in $L$ yield scalar
multiples of $[xy]$. If these are all $0$, then $L$ is abelian. Otherwise,
we can replace $x$ in the basis by a vector spanning the one-dimensional
space of multiples of the original $[xy]$, and take $y$ to be any other
vector independent of the new $x$. Then $[xy]=ax$ $(a\ne 0)$. Replacing
$y$ by $a^{-1}y$, we finally get $[xy]=x$. Abstractly, therefore, at most
one nonabelian $L$ exists (the reader should check that $[xy]=x$ actually
defines a Lie algebra.)
(1) If we replace $y$ by $a^{-1}y$, we should get
$a^{-1}[xy]=ax\rightarrow [xy]=a^2x$, shouldn't we?
(2) Why does $[xy]=x$ define a Lie algebra? If we take $y=x$, we get
$[xx]=x$. But aren't we supposed to have $[xx]=0$?