Is a connected algebraic group of dimension 2 necessarily commutative?

Question

There is a conclusion for a connected algebraic group $G$:

If dim$G \leq 2$, then $G$ is solvable.

I am wondering whether the stronger statement is true:

If dim$G \leq 2$, then $G$ is commutative.

If dim$G=0$, then $G=\{ e\}$. This is clearly commutative.

If dim$G=1$, then $G \cong G_a$ or $G_m$, thus is commutative.

If dim$G=2$, suppose that $x_1, x_2$ span the vector space containing $G$ over $K$. In $G$, the identity element $e =ax_1+bx_2$, $a,b \in K$ and are not both $0$. Then,

$x_1=ex_1=(ax_1+bx_2)x_1=ax_1^2+bx_2x_1$, (1)

$x_1=x_1e=x_1(ax_1+bx_2)=ax_1^2+bx_1x_2$, (2)

$x_2=ex_2=(ax_1+bx_2)x_2=ax_1x_2+bx_2^2$, (3)

$x_2=x_2e=x_2(ax_1+bx_2)=ax_2x_1+bx_2^2$. (4)

From (1) and (2), $b(x_1x_2-x_2x_1)=0$. Similarly, $a(x_1x_2-x_2x_1)=0$ from (3) and (4). So, if $x_1x_2 \neq x_2x_1$, then $a=b=0$, contradiction. And $G$ is commutative from the commutativity of $x_1$ and $x_2$.

Is this alright?

Thank you very much.