If infinitesimal transformations commute why don't the generators of the Lorentz group commute? If infinitesimal transformations commute as proved e.g. on this mathworld.wolfram page, why are the commutators for the generators of the Lorentz group nonzero?
 A: I'll use ordinary rotations as an example. The same reasoning applies for other Lorentz transformations, too.
Suppose that $G_x$ and $G_y$ generate rotations about the $x$- and $y$-axes, respectively. This means that $\exp(\theta G_x)$ is a rotation through angle $\theta$ about the $x$-axis, and $\exp(\phi G_y)$ is a rotation through angle $\phi$ about the $y$-axis. Those rotations don't commute with each other, so $G_x$ and $G_y$ must not commute with each other, either.
When the linked website (http://mathworld.wolfram.com/InfinitesimalRotation.html) talks about the "commutativity of infinitesimal transformations," it means that if the angles $\theta$ and $\phi$ both have infinitesimal magnitude $\epsilon\ll 1$, then the rotations $\exp(\theta G_x)$ and $\exp(\phi G_y)$ commute with each other to first order in $\epsilon$. This is true even though the generators $G_x$ and $G_y$ don't commute with each other, because the terms in the composite rotation $\exp(\theta G_x)\exp(\phi G_y)$ that involve products of $G_x$ and $G_y$ are of order $\epsilon^2\ll\epsilon$, so the non-commutativity of the generators $G_x$ and $G_y$ doesn't affect things at first order in $\epsilon$.
A: *

*Illustrative example: It is straightforward to prove that the Lie group
$$ SO(3)~:=~\{ M\in {\rm Mat}_{3\times 3}(\mathbb{R}) \mid M^tM=\mathbb{1}_{3\times 3}, ~\det(M)=1\} $$ 
of 3D rotations is generated by the corresponding Lie algebra
$$ so(3)~:=~\{ m\in {\rm Mat}_{3\times 3}(\mathbb{R}) \mid m^t=-m\} $$
of real $3\times 3$ antisymmetric matrices, which clearly do not all commute. 

*Concretely notice how the mathworld.wolfram page in eqs. (2)-(5) omits the second-order terms, whose difference reveals the commutator!
A: While 
\begin{align}
e^{\epsilon A} e^{\epsilon B}&=(1+\epsilon A+\textstyle\frac{1}{2}\epsilon^2 A^2+\ldots)(1+\epsilon B+\frac{1}{2}\epsilon^2 B^2+\ldots)\tag{1}\, ,\\
&= 1+ \epsilon (A+B)+ \frac{1}{2}
\epsilon^2 (A^2+AB +B^2)+\ldots 
\end{align}
we have 
\begin{align}
e^{\epsilon B} e^{\epsilon A}&=(1+\epsilon B+\frac{1}{2}\epsilon^2 B^2+\ldots)(1+\epsilon A+\textstyle\frac{1}{2}\epsilon^2 A^2+\ldots)\, ,\tag{2}\\
&= 1+ \epsilon (B+A)+ \frac{1}{2}
\epsilon^2 (A^2+BA +B^2)+\ldots 
\end{align}
so: 


*

*To order $\epsilon$, the transformations are the same,

*To order $\epsilon^2$ they are different.


Since the commutator $[A,B]$ involves products like $AB$ and $BA$, one must compare terms in $\epsilon^2$ in (1) and (2) to see that they differ by a commutator.
