TITLE: Does non-commuting $\mathfrak{g}$ imply non-abelian $G$?
QUESTION [10 upvotes]: Question 8.1 in Kristopher Tapp's introductory text on matrix groups asks to show that $SO(n)$ is non-abelian ($n>2$) by finding two elements of $so(n)$ that do not commute.
Why is this method valid?
At least in the text, he only proves that $[A,B]=0$ implies $e^A e^B = e^{A+B} = e^B e^A$, but does not mention whether the converse is true. The latter being true is the only way I could rationalize his statement.
Thanks!

REPLY [3 votes]: The Lie algebra structure on $\def\lie#1{\mathfrak{#1}}\lie{g}$ is defined by the following two-step procedure:

For a Lie group $G$, each element $g \in G$ acts by conjugation on $G$, $\def\on{\operatorname}\on{Ad}_g \colon h \mapsto ghg^{-1}$.  This map $G \to G$ has a differential map on the tangent space at the identity, $d\on{Ad}_g \colon TG_1 \to TG_1$.  By definition, $\lie{g} = TG_1$ and typically $d\on{Ad}_g$ is also written $\on{Ad}_g$, called the adjoint action of $g$ on $\lie{g}$.
The adjoint action is a Lie group action of $G$ on $\lie{g}$; mainly, the aggregate map $\on{Ad} \colon G \times \lie{g} \to \lie{g}$ is smooth.  It, too, has a differential $d\on{Ad} \colon T(G \times \lie{g})_{(1,0)} \to T\lie{g}_0$.  By general principles of differential topology, we have
$$T(G \times \lie{g})_{(1,0)}
 \cong TG_1 \times T\lie{g}_0
 \cong \lie{g} \times \lie{g}$$
where we have used the fact that $\lie{g}$ is a vector space and thus isomorphic to its own tangent space at any point.  Thus, $d\on{Ad}$ is some kind of bivariate map from $\lie{g}$ to itself; in fact, it's the Lie bracket.

It follows from the above definition that if $\on{Ad}_g \colon G \to G$ is the constant function $1$ for all $g \in G$, then ultimately, the Lie bracket is zero.  The former is the case whenever $G$ is abelian.  Taking the contrapositive, if $\lie{g}$ is nonabelian, then $G$ is necessarily as well.