Every 1- or 2-dimensional compact, connected Lie group is abelian I know that in the 1-dimensional case the conclusion follows immediately from the fact that there exists a maximal torus (which then has to be the group itself).
In the 2-dimensional case, we also know that there exists a maximal torus, and so the point is to show that it must have dimension 2 (or equivalently, that the dimension cannot be 1). I've been trying to prove this for quite a while, but unfortunately I'm stuck and really don't know what to do.
I understand that this is probably an easy question, but my background is in physics and I'm a beginner in these more abstract topics of mathematics; as such, any help would be greatly appreciated.
 A: The Lie algebra $\mathfrak g$ of a connected compact Lie group $G$ is unimodular: this means that for all $x\in\mathfrak g$ the map $\operatorname{ad}(x):\mathfrak g\to\mathfrak g$ has zero trace. 
Now let $G$ be a compact Lie group of dimension two. Its Lie algebra, by the result above, is unimodular. Now, there is exactly one Lie algebras of dimension two which is unimodular, and it is the abelian one. This means that the universal covering $\tilde G$ of $G$ is an abelian Lie group (indeed, this group $\tilde G$ is a simply connected Lie group with the same Lie algebra as that of the simply connected Lie group $\mathbb R^n$, so the groups $\tilde G$ and $\mathbb R^n$ are in fact isomorphic: this is the correspondence at work), and therefore $G$ is also abelian.
A: EDIT. The argument below can be shortened, following Mariano, if we start out with the universal cover directly. Suppose $H$ is a (noncommutative) connected $2$-dimensional compact Lie group. Its universal cover $\widetilde{H}$ carries a Lie group structure such that the cover $p:\widetilde{H}\to H$ is a morphism of groups, and its kernel is a discrete normal subgroup. Furthermore, $\widetilde{H}$ has to be isomorphic to the group $G$ of positive affine automorphisms of $\Bbb R$, but the only discrete normal subgroup of $G$ is the trivial group $\lbrace\mathrm{id}_{\Bbb R}\rbrace$, and the quotient, which is $G\simeq\Bbb R\rtimes\Bbb R^*_+$ itself, isn't compact, and so $G$ cannot cover the compact Lie group $H$, which is a contradiction.

The proof is divided in several steps.

In the first step we start with a compact connected surface
$\Sigma$ which we assume to be a Lie group. We establish that $\Sigma$
is a torus (topologically). This doesn't rule out that $\Sigma$ may be
noncommutative yet, but it narrows down the kind of space $\Sigma$ may
be. We use

*

*the classification of compact surfaces,

*facts about Lie groups such as parallelizability and commutativity of the fundamental group.

This rules out all but one (topological) surface : the torus.

Lie groups are orientable (and parallelizable), have empty boundary, and have commutative fundamental groups (being topological monoids). Closed orientable surfaces are classified by their genus, and the only genera $g$ that have commutative fundamental groups are $g=0$ and $g=1$. The sphere ($g=0$) isn't parallelizable, hence the only surface that can be a Lie group is the torus.

The second step is to show that the torus only carries commutative Lie group structures. So suppose, to the contrary, that the torus $\Bbb T^2$ is equipped with a non-commutative Lie group structure. The following will produce a contradiction.
The main ingredients of the proof are

*

*a group structure on a space $X$ can be lifted to its universal covering $\widetilde{X}$ in such a way that the cover $p:\widetilde{X}\to X$ becomes a morphism of topological groups (the only choice that one needs to make is that of the neutral element in the fiber $p^{-1}(1_X)$). The kernel $K$ of this morphism is a discrete normal subgroup of $\widetilde{X}$ (discreteness follows from $p$ being a cover). In the case at hand, where $X=\mathbb T^2$, the universal cover is the plane $\Bbb R^2$, and we will write $H$ for the Lie group obtained by lifting that of the torus to the plane. It is nonabelian.

*There are, up to isomorphism, only two $2$-dimensional Lie algebras: the trivial one, and that described below, which is nonabelian and is the Lie algebra of a certain explicitely describable (simply connected) Lie group $G$.

*This Lie group has no discrete normal subgroups.

*Finally, if $\mathfrak g,\mathfrak g'$ are the Lie algebras of simply connected (connected) Lie groups $G,G'$, then there is a one to one correspondence between Lie algebra morphisms $\phi:\mathfrak g\to \mathfrak g'$ and Lie group morphisms $\Phi:G\to G'$ whereby $\Phi$ corresponds to the Lie algebra morphism $\phi=d_e\Phi$

The universal cover of the torus is the plane $\Bbb R^2$, and the group structure on the torus lifts to a Lie group structure on the universal cover, which we shall write $H$, in such a way that the cover map $p:H\to\Bbb T^2$ is a map of Lie groups with discrete kernel $K$, and $T^2\simeq H/K$ as groups. So we have equipped $\Bbb R^2$ with a Lie non-commutative Lie group structure $H$. Its Lie algebra $\mathfrak h$ is (necessarily) isomorphic to the Lie algebra $\mathfrak g=\Bbb R X\oplus\Bbb R Y$ with the bracket $[X,Y]=X$, which is the Lie algebra of the positive affine group of the real line, i.e. the Lie group of all positive affine automorphisms of the line
$$G=\left\lbrace\phi:\Bbb R\to\Bbb R\mid\exists a\in\Bbb R^*_+,b\in\Bbb R,\forall x\in\Bbb R,\,\phi(x)=ax+b\right\rbrace$$
This isomorphism of Lie algebras extends to an isomorphism of Lie groups (since $\Bbb R^2$ is simply connected), and hence, as Lie groups,
$$H\simeq G$$
This cannot be. Indeed $G$ has no discrete normal subgroups other than the trivial group. This is because the conjugacy classes in $G$ are determined by the parameter $a$, and are thus non discrete: if $a,b,c,d\in\Bbb R$ with $a,c>0$, and $\phi,\psi$ are defined by the formulas $\phi(x)=ax+b$ and $\psi(x)=cx+d$, then $\psi^{-1}(x)=c^{-1}x-c^{-1}d$ and
$$\psi\phi\psi^{-1}=\lbrace x\mapsto ax-ad+cb+d\rbrace$$
so that if $\phi\neq \mathrm{id}_{\Bbb R}$, i.e., if $(a,b)\neq (1,0)$, then the conjugacy class of $\phi$ is

*

*if $a\neq 1$: all those maps $x\mapsto ax+\lambda$ where $\lambda\in\Bbb R$ is arbitrary

*if $a=1$: all of the maps $x\mapsto x+b'$ where $b'\neq 0$ has the same sign as $b$
So $G$ has no countably infinite normal subgroups, and thus has no quotient isomorphic to the torus.
As a consequence, any Lie group structure on the torus is abelian.
