maximal tori cover compact Lie group Let $G$ be a compact connected Lie group, $T$ be some maximal torus in $G$ (that is, inclusion-maximal connected abelian subgroup). Then the union of tori $gTg^{-1}$, $g\in G$, is the whole $G$. This is well-known (4.21 in Adams book). My question is rather methodological: is there any proof without use of algebraic topology? Adams presents A. Weil's proof, which uses some kind of Lefschetz fixed points theorem. 
(Yes, I am sorry but my motivation is mostly that I teach second year students this stuff.) 
 A: There are many ways of proving this, using all sorts of different methods. In the second edition my book "Lie groups, Lie algebras, and representations" (following Brocker and tom Dieck) I use the mapping degree theorem. If we fix a single maximal torus $T$ and consider the conjugation map $\Phi:T \times (K/T) \rightarrow K$ be given by $\Phi(t,[x])=xtx^{-1}$. If we can show that this map has nonzero mapping degree, we can conclude it is surjective, which is just what we are trying to show. In Section 11.5, I show that $\Phi$ has mapping degree equal to the order of the Weyl group. This approach also gives a proof of the Weyl integral formula by the same computation.
A: There are proofs that avoid algebraic topology: see for example Chapter 16 in Bump's Lie Groups or IV.5 in Knapp's Lie Groups Beyond an Introduction.
A: The claim in the question that maximal tori are the same as inclusion-maximal abelian subgroups is not correct. For example, the diagonal matrices with +1 or -1 on the diagonal form a maximal abelian subgroup of SO(n) that is not a torus. 
