There’s an amazing answer on Mathoverflow where Emerton gives a complete and very intuitive proof of the Bruhat decomposition
where is an -dimensional vector space, is the subgroup of upper triangular matrices, and is the subgroup of permutation matrices.
An an exercise, the reader is left to show that any two Borel subgroups contain a common torus. Since this step is the first point where one sees why the symmetric group is relevant, I’d like to record a proof here.
Proof. Let and be two chains of subspaces of such that for each . Say, just for this proof, that is –intersecting, if is minimal such that meets (or equivalently, ). Let be the function defined by if and only if is -intersecting.
Claim: is injective. Proof of claim. Suppose that and are -intersecting, where . By hypothesis, there exists and such that . Since is -dimensional, there exist scalars , , not both zero, such that . If then , a contradiction. And otherwise, since we have , again a contradiction.
Thus is a permutation. For each choose . (Necessarily .) The Borel subgroups corresponding to the two chains of subspaces clearly both contain the torus having as eigenvectors.
Edit. This proof gives a constructive algorithm to find vectors and a permutation of such that and for each . It is easy to see that is unique: suppose has the same property and . There exists such that . Then meets , contradicting the minimality of .
Algebraically the uniqueness of corresponds to the disjointness of the Bruhat decomposition. Emerton's post has a nice topological proof of this.
Whenever one has a matching problem it's natural to see if there's an easy reason why the condition in Hall's Marriage Theorem holds. This paper by R. Brualdi gives a fairly short proof of this fact. The key observation is the following: if for each , then given any , the set contains the subspace of spanned by . Of course a shorter proof follows from the proof in this post, since Hall’s condition is obviously necessary for a matching to exist.
Shameless plug: this joint paper with John Britnell gives an application of Hall’s Marriage Theorem to group theory where the use of Hall’s theorem appears to be essential. (At least, we know of no other way to prove the result.)