I have recently been looking at Hall's marriage theorem. One application of it is that given a finite group $G$ and a subgroup $H\leq G$, there is a left transversal of $H$ that is also a right transversal. I can see the theoretical importance of this, but am struggling to find any situations when one would actually use this. If anybody can enlighten me, that would be greatly appreciated.
After thinking of this a little, it happens to be easy to give a proof using Hall's marriage theorem.
Let $L$ be the set of left cosets of $H$. Let $R$ be the set of right cosets of $H$.
$L$ and $R$ are partitions of G, and the size of each of their elementes is equal to $|H|$.
Now let define the a bipartite graph with vertex set $(L+R)$ (where $L$ and $R$ are the bipartite sets). We will say that there exists an edge between $A\in L$ and $B \in R$ if and only if $A \cap B \not= \emptyset$.
Suppose that there is a set $S \subset L$ such that that $N(S) < S$ (The opposite of Hall's marriage theorem hypothesis).
It's easy to see that $|\cup N(S)| = |H||N(S)|$ and $|\cup S| = |H||S|$ (if you have troubles seeing this, remember that all the cosets have exactly the same size).
Then, our assumption implies that $|\cup N(S)| < |\cup S|$. But this is absurd.
This is saying that there is at least one element of $\cup S$ that is not covered by an element of $R$, but $L$ and $R$ are both partitions so for sure for each element of $|\cup S|$ there is someone in $R$ containing it.
Then, Hall's marriage theorem hypothesis holds. So there is a perfect matching M. As our edges represent non empty set intersections, its enough to take an element of each intersection in M to get a left transversal of $H$ that is also a right transversal of $H$.