Is there a "natural" way to define a group operation on the set of size-$n$ subsets of a finite set? It is easy to define a group operation on the set of all subsets of a given finite set S of size n: merely take the exclusive-or (disjoint sum) of the two sets.  This is associative, the empty set is the identity, and each element is its own inverse.
Can one somehow similarly define a group operation on the set of subsets of a given size k?
Of course, if k=1, then one could use addition modulo n...
Suppose k = 2, and n = 4.  Is there some cute way to define the "sum" of {a,b} and {c,d} to
be another subset {e,f} that satisfies the group axioms?
(I'm looking for "natural" or "combinatorial" solutions; one can clearly provide an index or rank to each of the C(n,k) subsets, and then define a group somehow on these ranks, e.g. by addition modulo C(n,k). But this is not the sort of solutions I'd like to see...)
 A: No there is not.
Let $S$ be the set of $k$-element subsets of an $n$-element set $A$, $1\le k<n$.
If there were a group law $\circ$ on $S$ that deserved being called natural, then it would be invariant under the permutations of $S$ that are induced by the group $\operatorname{Sym}(A)$ of permutations of $A$. Especially, this action of $\operatorname{Sym}(A)$ must leave the neutral element of $(S,\circ)$ fixed. Since $\operatorname{Sym}(A)$ acts transitively on $S$ and $|S|>1$, this is not possible.
A: For $n=4$ and $k=2$ this is a very artificial natural construction :)
Let $\{1,2,3,5\}$ be the four elements. Then identify a set $\{ a, b \}$  with $a+b \pmod 6$, this gives a bijection from the set of two elements subsets to $\mathbb Z /6 \mathbb Z$.
The "natural" definition is the following:
$$\{a,b\}+ \{a',b'\}= \{c,d\}$$
where $\{ c,d \}$ is the only two elements subset of $\{ 1,2,3,5\}$ such that
$$a+b+a'+b'=c+d \pmod 6 \,.$$
The idea can be extended to all pairs  $(n,k)$ for which you can find a set $\{ a_1,..,a_n \}$ such that any two distinct $k$-elements subsets have different sums $\pmod{\binom{n}{k}}$.
