Intuition strongly suggests that there exist $\left\lfloor\frac{\binom{n}{k}}{\lfloor\frac{n}{k}\rfloor}\right\rfloor$ independent sets in the complement of a Kneser graph each having $\lfloor\frac{n}{k}\rfloor$ vertices in it. Is this true. If true, how to establish it?


A construction of such a set of cliques in the Kneser graph $K(6,2)$ is as follows:
$$(12)(34)(56)$$
$$(13)(25)(46)$$
$$(14)(26)(35)$$
$$(15)(24)(36)$$
$$(16)(23)(45)$$
Thus, in this example we have $5$ disjoint triangles in the Kneser graph $K(6,2)$ which correspond to an equitable $5$ coloring of the complement graph $\overline{K}(6,2)$. Can such a construction be always done? I think this is related to the number of order $2$ elements in the symmetric group of order $n$. Thanks beforehand.



Independent sets in complement of Kneser graphs
According to [p. 8], Baranyai's theorem [B] implies that the vertex set of the Kneser graph $K(n,k)$ can be partitioned into $\left\lceil\frac{\binom{n}{k}}{\left\lfloor\frac{n}{k}\right\rfloor}\right\rceil$ cliques of size $\left\lfloor\frac{n}{k}\right\rfloor$.


*References*


[B] Zs. Baranyai, *On the factorization of the complete uniform hypergraph*, In: Eds. A. Hajnal, R. Rado, and V. T. Sós, Infinite and Finite Sets (Proc. Intern. Coll. Keszthely, 1973), Bolyai J. Mat. Társulat, Budapest & North-Holland, Amsterdam, 1975, 91–108.


[BP] Boštjan Brešar, Mario Valencia-Pabon, *[Independence number of products of Kneser graphs](https://lipn.univ-paris13.fr/%7Evalenciapabon/papers/kneser-revised-v2.pdf)*, (November 19, 2018).


