Extending the action $S_5$ on $2$-subsets of $\{1,\cdots,5\}$ to an action of $S_6$.

Question

The symmetric group $S_5$ acts on the set $\binom{5}{2}$ of ten $2$-subsets of $[5]=\{1,\cdots,5\}$. In The Finite Simple Groups (Wilson), problem 2.21 asks the reader to extend the group action $S_5\to S_{10}$ to an action $S_6\to S_{10}$.

There's probably a "hands-on" way to do this by writing down explicit cycle notations and relations and such, but is there a conceptual reason for this? Is this explained by some exceptional object, maybe like ${\rm Out}\,S_6$ somehow?

(For comparison, problem 2.24 asks to show things about a group generated by a couple of given permutations in $S_8$, without mentioning it's just ${\rm PSL}_2(\Bbb F_7)$ acting on the projective line $\Bbb F_7\Bbb P^1$, so hiding the story behind a problem seems like something the text would do.)

Answer 1

We need $10$ of something related to a set of size $6$ and need to build that from the $2$-subsets from a set of size $5$.

There are exactly $10$ different ways to partition a set of size $6$ into two parts of size $3$ each. We associate each of these $10$ partitions with a specific $2$-subset by choosing the half of the partition which contains $6$ and selecting the other two points.

For example,

$\{1,2\}$ becomes $\{\{1,2,6\},\{3,4,5\}\}$

This maintains the action of $S_5$ on a set of size $10$ and extends it to $S_6$.

Answer 2

An $S_6 \to S_{10}$ homomorphism shows up in the exceptional group $M_{24}$ on 24 points. Take an octad $O$ and dodecad $D$ which intersect in 6 points. (The details of the construction are in a later chapter of Wilson's book.) This partitions the 24 points as follows

(a) 6 points which lie in both $O$ and $D$

(b) 6 points which lie in $D$ but not $O$

(c) 2 points which lie in $O$ but not $D$

(d) 10 points which lie in neither $D$ nor $O$

The subgroup of $M_{24}$ which simultaneously stabilizes $D$ and $O$ is isomorphic to $S_6$, and maps each of the subsets (a) through (d) to itself. In particular, for (d), we get a homomorphism $S_6 \to S_{10}$.

It is not clear that this is an extension of the given $S_5 \to S_{10}$ in the problem, though.

An outer automorphism of $S_6$ also shows up here by considering (a) and (b).