I wanted to follow up on a brief comment Scott made in his last post about the outer automorphism of . An outer automorphism is (non-canonically) the same as finding two inequivalent actions of on 6-element sets.
Consider the set with the obvious action of . There are also six points in the projective line . Let’s consider the different ways of turning X into a projective line . There are 6! ways to do this, for example . But are all these ways really different? The projective line has a lot of automorphisms, so let’s identify two labelings if there is an automorphism of the projective line turning one into the other. Now how many equivalence classes are there? Without loss of generality, since the automorphism group acts 3-transitively we can assume that the labeling has , and . Thus there are only 6 equivalence classes of projective line structures! This is the other action we’re looking for.
I’ll give a quick computation after the cut which shows that this construction is practical for actual computations.
In order to turn the existence of two inequivalent -sets into an outer automorphism of we pick an identification of the two sets. Say we order the six projective lines as follows:
Now let’s see how the transpostion acts on these six lines. It sends to the line . Now we need to use an automorphism to put this into the standard form, namely . This turns the line into . So, is mapped to .
Similar calculations show that is mapped to ; and is mapped to . Hence under our outer automorphism we have .
Let’s check that this is sensible. An outer automorphism should send any conjugacy class to another conjugacy class of the same order. The number of transpositions in is while the number of products of three disjoint transpositions is . So we’re ok.