Exotic maps $S_5\to S_6$

This section says:

There is a subgroup (indeed, $6$ conjugate subgroups) of $S_6$ which are abstractly isomorphic to $S_5$,

At this point I'm thinking: certainly: the group of all permutations of $\{a,b,c,d,e,f\}$ that leave the letter $a$ fixed is isomorphic to $S_5$. And there are six groups like it, since one can choose any of the six letters as the one that will remain fixed. But the section continues:

There is a subgroup (indeed, $6$ conjugate subgroups) of $S_6$ which are abstractly isomorphic to $S_5$, and transitive as subgroups of $S_6$.

But the groups I identify above do not act transitively on $\{a,b,c,d,e,f\}$, so this must be about some other subgroups. What are they? Are they images of the six groups I mention above under an outer automorphism?

Solution 1:

I will show you in principle how this is done, but the actual mechanics are tedious. Let:

$$P_1 = \{e,(1\ 2\ 3\ 4\ 5),(1\ 3\ 5\ 2\ 4), (1\ 4\ 2\ 5\ 3),(1\ 5\ 4\ 3\ 2)\}\\ P_2 = \{e, (1\ 2\ 3\ 5\ 4),(1\ 3\ 4\ 2\ 5),(1\ 5\ 2\ 4\ 3),(1\ 4\ 5\ 3\ 2)\}\\ P_3 = \{e,(1\ 2\ 4\ 3\ 5),(1\ 4\ 5\ 2\ 3), (1\ 3\ 2\ 5\ 4), (1\ 5\ 3\ 4\ 2)\}\\ P_4 = \{e, (1\ 2\ 4\ 5\ 3),(1\ 4\ 3\ 2\ 5), (1\ 5\ 2\ 3\ 4), (1\ 3\ 5\ 4\ 2)\}\\ P_5 = \{e,(1\ 2\ 5\ 3\ 4),(1\ 5\ 4\ 2\ 3), (1\ 3\ 2\ 4\ 5), (1\ 4\ 3\ 5\ 2)\}\\ P_6 = \{e,(1\ 2\ 5\ 4\ 3), (1\ 5\ 3\ 2\ 4), (1\ 4\ 2\ 3\ 5), (1\ 3\ 4\ 5\ 2)\}$$

If we pick an element $\sigma$ of $S_5$, say $(1\ 2\ 3)(4\ 5)$, we find that:

$$\sigma P_1\sigma^{-1} = P_5\\ \sigma P_2\sigma^{-1} = P_3\\ \sigma P_3\sigma^{-1} = P_6\\ \sigma P_4\sigma^{-1} = P_2\\ \sigma P_5\sigma^{-1} = P_4\\ \sigma P_6\sigma^{-1} = P_1$$

That is, if $\phi:S_5 \to S_6$ is our (exotic) embedding, then:

$\phi((1\ 2\ 3)(4\ 5)) = (1\ 5\ 4\ 2\ 3\ 6)$ (note this indeed takes an element of order $6$ to an element of order $6$). It should be clear that if $\sigma \in S_5$ is a $5$-cycle, it fixes the sylow $5$-subgroup it belongs to, and permutes the rest, pick one at random, and verify it creates a $5$-cycle in $S_6$.

Note that we only need the $3$-cycles of $S_5$ to show this action is indeed transitive; for example, to send $P_1 \to P_5$, we can conjugate by the $3$-cycle $(3\ 5\ 4)$ (I deliberately chose the $5$-cycle generators to start with $(1\ 2\ \dots)$ to make this clear).

Solution 2:

Note that $S_5$ contains a subgroup of order $20$ (generated by, say, $(1,2,3,4,5)$ and $(1,3,4,2)$ ). The action of $S_5$ on the $6$ cosets of a subgroup of order $20$ provides a permutation representation of $S_5$ on $6$ points. And yes, an outer automorphism maps this sort of $S_5$ to the first type you were thinking of.