A lemma which admits as an easy corollary the structure of the automorphism group of the alternating group on $n\geq 7$ elements

Question

I was trying to read the following lemma which admit as an easy corollary the structure of the automorphism group of the alternating group on $n\geq 7$ elements.

Anyway there are two points that are really unclear to me.

First when it says

Therefore the `$3$-cycles' of $H$ can move at most four points.

Secondly when it says

Now the elements of $H$ corresponding to $(123)$ and $(124)$ in $A_{n-1}$ generate a subgroup isomorphic to $A_4$ (this is clear), and therefore map to cycles $(abc)$ (this is unclear to me).

Hope that someone could give me some hints about these problems.

Answer 1

For the first point you have to show that the three cycle acts trivially on the orbit of the subgroup $A_{n-4}$ of its centralizer that has size at most $n-4$. Since primitive permutation groups (other than those of prime order) have trivial centralizers in the symmetric group, that would follow if you could show that the action on that orbit was primitive. But it must, be because otherwise its point stabilizer would be properly contained in a larger proper subgroup, which would necessarily have index less than $n-4$ in $A_{n-4}$.

For the second point, the only essentially distinct images for the two $3$-cycles are (i) $(1,2,3),(1,3,2)$; (ii) $(1,2,3),(1,2,4)$; (iii) $(1,2,3),(3,4,5)$; and (iv) $(1,2,3),(4,5,6)$, and they only generate a subgroup isomorphic to $A_4$ in Case (ii).