I'm recently got interested by a series of very cool videos by Math Doctor Bob, specifically the ones about to compute automorphism groups. The one puzzling me a bit is this one: Automorphism of A4
Without forcing any of you to to watch the video, I'm just resuming here what I have not understood. Basically, he's able to estimate the order of the automorphism group of $A_4$ to 24 elements, establishing finally this must be isomorphic to $S_4$. Actually these are the facts:
1) $A_4$ is made up by the identity, 3 cycles of order 2 (like (1 2)(3 4)) and 8 cycles of order 3.
2) $A_4 = <( 1 2 3), (1 2)(3 4)>$, i.e. can be generated by a cycle of order 3 and one of order 2.
So fa, so good. Here he claims 2 interesting facts, which are the core business of this question indeed:
1) if we have automorphism, we have to preserve the order of elements. So a 3 cycles has to be carried to another 3-cycles and a couple to a couple (I condensed it a bit)
Assuming the above point, the results turns out naturally as a combinatorial reasoning, i.e.
you have 8 choices for a 3-cycle and 3 choices for a (2,2)-cycle, so a total of 24 choices
This made up the order of $Aut(A_4)$. With later reasoning is desuming this must be $S_4$, indeed.
What I'm struggling with is why an automorphism of a alternating group must preserve the order. I mean, these facts are clear to me:
1) $Inn(A_4) \lhd Aut(A_4)$ 2) $Inn(A_4)$ acts by conjugation and then must preserve cycle order (even this is the case in which the conjugacy classes split, however the order is guaranteed)
What I'm missing here (my bad) is it seems we are assuming $Aut(A_4)$ is made up only by $Inn(A_4)$, but this not since the $Out(A_4)$ is not trivial but being isomorphic to $\Bbb{Z}/2$. Since this latter involves an odd permutation, the order would be not guaranteed in that case.
I'm sure I'm wrong, but I don't know where.
thank for your precious support, as always.