Re: Aut A5 = S5
mareg_at_mimosa.csv.warwick.ac.uk
Date: 10/29/04
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Date: Fri, 29 Oct 2004 10:19:40 +0000 (UTC)
In article <Rabgd.10702$5i5.9729@newsread2.news.atl.earthlink.net>,
no@mail.com
U writes:
>Thanks for showing the way, but I'm still having some troubles.
>
>mareg@mimosa.csv.warwick.ac.uk wrote:
>>
>> OK, let's use Robin's argument on the 5 Sylow 2-subgroups. Aut(A_5) must
>> permute these 5 Sylow 2-subgroups.
>
>This is because automorphisms map sylow p-groups to sylow p-groups. Ok.
>
>> Each of them fixes a unique point, so
>> S_5 acts as S_5 on the 5 Sylow 2-subgroups.
>
>I think what you are trying to show here is that the group action of
>Aut(A5) on the 5 sylow 2-subgroups induces a homomorphism that maps
>Aut(A5) *surjectively* onto S5. But I'm not clear on how it is so.
>What do you mean by "each of them fixes a unique point?"
I just mean that each of the 5 Sylow 2-subgroups fixes a unique point in
the standard permutation action of A5 on 5 points. For example, the
Sylow 2-subgroup {identity, (1,3)(4,5), (1,4)(3,5), (1,5)(3,4)} fixes the
unique point 2. So you can think of Aut(A5) itself as permuting these
5 points. If f is in Aut(A5) maps Sylow 2-subgroup fixing 2 to Sylow 2-subgroup
fixing 4, then think of f as mapping 2 to 4. This induced action of Aut(A5)
on the five points defines a homomorphism phi: Aut(A5) -> S5, and as you sayd
below, you need to show that phi is injective.
>> Suppose f is in Aut(A_5).
>> Then by multiplying f by a suitable element of S_5, we can assume that
>> f fixes each of the Sylow 5-subgroups.
>
>I'm pretty lost here as to how an element of S5 can multiply with an
>element of Aut(A5).
Well, you said initially that you understood the fact that S5 was a subgroup
of Aut(A5), so that means exactly that you can multiply an element of S5
by an element of Aut(A5). But maybe you don't need to worry about that point,
since, as you say, what we really need to show is that phi is injective.
>> What you need to prove now is that
>> only the identity automorphism fixes all 5 Sylow 2-subgroups.
>
>So now you are going to show that the homomorphism Aut(A5) -> S5,
>induced by the group action, is *injective*.
Yes, exactly!
>>
>> You can probably do that by direct calculation, but a neater way is to note
>> that the subgroup N of Aut(A_5) which fixes all 5 Sylow 2-subgroups is
>> normal in Aut(A_5)
>
>Right. This is true of all kernels of group actions.
>
>> and has trivial intersection with A_5 (identified with
>> Inner(A_5)),
>
>I'm lost here. So, N and Inn(A5) are both normal in Aut(A5), why must
>these two normal subgroups intersect trivially?
Sorry, this was misleading. It is not the normality of these subgroups that
implies they have trivial intersection, it is their definition.
Remember that elements of N by definition fix all 5 Sylow 2-subgroups of A5.
The nontrivial elements of A5 clearly do not do this. For example the
element (3,4,5) of A5 maps Sylow 2-subgroup fixing 3 to Sylow subgroup fixing
4, etc. So they have trivial intersection.
>
>> so it must centralize A_5, which means it is acting trivially
>> on A_5.
This is where the normality comes in. By a standard commutator argument,
which you ought to know about, two normal subgroups of a group with trivial
intersection always centralize each other.
Derek Holt.
>> Derek Holt.
>>
>
>Now, this neater way of showing only the identity automorphism fixes all
>5 sylow 2-subgroups, also works for the other sylow p-subgroups, no? In
>another word, Aut(A5) acts also faithfully over the set of 10 sylow
>3-subgroups and over the set of 6 sylow 5 subgroups, correct? However,
>Aut(A5) is only isomorphic to a proper subgroup of S10 and S6.
>
>I'm sorry for asking such fundamental questions. This is my first take
>at algebra; I'm still feeling my way around the subject. Thanks for
>your patience.
>
>Kira
>
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