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Abstract Algebra - Group Theory

Anonymous Mon 29 Nov 18:05:07 2021 No.13904063 View Reply Original Report

Quoted By: >>13904084 >>13908369

Need help showing that the intersection of sylow p-groups is normal. I understand the proof for when there is only one sylow p-subgroup(subgroup of unique order is normal), but I am unsure what to do when there is more than one. Cheers.

Anonymous

Anonymous Mon 29 Nov 2021 18:17:18 No.13904084 Report

Quoted By: >>13905170

>>13904063
Whenever you have a definition of a subgroup that works for all groups and defines a unique subgroup, then that subgroup is always normal. Since H comes from such a definition, it's a normal subgroup. QED
A more tedious way to see this is to take g any element of G, and h belong to all sylow p subgroups. If P is a Sylow p subgroup, we want to show ghg^-1 belongs to P. Equivalently, h belongs to g^-1Pg, which is a conjugate of a Sylow p-subgroup, therefore a Sylow p-subgroup, and h belongs to it because it belongs to all Sylow p-subgroups.
It's up to you to decide which way you like better.

Anonymous

Anonymous Tue 30 Nov 2021 00:07:45 No.13905170 Report

Quoted By: >>13905448 >>13906171

>>13904084
>Whenever you have a definition of a subgroup that works for all groups and defines a unique subgroup, then that subgroup is always normal.
Source?
I don't disbelieve you.
Is there a way to convert a proof of the uniqueness of the construction into a homomorphism that has the constructed subgroup as a kernel?
It seems like there is some category theory black magic going on here.

Anonymous

Anonymous Tue 30 Nov 2021 01:52:45 No.13905448 Report

Quoted By:

>>13905170
One way to look at this is that a normal subgroup is one which "looks the same" from any perspective. When you perform a conjugation, you are in some sense changing the perspective with which you look at a group (think of a change of basis in linear algebra). Clearly, the intersection of all Sylow p-subgroups is not going to change after a conjunction, since the set of all Sylow p-subgroups is invariant under conjugation (i.e. the set of Sylow p-subgroups does not depend on the "perspective" with which we view the group).

Anonymous

Anonymous Tue 30 Nov 2021 08:24:12 No.13906165 Report

Quoted By:

https://math.stackexchange.com/questions/989782/intersection-of-normal-subgroups-proof#989850
Sorry, drunk phonefagging so no original proof. But basically boils down to left and right coset coinciding being invariant under intersection. Probably would be easier to express as homomorphism but it can be shown elegantly (or pedantically, if need be),

Anonymous

Anonymous Tue 30 Nov 2021 08:30:07 No.13906171 Report

Quoted By:

>>13905170
Conjugation is an automorphism. All such groups are invariant under automorphisms, and that includes conjugation. Thus they are characteristic subgroups, and a fortiori normal subgroups.

Anonymous

Anonymous Tue 30 Nov 2021 21:04:41 No.13908369 Report

Quoted By:

>>13904063
Use the fact that all p Sylow groups are conjugate to one another.

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