Proving Sylow p-Subgroup is Normal in Finite Group

  • Thread starter ehrenfest
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In summary, the conversation discusses the proof that a finite group with precisely one proper Sylow p-subgroup is a normal subgroup, implying that the group is not simple. It is mentioned that q is unnecessary in the proof, as its existence can be inferred from the existence of a proper Sylow p-subgroup. It is also noted that if |G| does not have another prime divisor, then it cannot be determined if G is simple or not.
  • #1
ehrenfest
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[SOLVED] Sylow p-subgroups

Homework Statement


Let G be a finite group and let primes p and q \neq p divide |G|. Prove that if G has precisely one proper Sylow p-subgroup, it is a normal subgroup, so G is simple.

EDIT: that should say "G is not simple"

Homework Equations


The Attempt at a Solution


I don't see the point of q. If G has precisely one proper Sylow p-subgroup, then you can conjugate with all the elements of the group and you cannot get of the subgroup or else you would have another Sylow-p-subgroup, right? So, it must be normal, right?
 
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  • #2
anyone?
 
  • #3
ehrenfest said:
I don't see the point of q. If G has precisely one proper Sylow p-subgroup, then you can conjugate with all the elements of the group and you cannot get of the subgroup or else you would have another Sylow-p-subgroup, right? So, it must be normal, right?

Yes. (Is it possible that you misread the question - if there is a normal subgroup, the group can't be simple.)
 
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  • #4
See the EDIT. But am I right about the part about q being unnecessary? You can induce the existence of q by the fact there is a proper Sylow p-subgroup, right?
 
  • #5
If |G| doesn't have another prime divisor, then we can't say for sure that G isn't simple (e.g. if |G|=p^2).
 
  • #6
morphism said:
If |G| doesn't have another prime divisor, then we can't say for sure that G isn't simple (e.g. if |G|=p^2).

But it must have another prime divisor if it has a proper p-Sylow subgroup.
 
  • #7
Ah, you're right. Missed the word proper.
 
  • #8
ehrenfest said:
See the EDIT. But am I right about the part about q being unnecessary? You can induce the existence of q by the fact there is a proper Sylow p-subgroup, right?

Yep.
 

Related to Proving Sylow p-Subgroup is Normal in Finite Group

1. What is a Sylow p-subgroup?

A Sylow p-subgroup is a subgroup of a finite group whose order is a power of a prime number p.

2. How is a Sylow p-subgroup proven to be normal in a finite group?

To prove that a Sylow p-subgroup is normal in a finite group, we use the Sylow Theorems, which state that if a group G has order pn, where p is a prime number and n is a positive integer, then G contains a subgroup of order pk for every k that divides n. Additionally, if pk is the largest power of p that divides n, then the subgroup of order pk is normal in G.

3. What is the significance of proving a Sylow p-subgroup to be normal in a finite group?

Proving a Sylow p-subgroup to be normal in a finite group is important because it allows us to better understand the structure and properties of the group. It also allows us to simplify computations and make connections to other areas of mathematics, such as group theory and number theory.

4. Can a finite group have more than one normal Sylow p-subgroup?

Yes, a finite group can have multiple normal Sylow p-subgroups. This is because the Sylow Theorems only guarantee the existence of one normal Sylow p-subgroup for each prime factor of the group's order, but it is possible for a group to have multiple prime factors.

5. Is there a direct method for proving a Sylow p-subgroup to be normal in a finite group?

No, there is no direct method for proving a Sylow p-subgroup to be normal in a finite group. The proof typically involves using the Sylow Theorems and constructing a normal subgroup through a series of steps and arguments. It is not a straightforward process and may require advanced mathematical techniques.

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