Sylow subgroup of some factor group

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Discussion Overview

The discussion revolves around the properties of Sylow p-subgroups in the context of finite groups, specifically examining whether the subgroup PN/N is a Sylow p-subgroup of the factor group KN/N, given certain conditions on the groups involved.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant proposes that PN/N is a Sylow p-subgroup of KN/N based on the isomorphism PN/N ≅ P/(P ∩ N) and the order calculations involving the indices of the groups.
  • Another participant questions the justification for PN/N being a p-group and requests clarification on the steps taken to derive the order of the groups involved.
  • Concerns are raised about the validity of the calculations presented, particularly regarding the application of the second isomorphism theorem and the implications of the orders of the groups.
  • A third participant points out that the question has been previously addressed in another forum, suggesting that the discussion may have been duplicated across platforms.
  • A later reply expresses frustration about cross-posting, indicating that it may lead to reduced engagement from the community in future discussions.

Areas of Agreement / Disagreement

Participants express differing views on the validity of the original claim and the calculations supporting it. There is no consensus on the correctness of the argument presented in the initial post.

Contextual Notes

Some participants highlight potential gaps in the justification of certain steps, particularly regarding the use of group orders and the implications of subgroup properties. The discussion remains focused on the technical aspects without resolving these uncertainties.

moont14263
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Hi. I have the following question:

Let G be a finite group. Let K be a subgroup of G and let N be a normal subgroup of G. Let P be a Sylow p-subgroup of K. Is PN/N is a Sylow p-subgroup of KN/N?

Here is what I think.

Since PN/N \cong P/(P \cap N), then PN/N is a p-subgroup of KN/N.

Now [KN/N:PN/N]=\frac{|KN|}{|N|} \frac{|N|}{|PN|}= \frac{|KN|}{|PN|}= \frac{|K||N|}{|K \cap N|} \frac{|P \cap N|}{|P||N|} = \frac{|K||P \cap N|}{|P||K \cap N|}= [K:P]\frac{|P \cap N|}{|K \cap N|}Since P is a Sylow p-subgroup of K, then p does not divide [K:P]. Also, p does not divide \frac{|P \cap N|}{|K \cap N|} as \frac{|P \cap N|}{|K \cap N|} \leq 1 because P \cap N is a subgroup of K \cap N. Therefore p does not divide [KN/N:PN/N].

Thus PN/N is a Sylow p-subgroup of KN/N.

Am I right?

Thanks in advance
 
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Looks broadly OK to me. I just have a few questions.
moont14263 said:
Since PN/N \cong P/(P \cap N), then PN/N is a p-subgroup of KN/N.
1. How do we know that ##PN/N## is a ##p##-group?

$$\frac{|KN|}{|PN|}= \frac{|K||N|}{|K \cap N|} \frac{|P \cap N|}{|P||N|}$$
2. This step seems to be doing a number of things, the validity of none of which is obvious to me. Can you justify it? I can get to the RHS from ##[KN/N:PN/N]##, but not by the route shown above. I would use the 2nd isomorphism theorem and the fact that, for finite groups, ##[A:B]= \frac{|A|}{|B|}##.

$$\frac{|K||P \cap N|}{|P||K \cap N|}=[K:P]\frac{|P \cap N|}{|K \cap N|}$$Since P is a Sylow p-subgroup of K, then p does not divide [K:P].
3. You can stop before the equals sign and just observe that, since P is a Sylow p-subgroup of K, p does not divide \frac{|K|}{|P|}, since the multiplicity of ##p## in the order of ##|P|## must equal the multiplicity of ##p## in the order of ##|K|##.
 
You are free to do what you want. But this is a cross-post. Asking the same question at several different sites. This means Bungo and Andrew both worked out an answer. If you believe it is your right to do this, the net result of persistent cross-posting is a strong possibility that folks will skip your posts next time they see them.

I already know I will. Bye.
 
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