Group Actions: Prime Divisors & Smallest Prime | Dummit & Foote

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Discussion Overview

The discussion revolves around a result from "Abstract Algebra" by Dummit and Foote regarding finite groups and their subgroups. Specifically, it addresses the claim that if \( G \) is a finite group of order \( n \) and \( p \) is the smallest prime dividing the order of \( G \), then any subgroup \( H \) of \( G \) whose index is \( p \) is normal. Participants focus on understanding a specific part of the proof related to the prime divisors of \( (p-1)! \).

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant questions why all prime divisors of \( (p-1)! \) are less than \( p \).
  • Another participant references the fundamental theorem of arithmetic, suggesting that no prime larger than \( p \) can divide \( (p-1)! \) since it consists of factors all less than \( p \).
  • A participant requests clarification on how this follows from the fundamental theorem of arithmetic.
  • Further explanation is provided that \( p \) does not appear in the expansion of \( (p-1)! \) because it is the product of numbers less than \( p \).
  • One participant presents a counterexample involving the number 6, arguing that while 6 cannot divide individual factors, it can divide their product, questioning if \( p \) could similarly divide products of factors in \( (p-1)! \).
  • Another participant counters that 6 is not prime and reiterates that \( p \) does not appear in the factorization of \( (p-1)! \), emphasizing the unique prime factorization.
  • Ultimately, a participant expresses understanding of the explanation provided.

Areas of Agreement / Disagreement

Participants engage in a debate regarding the properties of prime numbers and factorials, with some expressing confusion and others providing clarifications. The discussion does not reach a consensus, as differing viewpoints on the implications of the examples presented remain.

Contextual Notes

Participants reference the fundamental theorem of arithmetic and the nature of prime factorization, but the discussion does not resolve the initial query regarding the divisibility of products in relation to \( p \).

Maths Lover
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hi ,

this result is from text , Abstract Algebra by Dummit and foote .
page 120

the result says , if G is a finite group of order n , p is the smallest prime dividing the order of G , then , any subgroup H of G whose index is p is normal

and the text gave the proof of this result , but a part of this proof is not obivous for me !

this part is ,all prime divisors (p-1)! are less than p .

why is this true ?!

can anyone explain please ?
 
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all prime divisors (p-1)! are less than p

Do you mean that all prime divisors of (p-1)! are less than p? Are you familiar with the fundamental theorem of arithmetic? It follows immediately from what (p-1)! is that no prime larger than p can be a divisor. Write it out in full; is p in there anywhere?
 
Number Nine said:
Do you mean that all prime divisors of (p-1)! are less than p? Are you familiar with the fundamental theorem of arithmetic? It follows immediately from what (p-1)! is that no prime larger than p can be a divisor. Write it out in full; is p in there anywhere?

yes , I'm familiar with it !
can you explain how does this follows from the fundamental theorem of arithmetic ?
 
Maths Lover said:
yes , I'm familiar with it !
can you explain how does this follows from the fundamental theorem of arithmetic ?

It follows from the definition of the factorial. Write out the expansion of (p-1)!; p does not appear anywhere in the factorization. How could it? You're multiplying together a bunch of numbers less than p. None of them are going to multiply together to form p (it's prime).
 
Number Nine said:
It follows from the definition of the factorial. Write out the expansion of (p-1)!; p does not appear anywhere in the factorization. How could it? You're multiplying together a bunch of numbers less than p. None of them are going to multiply together to form p (it's prime).

take this example !
6 can't divide 3,4,5
but 6 can divide 3*4*5= 60

p can't divide any factor but maybe it can do this with some products of them like the example above ! why not ??
 
Maths Lover said:
take this example !
6 can't divide 3,4,5
but 6 can divide 3*4*5= 60

p can't divide any factor but maybe it can do this with some products of them like the example above ! why not ??

6 is not prime.
(p-1)! has a unique prime factorization. Write out the expansion of (p-1)!, as I said; p does not appear, nor is it a factor of any of the numbers that do appear. Again, review the fundamental theorem of arithmetic.
 
Number Nine said:
6 is not prime.
(p-1)! has a unique prime factorization. Write out the expansion of (p-1)!, as I said; p does not appear, nor is it a factor of any of the numbers that do appear. Again, review the fundamental theorem of arithmetic.

yes , i can understand it now :) thanx
 

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