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On Group Actions !

  1. Jan 28, 2013 #1
    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 plz ?
  2. jcsd
  3. Jan 28, 2013 #2
    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?
  4. Jan 28, 2013 #3
    yes , i'm familiar with it !
    can you explain how does this follows from the fundamental theorem of arithmetic ?
  5. Jan 28, 2013 #4
    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).
  6. Jan 28, 2013 #5
    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 ??
  7. Jan 28, 2013 #6
    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.
  8. Jan 29, 2013 #7
    yes , i can understand it now :) thanx
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