On Group Actions !


by Maths Lover
Tags: actions
Maths Lover
Maths Lover is offline
#1
Jan28-13, 07:01 PM
P: 62
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 ?
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Number Nine
Number Nine is offline
#2
Jan28-13, 07:26 PM
P: 771
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?
Maths Lover
Maths Lover is offline
#3
Jan28-13, 07:30 PM
P: 62
Quote Quote by Number Nine View Post
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 ?

Number Nine
Number Nine is offline
#4
Jan28-13, 09:13 PM
P: 771

On Group Actions !


Quote Quote by Maths Lover View Post
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).
Maths Lover
Maths Lover is offline
#5
Jan28-13, 09:20 PM
P: 62
Quote Quote by Number Nine View Post
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 ??
Number Nine
Number Nine is offline
#6
Jan28-13, 09:48 PM
P: 771
Quote Quote by Maths Lover View Post
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.
Maths Lover
Maths Lover is offline
#7
Jan29-13, 01:16 PM
P: 62
Quote Quote by Number Nine View Post
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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