On Group Actions !

Maths Lover

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 ?

Number Nine

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

yes , i'm familiar with it !
can you explain how does this follows from the fundamental theorem of arithmetic ?

Number Nine

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

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

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

yes , i can understand it now :) thanx