
#1
Jan2813, 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 (p1)! are less than p . why is this true ?!! can anyone explain plz ? 



#2
Jan2813, 07:26 PM

P: 771





#3
Jan2813, 07:30 PM

P: 62

can you explain how does this follows from the fundamental theorem of arithmetic ? 



#4
Jan2813, 09:13 PM

P: 771

On Group Actions ! 



#5
Jan2813, 09:20 PM

P: 62

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
Jan2813, 09:48 PM

P: 771

(p1)! has a unique prime factorization. Write out the expansion of (p1)!, 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. 



#7
Jan2913, 01:16 PM

P: 62




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