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Unformatted text preview: Jonathan Bergknoff Herstein Solutions Chapters 1 and 2 Throughout, G is a group and p is a prime integer unless otherwise stated. “ A ≤ B ” denotes that A is a subgroup of B while “ A E B ” denotes that A is a normal subgroup of B . H 1.3.14* (Fermat’s Little Theorem) – Prove that if a ∈ Z then a p ≡ a mod p . Proceed by induction on (positive) integer a . The theorem holds for a = 1 because 1 p = 1 ≡ 1 mod p . Suppose that k p ≡ k mod p and then compute, by the binomial theorem, ( k + 1) p = p X i =0 p i k i 1 p i = k p + 1 + p ( ... ) ≡ k + 1 mod p. In the last step, we used the induction hypothesis. This proves the result for positive integer a . The expansion ( k + 1) p = k p + 1 + p ( ... ) is justified by lemma 2 (below) which states that p  ( p n ) for n ∈ { 1 ,...,p 1 } . The case a = 0 is trivial: 0 p = 0 ≡ 0 mod p . Let a ∈ Z + as before. As was just proven, there exists c ∈ Z such that a p a = cp . Now, if p > 2 (odd), we have ( a ) p ( a ) = a p + a = ( c ) p so ( a ) p ≡  a mod p . On the other hand, if p = 2, we just need to see that “modulo 2” picks out even/odd parity. Regardless of being positive or negative, an even integer squared is even and an odd integer squared is odd. Therefore a 2 ≡ a mod 2 and the theorem is proven in its entirety. Lemma 1 – Let n ∈ Z + and r ∈ { ,...,n } . The binomial coefficient ( n r ) = n ! r !( n r )! is an integer. Proof: Proceed by induction on n . For n = 1, ( 1 ) = 1 and ( 1 1 ) = 1 are both integers, as claimed. Suppose that ( n 1 r ) is an integer for all r ∈ { ,...,n 2 } . Now n 1 r + n 1 r 1 = ( n 1)! r !( n r 1)! + ( n 1)! ( r 1)!( n r )! = ( n 1)! ( n r ) + r r !( n r )! = n r . Hence, when the above computation goes through, ( n r ) is a sum of integers and thus is, itself, an integer. However, the computation fails from the outset for r = 0 and r = n (we’re interested in things like ( r 1)! and ( n r 1)!), so those cases must be considered independently. We have n = n ! 0! n ! = 1 n n = n ! n !0! = 1 and the claim is proven. Lemma 2 – Let n ∈ { 1 ,...,p 1 } . Then p  ( p n ) . Proof: Intuitively, this lemma is true because the numerator has a factor of p and the denominator has no factors that cancel it (relying crucially on the primality of p ). By the fundamental theorem of arithmetic ( Z is a UFD), we can write the denominator as n !( p n )! = Q q a i i with q i the unique ascending list of prime divisors of n !( p n )! and a i their respective powers. As every factor of n ( n 1) ··· 2 · 1 · ( p n )( p n 1) ··· 2 · 1 is smaller than p (this fails if n = 0 or n = p ), p divides none of them and hence, as a prime, does not divide their product. Then none of the q i is p . The factor of p in the numerator is then preserved upon taking the quotient, and p  ( p n ) . Note that it makes no sense to talk about p dividing ( p n ) unless ( p n ) ∈ Z (lemma 1)....
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This note was uploaded on 11/14/2011 for the course MATH 367 taught by Professor Sdd during the Spring '11 term at Middle East Technical University.
 Spring '11
 sdd

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