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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 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 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 ∈ { 0 , . . . , n } . The binomial coefficient ( n r ) = n ! r !( n - r )! is an integer. Proof: Proceed by induction on n . For n = 1, ( 1 0 ) = 1 and ( 1 1 ) = 1 are both integers, as claimed. Suppose that ( n - 1 r ) is an integer for all r ∈ { 0 , . . . , 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 0 = 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 a i i with q i the unique ascending list of prime
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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). H 1.3.15 – Let m, n, a, b Z with ( m, n ) = 1 . Prove there exists x with x a mod m and x b mod n . As m and n are coprime, there exist integers c, d such that cm + dn = 1. We see that ( ac ) m + ( ad ) n = a , so that adn = a - acm a mod m . Similarly, ( bc ) m + ( bd ) n = b , so bcm = b - bdn b mod n . Set x = adn + bcm . Then we have x adn mod m a mod m x bcm mod n b mod n.
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