Algebra

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Math 150a: Modern Algebra Homework 6 Solutions GK1. Show that if G is a finite group and p is a prime number, then the number of elements of order p in G is divisible by p - 1. The result is certainly not true if p is not prime; be careful to explain where the argument uses this assumption. (Hint: This is a generalization of a midterm problem. Look at the solution to that problem and first try the case p = 5.) Solution: If G is a finite group, then there are a finite number of elements of order p . We will begin to count them off in groups of p - 1 (which will show that the number of elements of order p is divisible by p - 1). First, let’s check that the result still holds true when there are no elements of order p . Then 0 = ( p - 1 ) · 0. Now, given any one element g G of order p , all powers of g (except the identity) have order p . That is, since g p = 1; each of g 1 , g 2 , g 3 ,..., g p - 1 also have order p . This is because for 1 k p - 1 ( g k ) n = 1 g kn = 1 p | kn , because p is the order of g p | k or p | n , since p is prime p | n , since p is prime and 1 k p - 1 . Thus, the order of g k is p since ( g k ) p = ( g p ) k = 1 k = 1. [If p wasn’t prime, then g k would have order p only when ( k , p ) = 1 ( k relatively prime to p ).] To finish the proof, consider the set { g 1 , g 2 ,..., g p - 1 } . It either contains every element in G of order p , or it doesn’t. If not, then there is another h G of order p that’s not a power of g . Then we know that each of the p - 1 distinct powers of h also have order p , and the set { g 1 , g 2 ,..., g p - 1 , h 1 , h 2 ,..., h p - 1 } (again) either contains every element of G of order p or doesn’t. Continuing in this way, we will eventually count every element of G of order p , since there are a finite number of elements in G . ± GK2. One form of Lagrange’s theorem says that the order of every element g G divides the order of G . Cauchy’s theorem is a partial converse, that if a prime p divides the order of G , then there is an element of order p . a. Given two distinct primes p and q , find a group G whose order is divisible by both p and q , but which has no element of order pq . (Hint: It’s one of the standard groups of the course.) Solution: Given primes p and q , assume p < q . Consider the symmetric group S q . Then | S q | = q ! = q ( q - 1 ) ··· p ··· 2 · 1 , but S q has no element of order pq . Since p and q are both prime, any permutation of order pq must either be a pq -cycle or a product of a p -cycle and q -cycle (when written as a product of disjoint cycles). But S q acts on the numbers 1– q , and there are simply not enough numbers for either possibility. ± b. For each n , find a group G with n elements that has an element of order a for every divisor a | n . I.e. , it should be the same G for each a . (Hint: Again, it’s one of the standard groups of the course.) Solution: The cyclic group C n = { 1 , x , x 2 , x 3 ,..., x n - 1
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This homework help was uploaded on 02/01/2008 for the course MATH 150A taught by Professor Kuperberg during the Spring '03 term at UC Davis.

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SOLUTION 6 - Math 150a: Modern Algebra Homework 6 Solutions...

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