10 are roots of x 10 − 1, and the elements of order 15 are roots of x 15 − 1. Unfortunately, 6 + 10 + 15 = 31. This is too large to draw a conclusion about elements of order 30. The problem is caused by double counting. For example, the elements of order 3 are roots, both of x 6 − 1 and of x 15 − 1. When one eliminates the double counting, one sees that there must be elements of order 30. � It is fussy arithmetic to make a proof based on the method illustrated by these examples. We use a lemma about the orders of elements of an abelian group. Lemma 2. (a) Let u and v be elements of an abelian group G , of finite orders a and b , respectively, and let m be the least common multiple of a and b . Then G contains an element of order m . (b) Let G be a finite abelian group, and let m be the least common multiple of the orders of elements of G . Then G contains an element of order m . Note: The hypothesis that G be abelian is essential here. The symmetric group S 3 , which is not abelian, contains elements of orders 2 and 3 but no element of order 6. proof of the theorem. We’ll prove the theorem, assuming that Lemma 2 has been proved. Let m be the least common multiple of the orders of the elements of F × p . The lemma tells us that F × p contains an element � of order m . Therefore m divides the order of the group, which is p − 1, and m � p − 1. Also, since m is the least common multiple of the orders of the elements of F × p , the order of every element divides m . So every element of F × p is a root of the polynomial x m − 1. Since this polynomial has at most m roots, p − 1 � m .
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- Fall '13
- Prime number, Greatest common divisor, Divisor, Abelian group