{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# Case 2 n> 0 in this case z/n z = z n and so we see

This preview shows pages 33–35. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Case 2: n > 0. In this case, Z /n Z = Z n , and so we see that G ∼ = Z n . Moreover, by Theorem 4.17, za = 0 G if and only if n | z , and more generally, z 1 a = z 2 a if and only if z 1 ≡ z 2 (mod n ). The order of G is evidently n , and G consists of the distinct elements · a, 1 · a,..., ( n- 1) · a. From this characterization, we immediately have: Theorem 4.22 Let G be an abelian group and let a ∈ G . If there exists a positive integer m such that ma = 0 G , then the least such integer is the order of a . Moreover, if G of finite order n , then ord( a ) | n , and in particular na = 0 G . 28 Proof. The first statement follows from the above characterization. For the second statement, since h a i is a subgroup of G , by Theorem 4.13, its order must divide that of G . Of course, if ma = 0 G , then for any multiple m of m (in particular, m = n ), we also have m a = 0 G . 2 Based on the this theorem, we can trivially derive a classical result: Theorem 4.23 (Fermat’s Little Theorem) For any prime p , and any integer x 6≡ 0 (mod p ) , we have x p- 1 ≡ 1 (mod p ) . Moreover, for any integer x , we have x p ≡ x (mod p ) . Proof. The first statement follows from Theorem 4.22, and the fact that Since Z * p is an abelian group of order p- 1. The second statement is clearly true if x ≡ 0 (mod p ), and if x 6≡ 0 (mod p ), we simply multiply both sides of the congruence x p- 1 ≡ 1 (mod p ) by x . 2 It also follows from the above characterization of cyclic groups that that any subgroup of a cyclic group is cyclic — indeed, we have already characterized the subgroups of Z and Z n in Theorems 4.7 and 4.8, and it is clear that these subgroups are cyclic. Indeed, it is worth stating the following: Theorem 4.24 Let G be a cyclic group of finite order n . Then the subgroups of G are in one- to-one correspondence with the positive divisors of n , where each such divisor d corresponds to a cyclic subgroup of G d of order d . Moreover: • G d is the image of the ( n/d )-multiplication map (or ( n/d )-power map). • G d contains precisely those elements in G whose order divides d ; i.e., G d is the kernel of the d-multiplication map (or d-power map, for multiplicative groups). • G d ⊃ G d if and only if d | d . Proof. Since G ∼ = Z n , this follows immediately from Theorem 4.8, and the discussion in Exam- ple 4.17. We leave the details to the reader. 2 Example 4.36 Since m Z n is cyclic of order n/d , where d = gcd( m,n ), we have m Z n ∼ = Z n/d . 2 Example 4.37 Consider the group Z n 1 × Z n 2 . For m ∈ Z , then the element m ([1 mod n 1 ] , [1 mod n 2 ]) = ([0 mod n 1 ] , [0 mod n 2 ]) if and only if n 1 | m and n 2 | m . This implies that ([1 mod n 1 ] , [1 mod n 2 ]) has order lcm( n 1 ,n 2 ). In particular, if gcd( n 1 ,n 2 ) = 1, then Z n 1 × Z n 2 is cyclic of order n 1 n 2 , and so Z n 1 × Z n 2 ∼ = Z n 1 n 2 . Moreover, if gcd( n 1 ,n 2 ) = d > 1, then all elements of Z n 1 × Z n 2 have order dividing n 1 n 2 /d , and so Z n 1 × Z n 2 cannot be cyclic.cannot be cyclic....
View Full Document

{[ snackBarMessage ]}

### Page33 / 74

Case 2 n> 0 In this case Z/n Z = Z n and so we see that G...

This preview shows document pages 33 - 35. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online