Microsoft Word - Module III Finite Cyclic Groups.pdf

Microsoft Word - Module III Finite Cyclic Groups.pdf - 1...

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1 Module III n n n Finite Cyclic Groups and Conditions on n for to be Cyclic In Module II we determined that is a finite group of order (n) and that if a then ord (a) (n). Of course, if ord (a) = ( ϕ ϕ ϕ * * * Ζ Ζ ∈ Ζ { } 2 (n)-1 n n n n) then = 1, a, a ,..., a and we say that is cyclic. Our purpose is to determine those values of n for which is cyclic. In preparation for that discussion we b ϕ * * * Ζ Ζ Ζ egin with a general definition of a cyclic group and a result regarding the structure of an arbitary finite cyclic group. { } k Definition 1. Let G be a finite group. If a G such that G = (a) a k 0 then G is said to be cyclic and a is called a generator of G. Remark 1. Of course, if G = (a) then ord(a) = G Δ ( ) k Proposition 1. Let G be a finite group. (i) if a G and ord a = t then ord (a ) = t gcd (k, t) (ii) if G is cyclic and d ord (G) then G has d elements of order d. (iii) if G is cyclic an ϕ g d H G then H is cyclic. Moreover, if d ord(G) then exactly one subgroup of G having order = d. k k kt k t gcd(k, t) cm(k, t) gcd(k, t) k k ord(a ) k ord(a ) Proof (i) Observe that (a ) a = a = e because t cm(k, t). Thus ord(a ) t gcd(k, t). Next realize e = (a ) = a But then t k = lscript lscript k k k k ord (a ) and so k ord(a ) is a common multiple of k and t; whence cm(k, t) k ord(a ). cm(k,t) t Thus ord(a ) = . k gcd(k, t) lscript lscript
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2 t t (ii) Suppose is a generator. We want ord ( ) = d ord (G) But ord ( ) = ord G gcd (t, ord G) i.e. gcd (t, ord G) = . d ord (G) Now this holds if and only if gcd (t , d) = 1. d α α α ord(G) d ord (G) d But t < ord G t < d. ord(G) Conversely, suppose 1 a < d and set t = a d if gcd (a, d) = 1 then gcd (t , d) = 1 Thus 1-1 correspondence between the t's such that [ ] t G d t ord(G) and ord( ) = d and the a's d such that gcd(a, d) = 1 thereby implying that (d) such t's. (iii) Suppose is a generator of G, i.e. ord( ) = G . If d G then ord ( ) = α ϕ α α α d t t t t G gcd ( G , G ) Of course with t = G d H ( ) is cyclic and has H = d so for each divisior of ord G a cyclic subgroup of G having order d. By ii) H contains (d) elements of or α ϕ Δ t g der d. But d G also implies G contains EXACTLY (d) elements of order d. Thus H consists of all of the elements of G having order d. Now suppose H G such that H = d. It remains to show that H ϕ t t t = H . Consider a H so that ord (a) d. But then ord(a) G implies that G contains exactly (ord (a)) elements of ord (a) and ord (a) d = H implies that H contains (ord a) elements of ord(a). Hence H ϕ ϕ t t t contains all of the elements of G having ord (a); in particular a H . Thus H H . ( ) t t t t But H = H forces H= H . Hence every subgroup of G is cyclic and for each d G , H = is the unique subgroup of G having order d. α
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3 { } { } 9 9 9 1 2 3 4 9 1 Example1. Consider . Then (9) = 9 1- = 6 3 and = a gcd (a, 9) = 1 = 1, 2, 4, 5, 7, 8 . Now is cyclic since 2 = 2, 2 = 4, 2 = 8, 2 ϕ * * Ζ Ζ ∈Ζ Ζ 5 9 7 (mod 9), 2 5(mod 9) and it is easy to see that 5 is the other generator. But has two subgroups, one of order 2 and one of order 3. Since (2) = 1 there is one element of order 2 namely 8 so th ϕ * Ζ { } { } e unique subgroup of order 2 is 1,8 . Hence 1, 4, 7 is the sole subgroup of order 3. In summary we have
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