301 Abstract Algebra
2:00 - 2:50 MWF
Hentzel 432 Carver
hentzel@iastate.edu
Web Page
http://www.math.iastate.edu/hentzel/class.301

Text: A First Course in Abstract Algebra, sixth Edition 

by John B. Fraleigh

Course: Chapter 1,2,3,4,5 

The material we are covering today is on Page 75-82 

Section 1.5 beginning with "Cyclic Groups and Generators".   


Main Idea:  The subgroups of finite cyclic groups are the

obvious ones. 

Key Words: Cyclic Group, divisor, | G |, relatively prime


Goal:  If k divided | G |, then G has a unique subgroup of 

order k. 


Definition:  A group is called cyclic if it is generated

by one of its elements.  That is, if there exists a g in G

such that < g > = G.


The order of a set S is the number of elements in S.  The

order of S is indicated by | S |.



Let g be an element of a group G. By the generalized 

associative law, we define for positive n:

g^n = g g g g g g .... g   
     |__________________|
       n factors of g.

g^0 = I


g^(-n) = g' g' g' ... g'     
        |______________|
         n factors of the 
         inverse of n.


Theorem: For all integers m, n, we have  g^m g^n = g^(m+n).

Proof:  We have to consider the cases

               (a)    m or n is zero
               (b)    both m and n are positive
               (c)    both m and n are negative
               (d)    one of them is positive and one of them
                      is negative.
 
   

(a)  If m is zero, then  g^m g^0 = g^m e = g^m = g^(m+0) 

     The case for n = 0 is similar.

(b) If both m and n are positive, then the result is obvious 

because in the product g^m g^n one generates a string of g's 

of length m+n.

We will handle the cases when m and n can be negative by 

writing the negative exponents as -m, or -n with m and n positive.

(c) If both exponents are negative, then

    g^(-m) g^(-n) =  (g')^m (g')^n = (g')(m+n) = g^ -(m+n) = g^(-m-n) 

(d) If one exponent is positive and the other is negative:

                                 g^(n-m) if n>m  = g^(-m+n)   
g^(-m) g^(n) = (g')^m g^(n) =    I       if n=m  = g^(-m+n)
                                 g'(m-n) if n<n  = g^(-m+n)

In the above expression, we used the fact that a' cancels

off an a.  The other case is similar. 
             
 
We have considered all cases and in each case

a^m a^n = a^(m+n) for any integers m and n. 



Theorem:  Let G be a group and g an element of G.

Then <g> = { g^n | n is an integer }.

Proof:  Let S = { g^n | n is an integer. 

We have to show three things.  

               (1)  S is a group

               (2) S contains g.  

               (3) S is the smallest group containing g.


Part (1).  S is closed since g^m g^n = g^(m+n).  S has inverses since 

(g^n)' = g^(-n).  Thus S is a subgroup.  


Part (2)  S contains g since g^1 = g is in S. 


Part (3)  Suppose H is any subgroup containing g.  Since H

is closed, H contains all positive powers of g.  Since H

contains inverses, H contains all negative powers of g.

And H contains the identity so H contains the 0th power of g.




Theorem.  A cyclic group is always commutative.

Proof.  Let G be a cyclic group.  Then by definition

G = < g >.  Since all elements in G are powers of g,

g^m g^n = g^(m+n)  and g^n g^m = g^(m+n). So G is commutative.



Theorem.  Every subgroup of a cyclic group is cyclic.

Proof:  Let H be a subgroup of a cyclic group G = < g >,

Let d be the smallest integer such that g^d is in the

subgroup H.  Since g^d is in H, by closure of H

<g^d> c H.  Let us suppose that <g^d> =/= H.  Then

there must be some element h in H which is not in 

<g^d>.  Since G = < g >, h = g^n.

          n = q d + r where 0 <= r < d. 

if r = 0, then h  = g^n = (g^d)^q  is in < g^d >.

if r =/= 0, then g^n = g^(qd+r) = g^(qd) g^r = (g^d)^q g^r.
                 in H                          <----->   
                                                in H

But then g^r is also in H.  This contradicts that

d was the smallest positive power of g in H.  We must

conclude that H = <g^d>.




Theorem:  Let G be a finite cyclic group of order n.

Then for each divisor k of n, there is a unique subgroup

of order k.  


Proof.   We have two show two things.
 
(1)  There is a subgroup of order k.

(2)  There is only one subgroup of order k. 


Part (1):  If d k = n, and G = < g >, then g^n = e and

<g^d, (g^d)^2, (g^d)^3, (g^d)^4, ..., (g^d)^k=e> is a subgroup

of order k.


Part (2):  Suppose that H is a subgroup of order k.

We know that H = < g^a > where a is the smallest positive integer 

such that g^a is in H.  Then


H = { g^a, (g^a)^2, (g^a)^3, ..., (g^a)^k} and (g^a)^k = e.

Thus ak = n and a = d.  This shows that H = < g^d >. 



We state the following theorem with out proof:

Theorem:  If G = < g > is a finite cyclic group of order n, then

< g^x > = G  when x and n are relatively prime.

< g^x > = < g^d > where d is the greatest common divisor of x and n.




Assignment:

(1)  List all the additive subgroups of Z12.


(2)  List all the additive subgroups of Z19


(3)  How many additive subgroups are there of Z100.


(4)  Find integers x and y such that

         x 13 + y 17 = 1.