Abstract Algebra/Group Theory/Cyclic groups

< Abstract Algebra‎ | Group Theory
  • A cyclic group generated by g is


  • where
  • Induction shows:

A cyclic group of order n is isomorphic to the integers modulo n with additionEdit

TheoremEdit

Let Cm be a cyclic group of order m generated by g with  

Let   be the group of integers modulo m with addition

Cm is isomorphic to  

LemmaEdit

Let n be the minimal positive integer such that gn = e

 
Proof of Lemma
Let i > j. Let i - j = sn + r where 0 ≤ r < n and s,r,n are all integers.
1.  

2.   as i - j = sn + r, and gn = e
3.  

4.   as n is the minimal positive integer such that gn = e
and 0 ≤ r < n

5.   0. and 7.
6.  

ProofEdit

0. Define    
Lemma shows f is well defined (only has one output for each input).
f is homomorphism:
 
f is injective by lemma
f is surjective as both   and   have m elements and f is injective

Cyclic groupsEdit

In the previous section about subgroups we saw that if   is a group with  , then the set of powers of  ,   constituted a subgroup of  , called the cyclic subgroup generated by  . In this section, we will generalize this concept, and in the process, obtain an important family of groups which is very rich in structure.

Definition 1: Let   be a group with an element   such that  . Then   is called a cyclic group, and   is called a generator of  . Alternatively,   is said to generate  . If there exists an integer   such that  , and   is the smallest positive such integer,   is denoted  , the cyclic group of order  . If no such integer exists,   is denoted  , the infinite cyclic group.

The infinite cyclic group can also be denoted  , the free group with one generator. This is foreshadowing for a future section and can be ignored for now.

Theorem 2: Any cyclic group is abelian.

Proof: Let   be a cyclic group with generator  . Then if  , then   and   for some  . To show commutativity, observe that   and we are done.

Theorem 3: Any subgroup of a cyclic group is cyclic.

Proof: Let   be a cyclic group with generator  , and let  . Since  , in particular every element of   equals   for some  . We claim that if   the lowest positive integer such that  , then  . To see this, let  . Then   and   for unique  . Since   is a subgroup and  , we must have  . Now, assume that  . Then   contradicts our assumption that   is the least positive integer such that  . Therefore,  . Consequently,   only if  , and   and is cyclic, as was to be shown.

As the alert reader will have noticed, the preceding proof invoked the notion of division with remainder which should be familiar from number theory. Our treatment of cyclic groups will have close ties with notions from number theory. This is no coincidence, as the next few statements will show. Indeed, an alternative title for this section could have been "Modular arithmetic and integer ideals". The notion of an ideal may not yet be familiar to the reader, who is asked to wait patiently until the chapter about rings.

Theorem 4: Let   with addition defined modulo  . That is  , where  . We denote this operation by  . Then   is a cyclic group.

Proof: We must first show that   is a group, then find a generator. We verify the group axioms. Associativity is inherited from the integers. The element   is an identity element with respect to  . An inverse of   is an element   such that  . Thus  . Then,  , and so  , and   is a group. Now, since  ,   generates   and so   is cyclic.

Unless we explicitly state otherwise, by   we will always refer to the cyclic group  . Since the argument for the generator of   can be made valid for any integer  , this shows that also   is cyclic with the generator  .

Theorem 5: An element   is a generator if and only if  .

Proof: We will need the following theorem from number theory: If   are integers, then there exists integers   such that  , if and only if  . We will not prove this here. A proof can be found in the number theory section.

For the right implication, assume that  . Then for all  ,   for some integer  . In particular, there exists an integer   such that  . This implies that there exists another integer   such that  . By the above-mentioned theorem from number theory, we then have  . For the left implication, assume  . Then there exists integers   such that  , implying that   in  . Since   generates  , it must be true that   is also a generator, proving the theorem.

We can generalize Theorem 5 a bit by looking at the orders of the elements in cyclic groups.

Theorem 6: Let  . Then,  .

Proof: Recall that the order of   is defined as the lowest positive integer   such that   in  . Since   is cyclic, there exists an integer   such that   is minimal and positive. This is the definition of the least common multiple;  . Recall from number theory that  . Thus,  , as was to be proven.

Theorem 7: Every subgroup of   is of the form  .

Proof: The fact that any subgroup of   is cyclic follows from Theorem 3. Therefore, let   generate  . Then we see immediately that  .

Theorem 8: Let   be fixed, and let  . Then   is a subgroup of   generated by  .

Proof: We msut first show that   is a subgroup. This is immediate since  . From the proof of Theorem 3, we see that any subgroup of   is generated by its lowest positive element. It is a theorem of number theory that the lowest positive integer   such that   for fixed integers   and   equals the greatest common divisor of   and   or  . Thus   generates  .

Theorem 9: Let   and   be subgroups of  . Then   is the subgroup generated by  .

Proof: The fact that   is a subgroup is obvious since   and   are subgroups. To find a generator of  , we must find its lowest positive element. That is, the lowest positive integer   such that   is both a multiple of   and of  . This is the definition of the least common multiple of   and  , or  , and the result follows.

It should be obvious by now that   and  , and   and   are the same groups. This will be made precise in a later section but can be visualized by denoting any generator of   or   by  .

We will have more to say about cyclic groups later, when we have more tools at our disposal.