# Cyclic Groups

Now we return to the case of , a cyclic group. Clearly, . If is finite, then of course must be finite, say . Then recalling our discussion of elements of finite order (see Property 10 from Chapter 2),

Thus we see in this case.

Next, let and be two infinite cyclic groups. Consider the mapping given by (). This mapping is well-defined since is an infinite cyclic group generated by (all powers of are distinct since is infinite) and f is onto . Also, so preserves the operation. Finally if then . Since is infinite cyclic, generated by , however, this implies ; hence . Thus is an isomorphism mapping onto and so . We have shown that any two infinite cyclic groups are isomorphic. We note that since , the additive group of integers, is infinite cyclic generated by , we have proven any infinite cyclic group is isomorphic to .

Next suppose that and are finite cyclic groups such that . Then , and . Define again by . To show that is well-defined, we note that if , with say, then and therefore (again see Property 10 Chapter 2). Therefore , since , and . Thus is well-defined. The rest of the justification needed to show is an isomorphism of ontp is left as an exercise (see exercise 1 for this section). Hence . This shows that any finite cyclic group of order is isomorphic to (see Example 5.1.4).

As our next consideration, we wish to determine the subgroups of a cyclic group. Suppose that is a cyclic group and let be any subgroup of . Let be the smallest positive integer such that ; such a exists because must contain at least one positive power of . For if and then either or . If , we have a positive power of in , i.e., . If , then and again we have a positive power of in , i.e., . We claim that . For suppose that . Then for some integer since . By the Division Algorithm, , . Then and since . This implies by the minimality of , that and . So , i.e., . Therefore we have shown that any subgroup of a cyclic group is also cyclic.

Now suppose that is an infinite cyclic group. We claim that has infinitely many subgroups. If is any positive integer, then . Let be positive integers. If then we could write for some and and for some . Hence . But , thus , and . Again since has infinite order and since and are positive, we must have . The only way to avoid this contradiction is to conclude that the hypothesis is false. In other words, we have shown that if then . This proves the claim.

Suppose next that , and . If is a subgroup, then we know, , where is the smallest positive integer such that . We claim that in this case . For , so by our earlier argument on the minimality of (in the paragraph above where we showed any subgroup of a cyclic group is also cyclic). Conversely if is a positive integer such that , then . Moreover if and are positive integers such that and and then . For otherwise (i.e., assume instead that ), we must have and . Thus , but so , so . Similarly and so since we must have , a contradiction.

Summarizing, we see that for a finite cyclic group , with , then any subgroup of is of the form , where and . Moreover, for each positive dividing , there is such a subgroup. Also distinct divisors determine distinct subgroups. Finally since for a positive divisor of one has if and only if if and only if , we conclude from Property 10 that , as runs through the positive divisors of . Of course, if goes through all the positive divisors of , so does . Also note that the trivial subgroup . Thus in the case of cyclic groups of finite order, the converse of Lagrange's theorem is true, viz, for each , , there exists a (cyclic) subgroup of order . Moreover, in this case there is precisely one such subgroup (see exercise 2 for this section).

The converse of Lagrange's Theorem is not true in general, i.e., if and , , there need not exist a subgroup of order . The smallest example is the group of order ; it turns out that has no subgroup of order . This will be left as an exercise (see exercise 4 for section 6.2) after more tools are developed to handle it efficiently. We shall meet further instances of this and will point it out for specific cases later. We point out now that it was no accident that the divisor was taken to be composite, i.e., not a prime, for prime divisors or for prime power divisors for that matter, there must exist subgroups of such orders. These matters will be attended to when we discuss the Sylow theorems.

Let us now summarize our results for cyclic groups.

Theorem 5.2.1
(a)
Any two infinite cyclic groups are isomorphic (to ).

(b)
Any two finite cyclic groups of the same order are isomorphic (to some , - see Example 5.1.4).

(c)
Any subgroup of a cyclic group is cyclic. In particular if and , then , where is the smallest positive integer such that .

(d)
For an infinite cyclic group , and for each positive , , and distinct positive 's determine distinct subgroups.

(e)
If is a finite cyclic group of order , then the in part (c) divides . Moreover to each positive divisor of there is one and only one cyclic subgroup of that order.

We shall see presently that (e) really characterizes finite cyclic groups, but first we establish an extremely useful theorem, which will be used many times. L

Theorem 5.2.2   Let be an arbitrary group and let a such that . Then .

Proof: Let . Then which implies that (see Property 10 of elementary properties of groups, in chapter 2). Now write (see Corollary 1.2.7)

where . Hence,

or . because of Theorem 1.2.8 and the fact that , we have . But , so . Now . Thus . Therefore combining these two results, .

We apply the theorem immediately to the case of a finite cyclic group of order . Let . By the theorem . Thus (and is consequently also a generator of ) if and only if . In other words, if is a cyclic group of order , then has generators, where is the Euler -function. Summarizing, we have the following result.

Corollary 5.2.3   Suppose and . Then is a generator of if and only if and are relatively prime. Thus there are generators of .

Using the same notation as before, for a cyclic group with , let us see which elements, , of are of order . Again by Theorem 5.2.4, if and only if . Thus . In which case, we can write , , where . Thus the elements of order are those of the form , where . and since . There are then, of course, elements of order . Since , we have

 (5.1)

an interesting number theoretic relationship involving the -function.

We shall make use of (5.1) directly in the following theorem. The author is indebted to W. Wardlaw for pointing this theorem out to him. (Also see [R].)

Theorem 5.2.4   If and if for each positive such that , has at most one cyclic subgroup of order , then is cyclic (and consequently, has exactly one cyclic subgroup of order ).

Proof: For each , , let = the number of elements of of order . Then

Now suppose that for a given . Then there exists an of order which generates a cyclic subgroup, , of order , of . We claim all elements of of order are in . Indeed, if with and , then is a second cyclic subgroup of order , distinct from . This contradicts the hypothesis, so the claim is proven. Thus, if , then . In general, we have , for all positive . But , by our previous work. It follows, clearly, from this that for all . In particular, . Hence, there exists at least one element of of order ; hence is cyclic. This completes the proof.

Corollary 5.2.5   If in a group of order , for each , the equation has at most solutions in , then is cyclic.

Proof: The hypothesis clearly implies that can have at most one cyclic subgroup of order since all elements of such a subgroup satisfy the equation. So Theorem 5.2.4 applies to give our result.

We have given a few applications along the way in this section and the preceding one of some of our results to number theory. We end this chapter with one further application. In particular, we claim that Theorem 4.3.6 is a generalization of Theorem 1.2.11, i.e., . Let us see how this result follows from Theorem 4.3.6. We shall apply the theorem to the case where is a cyclic group of finite order. This frequently, as we shall see in other instances, yields a number theoretic result as a special case of a group theoretical result. Let , be arbitrary positive integers. Choose a positive integer such that and . Then take with . Let and . Then from Theorem 5.2.2, , and . It is not difficult to show (see exercise 3 for this section) that

Thus Theorem 5.2.2 also implies that

Then Theorem 4.3.6 tells us that

which implies that as desired.

Subsections

David Joyner 2007-08-06