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.

- (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

**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.

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

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].)

**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

**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.

David Joyner 2007-08-06