* Cyclic groups* are groups in which every
element is a power of some fixed element. (If the group is abelian
and I'm using + as the operation, then I should say instead that
every element is a *multiple* of some fixed element.) Here are
the relevant definitions.

* Definition.* Let G be a group, . The * order* of g is the
smallest positive integer n such that . If there is no
positive integer n such that , then g has * infinite order*.

In the case of an abelian group with + as the operation and 0 as the identity, the order of g is the smallest positive integer n such that .

* Definition.* If G is a group and , then the * subgroup generated
by* g is

If the group is abelian and I'm using + as the operation, then

* Definition.* A group G is *
cyclic* if for some . g is a * generator* of .

If a generator g has order n, is * cyclic of order n*. If a generator g has infinite
order, is * infinite
cyclic*.

* Example.* (* The integers and
the integers mod n are cyclic*) Show that and for are cyclic.

is an infinite cyclic group, because every element is a multiple of 1 (or of -1). For instance, .

(Remember that " " is really shorthand for --- 1 added to itself 117 times.)

In fact, it is the only infinite cyclic group up to *
isomorphism*.

Notice that a cyclic group can have more than one generator.

If n is a positive integer, is a cyclic group of order n generated by 1.

For example, 1 generates , since

In other words, if you add 1 to itself repeatedly, you eventually cycle back to 0.

Notice that 3 also generates :

The "same" group can be written using multiplicative notation this way:

In this form, a is a generator of .

It turns out that in , every nonzero element generates the group.

On the other hand, in , only 1 and 5 generate.

* Lemma.* Let be a
finite cyclic group, where g has order n. Then the powers are distinct.

* Proof.* Since g has order n, g, , ... are all different from 1.

Now I'll show that the powers are distinct. Suppose where . Then and , contrary to the preceding observation.

Therefore, the powers are distinct.

* Lemma.* Let be
infinite cyclic. If m and n are integers and , then .

* Proof.* One of m, n is larger --- suppose
without loss of generality that . I want to show
that ; suppose this is false, so . Then , so g has finite order. This
contradicts the fact that a generator of an infinite cyclic group has
infinite order. Therefore, .

The next result characterizes subgroups of cyclic groups. The proof uses the Division Algorithm for integers in an important way.

* Theorem.* Subgroups of cyclic groups are
cyclic.

* Proof.* Let be a
cyclic group, where . Let . If , then H is cyclic with generator
1. So assume .

To show H is cyclic, I must produce a generator for H. What is a
generator? It is an element whose powers make up the group. *A
thing should be smaller than things which are "built from"
it* --- for example, a brick is smaller than a brick building.
Since elements of the subgroup are "built from" the
generator, the generator should be the "smallest" thing in
the subgroup.

What should I mean by "smallest"?

Well, G is cyclic, so everything in G is a power of g. With this discussion as motivation, let m be the smallest positive integer such that .

Why is there such an integer m? Well, H contains something other than
, since . That "something
other" is either a positive or negative power of g. If H
contains a positive power of g, it must contain a *smallest*
positive power, by well ordering.

On the other hand, if H contains a negative power of g --- say , where --- then , since H is closed under inverses. Hence, H again
contains positive powers of g, so it contains a *smallest*
positive power, by Well Ordering.

So I have , the smallest positive power of g in H. I claim that generates H. I must show that every is a power of . Well, , so at least I can write for some n. But by the Division Algorithm, there are unique integers q and r such that

It follows that

Now , so . Hence,
, so . However, was the *smallest positive
power of g lying in H.* Since and , the only way out is if . Therefore, , and .

This proves that generates H, so H is cyclic.

* Example.* (* Subgroups of the
integers*) Describe the subgroups of .

Every subgroup of has the form for .

For example, here is the subgroup generated by 13:

* Example.* Consider the following subset of :

(a) Prove that H is a subgroup of .

(b) Find a generator for H.

(a) First,

If , then

If , then

Hence, H is a subgroup.

(b) Note that . I'll show that .

First, if , then

Therefore, .

Conversely, suppose . I must show .

The idea is to write 2 as a linear combination of 30, 42, and 70. I'll do this in two steps.

First, note that , and

(You can do this by juggling numbers or using the Extended Euclidean algorithm.) Now , and

Plugging into the last equation, I get

Now multiply the last equation by n:

This shows that .

Therefore, .

* Lemma.* Let G be a group, and let have order m. Then if and only if m
divides n.

* Proof.* If m divides n, then for some q, so .

Conversely, suppose that . By the Division Algorithm,

Hence,

Since m is the smallest positive power of g which equals 1, and since , this is only possible if . Therefore, , which means that m divides n.

* Example.* (* The order of an
element*) Suppose an element g in a group G satisfies . What are the possible values for the order of g?

The order of g must be a divisor of 45. Thus, the order could be

And the order is certainly not (say) 7, since 7 doesn't divide 45.

Thus, the order of an element is the *smallest* power which
gives the identity the element in two ways. It is *smallest*
in the sense of being *numerically* smallest, but it is also
*smallest* in the sense that it *divides* any power
which gives the identity.

Next, I'll find a formula for the order of an element in a cyclic group.

* Proposition.* Let be a cyclic group of order n, and let . Then has order .

* Proof.* Since divides m, it
follows that is an integer. Therefore, n
divides , and by the last lemma,

Now suppose that . By the preceding lemma, n divides , so

However, , so divides k. Thus, divides any power of which is 1, so it is the order of .

In terms of , this result says that has order .

* Example.* (* Finding the order
of an element*) Find the order of the element in the cyclic group . (Thus, G is cyclic of order 38 with generator a.)

In the notation of the Proposition, and . Since , it follows that has order .

* Example.* (* Finding the order
of an element*) Find the order of the element .

In this case, I'm using *additive* notation instead of
multiplicative notation. The group is cyclic with order , and the element
corresponds to in the Proposition --- so .

, so the order of 18 is .

Next, I'll give two important Corollaries of the proposition.

* Corollary.* The generators of are the elements of which are relatively prime to n.

* Proof.* If is a generator, its order is n. The Proposition says
its order is . Therefore, , so .

Conversely, if , then the order of m is

Therefore, m is a generator of .

* Example.* (* Finding the
generators of a cyclic group*) List the generators of:

(a) .

(b) , where p is prime.

(a) The generators of are 1, 5, 7, and 11. These are the elements of which are relatively prime to 12.

(b) If p is prime, the generators of are 1, 2, ..., .

* Example.* (a) List the generators of .

(b) List the elements of the subgroup of .

(c) List the generators of the subgroup of .

(a) The generators are the elements relatively prime to 9, namely 1, 2, 4, 5, 7, and 8.

(b)

(c) is cyclic of order 9, so its generators are the elements corresponding to the generators 1, 2, 4, 5, 7, and 8 of . Since , I can just multiply these generators by 3.

Thus, the generators of are 3, 6, 12, 15, 21, and 24.

* Corollary.* A finite cyclic group of order n
contains a of order m for each positive integer m which divides n.

* Proof.* Suppose G is a finite cyclic group of
order n with generator g, and suppose . Thus, for some p.

I claim that generates a subgroup of order m. The preceding proposition says that the order of is . However, , so . Therefore, has order

In other words, generates a subgroup of order m.

In fact, it's possible to prove that there is a *unique* a
subgroup of order m for each m dividing n.

Note that for an *arbitrary* finite group G, it isn't true
that if , then G contains a cyclic subgroup of
order n.

* Example.* (* Subgroups of a
cyclic group*) (a) List the subgroups of .

(b) List the subgroups of .

(a) contains subgroups of order 1, 3, 5, and 15, since these are the divisors of 15. The subgroup of order 1 is the identity, and the subgroup of order 15 is the entire group.

The last result says: If n divides 15, then there is a subgroup of
order n --- in fact, a *unique* subgroup of order n.

Since is cyclic, these subgroups must be cyclic. They are generated by 0 and the nonzero elements in which divide 15: 1, 3, and 5.

* Lagrange's theorem* (which I'll discuss later)
says that in any finite group, the order of a subgroup must divide
the order of the group. In this context, Lagrange's theorem says if H
is a subgroup of order n, then n divides 15.

Putting these results together, this means that you can find
*all* the subgroups of by taking
(the trivial subgroup), together with the cyclic
subgroups generated by the nonzero elements in which divide 15: 1, 3, and 5.

1 generates .

5 generates a subgroup of order 3:

3 generates a subgroup of order 5:

(b) Since the divisors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24, the subgroups of are:

The subgroup generated by 3 has order 8:

* Example.* (* A product of
cyclic groups*) Consider the group

Show that is cyclic by finding a generator.

The operation is componentwise addition:

It is routine to verify that this is a group, the *
direct product* of and .

The element has order 6:

Hence, is cyclic of order 6.
More generally, if , then is cyclic of order . Be careful! --- is *not* the same as !

Copyright 2018 by Bruce Ikenaga