- A cyclic group generated by g is

- where

- Induction shows:
![{\displaystyle g^{m+n}=g^{m}\ast g^{n}{\text{ and }}g^{mn}=[g^{m}]^{n}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0e5441361ebe963c2c83f80f9bf87012ea079bcf)
A cyclic group of order n is isomorphic to the integers modulo n with addition
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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

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

- 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
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.