Fundamental theorem of cyclic groups
Encyclopedia
In abstract algebra
, the fundamental theorem of cyclic groups states that every subgroup
of a cyclic group
is cyclic. Moreover, the order
of any subgroup of a cyclic group
of order 
is a divisor of 
, and for each positive divisor 
of 
the group 
has exactly one subgroup of order 
.
be a cyclic group for some 
and with identity 
and order 
, and let 
be a subgroup of 
.
We will now show that
is cyclic. If
then 
.
If
then since 
is cyclic every element in 
is of the form 
, where 
is an integer. Let 
be the least positive integer such that 
.
We will now show that
. It follows immediately from the closure property that 
.
To show that
we let 
. Since 
we have that 
for some positive integer 
.
By the division algorithm
,
with 
, and so
, which yields 
.
Now since
and 
, it follows from closure that 
.
But
is the least positive integer such that
and 
,
which means that
and so
.
Thus
.
Since
and 
it follows that 
and so 
is cyclic.
We will now show that the order of any subgroup of
is a divisor of 
.
Let
be any subgroup of 
of order 
. We have already shown that
, where 
is the least positive integer such that
. We know that 
and therefore we can write
, with 
.
Since
,
we must have :
, since 
is the smallest positive integer such that
.
It follows that
for some integer 
. Thus 
.
We will now prove the last part of the theorem. Let
be any positive divisor of 
. We will show that
is the one and only subgroup of
of order 
. Note that
has order
.
Let
be any subgroup of 
with order 
. We know that
,
where
is a divisor of 
. So
and 
.
Consequently
and so 
,
and thus the theorem is proved.
be a cyclic group, and let 
be a subgroup of 
. Define a morphism 
by 
. Since 
is cyclic generated by 
, 
is surjective. Let 
. 
is a subgroup of 
. Since 
is surjective, the restriction of 
to 
defines a surjective homomorphism from 
onto 
, and therefore 
is isomorphic to a quotient of 
. Since 
is a subgroup of 
, 
is 
for some integer 
. If 
, then 
, hence 
, which is cyclic. Otherwise, 
is isomorphic to 
. Therefore 
is isomorphic to a quotient of 
, and they are commonly known to be cyclic.
which is generated by n elements. Then each submodule of M can also be generated by n elements (and possibly fewer). This result implies the fundamental theorem of cyclic groups by observing that the ring
of integers satisfies these conditions, and a cyclic group is precisely a left 
-module which is generated by one element. (Its submodules are its subgroups.)
Abstract algebra
Abstract algebra is the subject area of mathematics that studies algebraic structures, such as groups, rings, fields, modules, vector spaces, and algebras...
, the fundamental theorem of cyclic groups states that every subgroup
Subgroup
In group theory, given a group G under a binary operation *, a subset H of G is called a subgroup of G if H also forms a group under the operation *. More precisely, H is a subgroup of G if the restriction of * to H x H is a group operation on H...
of a cyclic group
Cyclic group
In group theory, a cyclic group is a group that can be generated by a single element, in the sense that the group has an element g such that, when written multiplicatively, every element of the group is a power of g .-Definition:A group G is called cyclic if there exists an element g...
is cyclic. Moreover, the order
Order (group theory)
In group theory, a branch of mathematics, the term order is used in two closely related senses:* The order of a group is its cardinality, i.e., the number of its elements....
of any subgroup of a cyclic group
of order is a divisor of , and for each positive divisor of the group has exactly one subgroup of order .Proof
Let be a cyclic group for some and with identity and order , and let be a subgroup of .We will now show that
is cyclic. If then .If
then since is cyclic every element in is of the form , where is an integer. Let be the least positive integer such that .We will now show that
. It follows immediately from the closure property that .To show that
we let . Since we have that for some positive integer .By the division algorithm
Division algorithm
In mathematics, and more particularly in arithmetic, the usual process of division of integers producing a quotient and a remainder can be specified precisely by a theorem stating that these exist uniquely with given properties. An integer division algorithm is any effective method for producing...
,
with , and so, which yields .Now since
and , it follows from closure that .But
is the least positive integer such that and ,which means that
and so.Thus
.Since
and it follows that and so is cyclic.We will now show that the order of any subgroup of
is a divisor of .Let
be any subgroup of of order . We have already shown that, where is the least positive integer such that
. We know that and therefore we can write, with .Since
,we must have :
, since is the smallest positive integer such that.It follows that
for some integer . Thus .We will now prove the last part of the theorem. Let
be any positive divisor of . We will show thatis the one and only subgroup of
of order . Note thathas order
.Let
be any subgroup of with order . We know that,where
is a divisor of . So and .Consequently
and so ,and thus the theorem is proved.
Proof by homomorphism with integers
Let be a cyclic group, and let be a subgroup of . Define a morphism by . Since is cyclic generated by , is surjective. Let . is a subgroup of . Since is surjective, the restriction of to defines a surjective homomorphism from onto , and therefore is isomorphic to a quotient of . Since is a subgroup of , is for some integer . If , then , hence , which is cyclic. Otherwise, is isomorphic to . Therefore is isomorphic to a quotient of , and they are commonly known to be cyclic.Converse
The following statements are equivalent.- A group G of order
is cyclic. - For every divisor
of 
a group G has exactly one subgroup of order 
. - For every divisor
of 
a group G has at most one subgroup of order 
.
Generalization
Suppose that R is a right Ore domain in which every left ideal is principal, and let M be a left R-moduleModule (mathematics)
In abstract algebra, the concept of a module over a ring is a generalization of the notion of vector space, wherein the corresponding scalars are allowed to lie in an arbitrary ring...
which is generated by n elements. Then each submodule of M can also be generated by n elements (and possibly fewer). This result implies the fundamental theorem of cyclic groups by observing that the ring
of integers satisfies these conditions, and a cyclic group is precisely a left -module which is generated by one element. (Its submodules are its subgroups.)