All we've studied so far is treating a group as a
complete black box, discussing theories applicable
to all kinds of groups, and might be really dull.
In this text, we discuss a class of group called
cyclic group, based on which we introduce an
effective method called cyclic subgroups for
ripping down a group and inspecting what kind of
structure can a group possesses.
We will apply the method of cyclic subgroups to
finite abelian groups, which is a super class
of cyclic group, and see how finite abelian groups
and cyclic groups are interconnected structurally.
Cyclic groups
A group is cyclic when all elements result
from repetitive group operations or inverse
operations on a single element. Or formally,
.
The identity and the inverse element
are also included in such notation.
In a cyclic group, the element that generates
all other elements is called the generator.
We usually use the notation to
represent such a group.
We can easily pick up some examples for cyclic groups:
the group is cyclic since any
integer can be generated from repetead sum of ,
and similarily can
be generated from .
The order of element in
is the smallest positive number such that .
It's only possible that
otherwise there will be duplicate or missing
elements in , by
. It's
possible that never return to when the
group has infinite order, like .
Properties
Every cyclic group is abelian,
consider group operation on , obviously
. This shows
us how the associative property of a group and the
fact that elements are derived from repetive group
operation on are tied with its abelian property.
Every subgroup is also
cyclic. When , it is generated by .
Otherwise since all subgroup elements can be written
as , we are able to find the smallest positive
integer such that ,
and in this case .
To prove, assume there's which can't be
generated from . By the group's closed property,
we multiply by multiple times until we
derive that . However given that
, is not the smallest positive
integer and is a contradiction. For the case of
, we just need to invert it into
and then treat it as the case we've discussed above.
The order of subgroups
Now we know every subgroup of is
in the form of . Consider
the subgroups of
,
it won't be hard to verify that
while
,
so subgroups generated from different generators are
not necessarily distinct, and we care about each
exact cases of the subgroups to be distinct.
Let , consider the subgroup
, every element generated would
be . Consider
the Bezout's identity of :
Obviously all possible values of can be and
only be derived from multiplying the Bezout's identity
by some integer so that .
They are in the form of , and there are
of them. So finally:
Conversely, we claim that has a
subgroup of order iff , and the subgroup
is unique. For the existence, By Lagrange's theorem we
know has a subgroup of order
only if , and from the theory above, we know
, so we have
,
and the subgroup of order exists if .
For the uniqueness, the element generator of
subgroup of order fulfils ,
implying and
.
Finally, by the amount of elements we have
,
so the subgroup of order is unique.
Isomorphisms and decompositions
For cyclic group of infinite order,
consider the map
defined by , it's easy to verify the map
is injective and surjective, and thus group isomorphism.
Similarily for cyclic group of
finite order , consider the map
,
defined by , it is also a
group isomorphism. So:
So we just need to study the properties of
or in order to derive all
properties of an arbitrary cyclic group.
Consider the group ,
by the Chinese remainder theorem, there's a bijective map
defined by whose
inverse map is defined by
where
are the Bezout's coefficients for .
By , the identity is preserved under
the map and the map is a group isomorphism. So:
Please notice the decomposition can be done recursively
until the direct product of prime powers, which we've
not proved whether it can be decomposed yet. That is,
for distinct primes :
Now let's prove that
where is prime can't be decomposed into direct
products of smaller groups furtherly.
Let's show that if ,
then must also be cyclic. This is due to we
can convert the external direct product into internal
direct products of and
where
is the
corresponding group isomorphism. Both the internal
direct product are subgroup of and
thus cyclic, and isomorphic to pairwisely.
Then let's show
cannot be decomposed into
.
Assume there's such group isomorphism
,
and where must be generator
of .
It won't be hard to notice that must also be
generators of each groups, otherwise number of the
elements generated from will be smaller than
and contradiction to is group isomorphism.
Noted that
so there are at most elements that
could be generated by , given that
, it
fails to generate all elements in
.
Now we've proved there's no group isomorphism for
such decomposition to become feasible, regardless of
the choice of elements to be generator in the direct
product component groups.
So if we want to decompose
into direct products of smaller groups, they can only
be cyclic and possible decompositions can only be
.
However none of the decomposition is feasible because
. Finally we can sleep
on terminating the decomposition of cyclic groups at
.
Cyclic subgroups
Consider a group without any special property
previously known to us. For arbitrary element
, we can take from
the group since the identity and inverse element of
must be in the group, and their binary operations
must be closed in group. We put elements taken in
this way into a set , and it's easy to
verify :
- .
- .
- .
Obviously is cyclic and generated by ,
so we denote as and
nominate it as the cyclic subgroup.
To our astonishment, the general feasibility of
constructing cyclic subgroup reveals to us that
cyclic is not just characterization of specific
type of groups, but a profound property of some
indispensible substructure imbued in every group.
Groups of prime order are cyclic
For a finite group with prime order , it
is only possible that:
Since , it is always possible to choose
some element that is not identity to construct
a cyclic subgroup .
By Lagrange's theorem, divides
, and since the only
possibility is . This
implies and must be
cyclic group of order .
For cyclic group ,
there's no non-trivial subgroup of it so it is
indeed a simple group.
Cauchy's theorem of finite abelian groups
Remeber the Lagrange's theorem states that when
, must divide , however there's
no guarantee for existence of of certain order.
The Cauchy's theorem sheds a light on determining
the existence of of certain order: for some
prime dividing , there must be a cyclic
subgroup of order .
Despite the full proof of general case of Cauchy's
theorem requires extensive knowledge, we are able to
prove a special case of Cauchy's theorem that is
applicable to finite abelian groups, with knowledge
and tools we've learned so far.
Given a group ,
which is a finite abelian group of order
where is prime and
, assume for any ,
does not divide .
It won't be hard to notice
must divide in this case, which means for all
, . Consider the map
defined by
,
by applying 's abelian property, we derive:
So is a group homomorphism when is abelian.
It won't be hard to notice that the element
maps to each distinct elements in so is
surjective. By the first isomorphism theorem, there's:
So and is a
contradiction to divides . Thus the setting
of does not divide is
infeasible and there must be a cyclic subgroup whose
order is divisible by . Furtherly by properties
of cyclic groups, under such cyclic subgroup there
must furthermore be a cyclic subgroup of , which
concludes our proof.
Finite abelian groups
Cyclic groups of finite order, which we've been
discussing above, are obviously finite and abelian.
However, all finite abelian groups are not
necessarily cyclic. Consider the group
which is sometimes called the Klein four-group,
it is finite and abelian but not cyclic.
However, in this section, we will derive a
decomposition of any finite abelian groups, using
the knowledge and tools we've learned so far.
This will reveal an astonishing affinity between
decomposition of finite abelian groups and
decomposition of finite cyclic groups.
Let's denote arbitrary finite abelian group of
order as . We know nothing about
except for its size and abelian property.
Decomposition into coprime components
Given finite abelian group such that
, there must be:
The are finite abelian group of
specified order, constructed from the method we
propose below.
Given that , consider the Bezout's
identity such that , there's:
Let's construct two sets
and
,
by iterating through all elements in .
And it's easy to show :
- .
- .
- .
Please notice relies on
, which depends on 's
abelian property to repermutate elements in
product. Similarily we can show .
Noted that because
.
The product of elements from are also
commutative deriving from 's being
abelian. So if ,
is the internal direct product of .
Consider the order of elements in , obviously
so
must divide .
Just like what we've done in the proof of Cauchy's
theorem of finite abelian groups , let
and consider the map
defined by
,
which is a group homomorphism and there's
,
and . Similarily
we can derive
and thus , so it is only
possible that and
is the internal direct product of .
Finally, is finite and direct product of
requires . Given that
, the only
possibility remaining is . So
and we can continue
on the decomposition of the finite abelian group.
Decomposition into prime power components
So just like the decomposition of finite cyclic
group, we are able to decompose a finite abelian
group into direct product of smaller finite
abelian groups of prime power order.
However, unlike the finite cyclic group
of prime power
order which cannot be decomposed, finite
abelian group can be furtherly
decomposed under certain circumstance. Like
both cyclic group
of order and Klein four group
are finite abelian group . The question
remains as what are the possible structures can
a finite abelian group be?
According to the Cauchy's theorem of finite
abelian groups, must have a cyclic
subgroup of order . For example,
has
of order , while
has ,
and
of order .
In fact, we are going to prove that
is cyclic if there's only one distinct cyclic
subgroup of order in it.
The case is trivial when or , so
we just need to concern about the case of .
Let's iterate through the set of and
choose any element such with maximal order
(which means
).
The element is of order since
and if exists
such that , the order of
is and is a contradiction.
Since ,
makes sense and has
order . When ,
and we conclude the
proof. When , noted that the quotient group
is also a finite abelian group of cosets, whose
factor is divisible by , by the Cauchy's theorem
of finite abelian groups, there exists some
such that
while
.
Let ,
consider the possible configurations of :
- If is divisible by and , then
, which implies
since there's only one distinct cyclic subgroup
of order . This leads to a contradiction that
.
- If is not divisible by , then
and the order of is
, and .
This leads to a contradiction that
.
So there's no feasible configuration for such
to exist and it is impossible for to hold.
So when there's only one distinct cyclic subgroup
of order , must be cyclic.
What about removing the constraint of single
distinct cyclic subgroup of order in ?
We adopt the same settings as above that is
some element of maximal order in .
But this time, we would like to prove
can be decomposed by:
This is an internal direct product form because
by the second isomorphism theorem there's
,
which requires .
The proof is done with mathematical induction.
For the case of , it is a cyclic
group with generator , , and
.
For the case of , we know the statement
holds for every .
When there's only one distinct cyclic subgroup
of order , is cyclic group with
generator whose order is , and
.
When there're multiple distinct cyclic subgroups
of order , we know
is one of such cyclic subgroup, and let there
be another cyclic subgroup
of order .
To show
,
assume there's some
,
then . Given that the order of
is , must divides and is
also in . Since
every element in cyclic group is generator,
,
which is a contradiction to the distinction of
from .
Consider the quotient group
, obviously there's
.
The order of is , since
if its order is , then there'll be
, which
is a contradiction to
.
The order of is also maximal
in , since for any
,
whose order is , there's
,
the order of must divides
and thus also divides .
The quotient group
is a finite abelian group of order , with
element of
maximal order . We know the statement holds for
arbitrary finite abelian group of order
so there's subgroup of quotient group
of order such that
.
By the correspondence theorem, with the common
normal subgroup , there's
,
and since
,
.
Finally by the second isomorphism theorem
,
and is an internal direct product of
and .
By now, we've proved the statement that a finite
abelian group can be rewritten as
internal product form
,
where is the element with maximal order
in and is a subgroup
with order . You might also express the
statement in its external product form:
In practice, we will first inspect whether
there're multiple cyclic subgroups of order
in , and halt the decomposition when
there's just single, since will be
cyclic group of order then and cannot be
decomposed any furtherly.
Fundamental theorem of finite abelian groups
Summing up the rationale we've discussed about
decomposition of finite abelian group, we can draw
a conclusion that for finite abelian group
where are primes and
not necessarily distinct:
This is a generalization of decomposition of
finite cyclic groups and other finite abelian
groups which are not cyclic.
Meantime, it won't be hard to prove that
is cyclic iff are
distinct. This is done by showing the order of
is ,
iff are distinct, in
the evaluation of their least common multiple.
Conclusion
In this text, we've studied a basic class of group
called cyclic group, by discussing its properties
and decomposition.
Then, we introduced a cyclic group based technique
called cyclic subgroup, which is nearly applicable
to arbitrary group and an important way for
peepholing and proving properties of the group.
Finally, we studied the finite abelian group which
is a super class of finite cyclic group, by applying
cyclic subgroup and techniques we've learned in the
previous texts. As a consequence, we've succesfully
generalized the structures of all finite abelian
groups and made clarified how finite cyclic groups
are interconnected with finite abelian groups.