Besides studying groups by decomposing them into
substructures, it's also possible to study group by
building a special map between two groups that
preserves some group structure.
Consider the homogeneous matrix group
under multiplication and the
non-zero real number under multiplication
. It's possible to draw a map
defined by , and
there's when
, which means
the elements' multiplication is preserved under such
map. And such map is common in mathematical world.
In this text, you will soon find out the possibility
of drawing a map between two groups reveals some
non-trivial structural similarities between them,
which can be utilized as a powerful tool to create
solutions and proofs to group theory problems.
Group homomorphisms
In group theory, among possible maps between elements
of two groups and , we're
especially interested in maps such that
holds. Such maps are called group homomorphisms
from to .
Since a group homomorphism is a map in the first place,
when working with a group homomorphism, you'll have to
keep its map instincts in mind, and consider whether it
has injectivity and surjectivity.
The most ideal case is that group homomorphism is
injective and surjective, in which case the elements
are just renamed correspondingly between two groups.
So you will see there's a one-to-one correspondence
between that
, so the group
structure is kept identical. Such group homomorphism
is called a group isomorphism. We also denote
such relation between two isomorphic groups as
.
However, in order to get anywhere further, we'll have
to do some algebraic reasoning on the homomorphism.
Element constrains
Although a map in the first place, a group homomorphism
is constrained by the group structure
of and , and the requirement of
.
So a group homomorphism is not on the loose and should
have some part of it restrained.
Let's substitute the identity of in, we get
.
So is the element that fulfils in
group and is the identity of ,
which means a group homomorphism always map identity
to identity.
Let's select , we get
.
So , which means a group
homomorphism should always map the inverse element to
a image that is inverse of the element's image.
Let's select , and we will get
,
,
which means the binary operation of image of 's
element is still in the image of . Noted that it
won't be hard to show and
, so by the
element constrain of a subgroup .
This is also applicable to subgroup of , by
trimming the domain and codomain of to ,
and repeat the reasoning of .
So when the group homomorphism is injective, noted
that there's a bijection between and .
And since is a subgroup of , we will
get .
Other group homomorphism might seem wild since there's
not yet been applicable properties to restrain their
behaviour. But we will see how these "wild" group
homomorphisms are constrained generally under the
first isomorphism theorem.
First isomorphism theorem
Will it sound a little bit crazy when the structure
of the homomorphically mapped group
completely depends on the kernel of the homomorphism
on
the group, rather than concrete map definition of
the homomorphism? Or more precisely, the relation
between the kernel and the image is defined by:
The statement leads to possible homomorphisms
between and completely depends on the
structure of and . And if or
is finite, the amount of possible homomorphism is
also finite.
This sounds even more crazy, but once you've got
the point of its proof, you will get a step closer
to the beauty of abstraction and symmetry.
First, let's show that . For the
identity, obviously so
. For the inverse element, noted that
and , so and
. Finally given two element
, ,
so and so does .
So there's in the first place.
Second, let's show that .
,
,
so ,
and there's .
Finally, since , there
exists a quotient group , and we are
about to show , which
means we need to find a group homomorphism
which is an
isomorphism. It won't be hard to find that
,
so let's begin with .
To show it's group homomorphism, noted that
,
consider
,
so is indeed a group homomorphism.
To show it's injective, let's choose some coset such
that while
.
Multiplying by yields
,
so and
, which is a contradiction
to .
To show it's surjective, let's assume there's some
such that
.
Noted that so
it is a contradiction to such coset's non-existence.
By now, we've shown there exists a group isomorphism
defined by
, which concludes
our proof of .
The first isomorphism theorem of group shows some
extra symmetries of group homomorphism constrained
by the group structure: the group homomorphism uses
the kernel on the group as the mould,
putting together every elements
(based on the coset formed by , of course)
and forges them into a pinpoint in .
So if a group homomorphism wants to be injective,
it must only put together exactly one element in
each coset, so , in which case the
only possibility is .
More specifically, consider the normal subgroup
, the map
defined by
is group homomorphism since
,
the kernel of which is since
.
Such group homomorphism is also called the
natural homomorphism or canonical homomorphism
of . Under the first isomorphism theorem, all
group homomorphisms such that
is equivalent to the natural homomorphism of .
One must take extra attention to a pitfall that
merely having
does not necessarily mean is bijective.
You might have seen examples of infinite sets mapping
to itself: given non-zero complex number
,
which forms a group of multiplication. Consider
defined by , its codomain is
still since the range of
remains unchanged, while
, which means is
not bijective.
Direct products
Given two group and , their
external direct product is defined as
group such that
.
Let's prove is a group:
- (Closed) Noted that and
, so
.
- (Identity) Consider , obviously
,
so is the identity.
- (Associative)
,
,
so binary operation is associative.
- (Invertible)
,
,
so binary operation is invertible.
On the other hand, consider another group
constructed in similar way
as above. Obviously there exists an group
isomorphism
defined by . In this way
there's . So when we
decompose a group into external direct product of
a set of groups, their order does not matter.
Given two subgroup of ,
define ,
when there're and
,
their internal direct product exists, is
group , and :
- Since , by definition
.
- , there exists
which is the inverse
element for .
- ,
.
When is an external direct product
and is an internal direct product,
consider a map
defined by . Since
,
it is indeed a group homomorphism. Consider the
kernel , assume there exists some
, due to
there's , and
since is a group, there's ,
similarily there's . Which means
and is a
contradiction to . So
, and according to the
first isomorphism theorem, is a group
isomorphism, and the connection between external
and internal direct product can be restated as:
Conversely, how do we find the internal direct
product when we know ?
Denote the associated group isomorphism as
, obviously there're
and
.
Let's extract some properties from them:
- Since ,
.
- Since ,
and .
Similarily we have ,
and , so .
- Obviously ,
.
So obviously by utilizing
and
we can
conveniently find the internal direct product
corresponding to an external direct product. This
shows us the connection and unity between the two
types of direct products.
Someone might be lazily denoting the existence of the
internal direct product as .
However here comes another common pitfall that being
does not necessarily
mean is an internal direct product of
, given that it is not necessarily the case
.
So we will always claim that " is the
internal direct produt of " instead of
flawed description "".
Again, it is time to ring an alarm that clarifying
which group isomorphism the current context is talking
about is at uttermost priority.
Factorizing a group into direct product is an even
stronger decomposition of group structure than
normal subgroup and quotient group decomposition:
Elements from different internal direct product
component are commutative, and operations on them
do not affect each other, which enables us to
study the behaviours of its every direct product
components first and them combine them together to
describe the group's behaviour.
Second isomorphism theorem
As you might have noticed, when we slack the constrain
for to be direct product of that
,
to , though
no longer a direct product, remains to be
a group. This is done by showing :
- Since , by definition
.
- , consider
its inverse element ,
by definition
,
so .
- ,
,
.
So and is a group,
moreover, it is obvious that
and
exists. However,
defined by is not even a
homomorphism, and is not a direct product.
From the example above, it won't be hard to find out
that a group has direct product decomposition
is just an extremely
ideal case of a group has normal subgroup
decomposition . Is it
possible to generalize both cases in one formula?
To do so, consider their intersection ,
let's show :
- Since , by definition
.
- , since is group,
so that , similarily
, so .
- , since is group,
, similarily , so
.
Since and is a group, we've
proved that and
is also a group. What's more, consider any
, since
, by the closed property
, and since
,
by the normal property of ,
and
. Now we have another
associated quotient group .
Consider a map defined by
, since ,
,
so is a group homomorphism. Noted that
iff
, and by group 's
closed property there will also be ,
so . By first isomorphism theorem:
The second isomorphism theorem, which is the statement
we've deduced above, show the relationship between the
product with a normal subgroup , and the
normalizing intersection with the normal
subgroup. As you might have noticed, the theorem is in
the form of .
Consider the normal subgroup of ,
where . Given that
, apply the second
isomorphism theorem and we get
. However, is
not a direct product of the subgroups since
, which
means the binary operation between elements from the two
subgroups are not commutative.
Consider a special case that
.
Apply the second isomorphism theorem and we get both
and .
Assume there's
,
right multiply by and we get
,
by the condition that
there must be . And
since , ,
there's . So and
is the internal direct product of .
Conversely, when is the internal direct
product of , there's , and
thus implies
and .
It won't be hard to show when at least one of
is finite , combined with
are equivalent to
is internal direct product of . Let's assume
is finite,
is only possible when , and thus
falls back to the special case we've discussed above.
Third isomorphism theorem
Given a normal subgroup ,
consider some (not necessarily normal) subgroup
, it is intuitive that
. Conversely, consider some subgroup
, it is also intuitive that
. However, intuition without rigid
proof is the root of all mistakes, so let's begin
with drawing some proof.
For any and ,
let's show that :
- (Identity) .
- (Inverse) .
- (Closed) .
For any subgroup , let's show
that :
- (Identity) .
- (Inverse) ,
such that .
- (Closed) .
Finally, we would like to show the map (not a
homomorphism) is bijective. Since
by that , the map
should be surjective, all remaining to be done is
to show is injective.
Assume there're such
that . For each ,
, so
and , conversely we can get
and . Which concludes
our proof of 's injectivity.
So we've proved the map is
bijective, when there's a normal subgroup
partitioning the group into quotient group ,
each super group uniquely
corresponds to a subgroup .
This is also called the correspondence theorem.
More specifically, consider the case that
and ,
consider the map
( must be on the right since might
have more elements and might be smaller)
defined by , since
,
is a group homomorphism. And obviously
,
is the kernel of such homomorphism.
By applying the first isomorphism theorem we get:
The third isomorphism theorem, which we've proved
above, alongside with the correspondence theorem we've
proved earlier, describe the symmetry of quotient
group hierarchy over the same normal subgroup.
Conclusion
In this text, we introduce the concept of group
homomorphism, isomorphism and direct product,
alongside with the three isomorphism theorems.
The first isomorphism theorem tells us that it is
only possible to build homomorphisms from group
to when there's some quotient group
isomorphic to some subgroup .
And all group homomorphism is fundamentally equivalent
to their kernel's natural homomorphism, where their
kernel is some normal subgroup of the group.
The second isomorphism theorem is a generalization
of group decomposition into quotient group with normal
subgroup or direct products. Specifically, it states
will become a direct product when and
normalize each other and their intersection
.
The third isomorphism theorem, alongside with the
correspondence theorem, describe the symmetry of
quotient group hierarchy over the same normal subgroup.
The group homomorphism, isomorphism and direct
product arm us with another powerful tool for
studying group theory, and their intrinsic connection
with normal subgroup and quotient group is a
show-off of how symmetry of group will constrain and
affect other components in mathematical world.