[transformation_group]
Composition can be thought of as an operation . If we fix one of the positions, we have
- Pre-composition , also called right multiplication
- Post-composition , also called left multiplication
The composition of maps/functions is associative:
- If are bijections, the composition is a bijection
- The identity map is a bijection and is the identity element for the composition operation :
- The inverse map is the inverse element for the composition operation :
So we have the concept of transformation groups: the group consisting of all bijections from to itself, and subgroups of , which are usually groups consisting of bijections that preserve some structure on , for example
- has a linear structure, and the bijections that preserve this linear structure form , which is a subgroup of
Similar to how the power set is denoted as , can be denoted as , because the bijections from to itself form the permutation group , and the number of elements in the set is
[binary_operation_group]
There is also the concept of binary operation groups: Example Addition of real numbers, addition in vector spaces
Addition is associative
- Zero is the identity element of the addition operation
- Additive inverse is the inverse element of the addition operation
Similar to curry โ equivalent to โ, transformation and operation group can mutually isomorphically turned into each other
Taking composition as a binary operation on , a transformation group can be isomorphically turned into an operation group
An operation group can be isomorphically turned into a transformation group, for example
-
Left operation as a bijection from to itself
- , by associativity
-
Right operation as a bijection from to itself
- , by associativity
Multiplication of non-zero real numbers also forms a group
Multiplication of non-zero octonions is not associative: generally
So this kind of non-associative operation group is not isomorphic to the associative transformation group of formed by :
[group_homomorfsm]
Def Group homomorphism
This implies
-
by
-
by
Example Homomorphism from to :
[group_isomorfsm] Def Group Isomorphism: is a bijection and are group homomorphisms
Example Isomorphism from to , the inverse map is a homomorphism
Prop If is a homomorphism and is a bijection, then is an isomorphism
Proof
Need to prove is a homomorphism
From surjectivity, forall , there exist such that
From homomorphism
From invertibility
So
[subgroup]
A subgroup of group is defined as
- Subset
- Closed under binary operation
- Closed under inverse operation
equivalently, the identity embedding is group homomorphism
Example The multiplicative group on has a subgroup
Prop are subgroups ==> is a subgroup
Let be a group homomorphism, be a subgroup, then is a subgroup
is a subgroup, thus is a subgroup
[group_kernel] Def Kernel of a group homomorphism
is injective <==>
Suppose is a subgroup, then is a subgroup
[group_action] Def Group action := a homomorphism from a group to the bijective automorphism group of , also called a representation
Or a homomorphism to the image group
Group action can also be written in the following form
And satisfies
Usually is omitted and written as
[orbit] :=
Example acts on , orbit
[isotropy] :=
Example acts on , isotropy = rotation about the axis where lies, which is an embedded
is a subgroup of
Orbit after changing the orbit base point. forall ==>
Proof
Is a bijection. (Invertible.) So
[decomposition_into_orbit]
Proof
Contrapositive
Only need to prove <==
But we have already proven
Example , different orbits are spheres of different radii
Set of orbits :=
We can give the additive decomposition of the acted-upon space as
Isotropy after changing the orbit base point
Mapping
- Homomorphism
- Bijection
[isotropy_in_same_orbit_is_isom] Thus it is a group isomorphism from to itself, and restricting it to gives an isomorphism between subgroups. In other words, if are in the same orbit , then the isotropy groups are isomorphic
can also be written as
Using the inverse image of acting on , can be decomposed
Calculate the inverse image of
-
is generally not a group. For example, when , , thus , because
-
because is a bijection, and thus restricted to is a bijection
-
[orbit_istropy_product_decomposition] The orbit and isotropy form a product decomposition of the group on the set:
For every , select an such that (Axiom of Choice)
Thus there exists a bijection
This implies
It also implies
[conjugate_action] Conjugate action, similar to change of coordinates
Example
- Representation of linear mappings under different bases
- Representation of manifold mappings under different coordinates
It can be considered that forms an action of on itself
Proof
Of conjugate action
- orbit called [conjugate_class]
- isotropy called centralizer of
Example The conjugate-class of a permutation is the cycle
The isotropy of the conjugation action gives that commutes with
where is called the commutator of the group [commutator]
[action_surjective] alias [action_transitive] := The following definitions are equivalent
- is the surjective map
Example acting on is not transitive. acting on is transitive
[action_injective] alias [action_free] := The following definitions are equivalent
- Every orbit is a copy of
[action_faithful] := The following definitions are equivalent
- The group homomorphism of the group action is injective
Proof
if
if
if the group homomorphism of the group action is injective
if
Prop action-free ==> action-faithful
[coset]
Given a subgroup of , we can define the coset
Left coset
Right coset
Left/Right multiplication gives a group action of on
Prop
- The Right/Left cosets are the corresponding orbits
- Left/Right cosets form a partition of : forall , either , or
- The cardinalities of Left/Right cosets are equal:
- For every , the isotropy is , which means , , and thus it is action-free
is the set of orbits
[action_on_coset]
The group can act on
Since is a bijection, thus can map to the entire , thus this action is action-transitive
The isotropy of is , because is a subgroup, this is equivalent to
- ==> the map is a bijection on ==>
- ==> ==>
Therefore, the isotropy of ,
There is product decomposition
[product_group] Let be groups, then is also a group, with multiplication defined as
[subgroup_coset_sub_quotient_decomposition]
Def Subset multiplication operation: forall
Prop satisfies associativity
Specifically, the multiplication operation of cosets in ,
Prop
We know there is a set product decomposition , where is a subgroup. If we want it to become a product-group decomposition under the coset multiplication defined above, we need the coset multiplication to form a group
There are the following equivalent propositions
- , the meaning is, for two elements in coset/orbit, their multiplication is in the coset that lies in.
Now become group, called the [quotient_group] , is group homomorphism, - For every , the left and right cosets are the same
- is a normal subgroup [normal_subgroup] or called an invariant subgroup [invariant_subgroup] , the conjugation group action preserves , and thus can be restricted to to form a group action. , in fact
However, this decomposition is generally not a product-group decomposition
If is a commutative group [commutative_group] alias [abelian_group] , then all its subgroups are also commutative groups and are normal subgroups
For , if there exists such that , then the order of is defined as
Example
Cyclic group
Take
Then
The order of elements in is , while the order of is
Decomposition should be understood as being a group homomorphism embedding and being a group homomorphism covering
The naturalness of this decomposition also depends on whether you think that inherited multiplication on the cosets is a good construction
[simple_group] Group that have no normal subgroup other than and is called simple group
Example let be a finite group, let . Then is a finite set and is a subgroup. There exists a minimum such that , thus . Let group act on the coset space , isotropy , so or is divisible by
Prop is a subgroup of , the set of cosets is isomorphic to the orbit
Proof We construct a bijection. let , pick such that . Consider the mapping
- Injective: ==> ==>
- Surjective: Suppose , take such that , then just take
[Burnside_lemma] Define as the fixed point set of , then there exists a bijection
This implies