Homomorphisms
Group Homomorphism
Let
and be groups. A homomorphism is a map that satisfies for all .
Proposition
If
is a group homomorphism, then the identity of is mapped to the identity of .
Kernel & Image
Let
be a group homomorphism. The kernel and the image of are defined by
Proposition
A homomorphism
is injective if and only if .
Proof If
Normal Subgroup
Conjugate
Let
be a group. For any , we say and and conjugate if there exists such that .
Normal Subgroup
Let
be a group. A subgroup is called normal if for every . We denote a normal subgroup by .
e.g. Let
Proposition
The followings are true:
is normal iff the right coset coincides with the left coset for every . - We required in the definition that
, but it follows that . - Every subgroup of an abelian group is normal.
Definition
Def Simple Group A group is simple if there’s no non-trivial normal subgroups.
Normaliser
Let
be a group and . The normaliser is the subgroup of that In other words, it is the largest subgroup of in which is normal.
Proposition
.
Isomorphisms
Group Isomorphism
A group homomorphism
is called isomorphism if it has an inverse, that is a homomorphism such that and .
Proposition
Prop A homomorphism
is an isomorphism if and only if it is bijective.
Proposition
Prop Groups of prime order are cyclic.
Proposition
Prop Any two cyclic group of the same order are isomorphic.
Definition
Def Lambda-Map For every
define a map by left multiplication with :
Proposition
Prop
.
Proposition
Prop
is the inverse of .
Cayley’s Theorem
The map
given by is an injective group homomorphism.
Proof Clearly
Corollary
Corollary Every finite group is isomorphic to a subgroup of a symmetric group
. Proof For any finite group , the map forms an isomorphism, thus for some .
Definition
Def Automorphism An automorphism is an isomorphism from a group
to . Denote as the set of all automorphisms on .
Conjugation
Definition
Def Conjugation Let
be a group and . The map is called conjugation by . We say that the elements and are conjugate.
Proposition
Prop Let
be a group and . Then is an isomorphism. Proof We first check that is a homomorphism: The map is injective because . And is surjective since for all , we have .