Endomorphism kernel

This article defines a subgroup property: a property that can be evaluated to true/false given a group and a subgroup thereof, invariant under subgroup equivalence. View a complete list of subgroup properties[SHOW MORE]

|Get subgroup property lookup help |Get exploration suggestions
VIEW RELATED: Subgroup property implications | Subgroup property non-implications | Subgroup metaproperty satisfactions | Subgroup metaproperty dissatisfactions | Subgroup property satisfactions |Subgroup property dissatisfactions
VIEW FACTS THAT:use, or start with, a subgroup satisfying the property|prove, or show that, a subgroup satisfies the property

BEWARE! This term is nonstandard and is being used locally within the wiki. [SHOW MORE]

For its use outside the wiki, please define the term when using it. If you are aware of an equivalent standard term, please leave a comment on the talk page
VIEW: Definitions built on this | Facts about this: (facts closely related to Endomorphism kernel, all facts related to Endomorphism kernel) |Survey articles about this | Survey articles about definitions built on this
VIEW RELATED: Analogues of this | Variations of this | Opposites of this |
Learn more about terminology local to the wiki | view a complete list of such terminology

Definition

Equivalent definitions in tabular format

No.	Shorthand	A subgroup of a group is termed an endomorphism kernel if ...	A subgroup $H$ of a group $G$ is termed an endomorphism kernel in $G$ if ...
1	normal, quotient isomorphic to subgroup	it is normal in the whole group and its quotient group is isomorphic to some subgroup of the whole group.	$H$ is a normal subgroup of $G$ and there is a subgroup $M$ of $G$ such that the quotient group $G/H$ is isomorphic to $M$.
2	endomorphism kernel	there is an endomorphism of the whole group whose kernel is precisely the subgroup.	there is an endomorphism $\sigma$ of $G$ such that the kernel of $\sigma$ is $H$

Metaproperties

Metaproperty name	Satisfied?	Proof	Statement with symbols
transitive subgroup property	No	endomorphism kernel is not transitive	It is possible to have group $H\leq K\leq G$ such that $H$ is an endomorphism kernel in $K$ and $K$ is an endomorphism kernel in $G$ but $H$ is not an endomorphism kernel in $G$.
intermediate subgroup condition	No	endomorphism kernel does not satisfy intermediate subgroup condition	It is possible to have groups $H\leq K\leq G$ such that $H$ is an endomorphism kernel in $G$ but is not an endomorphism kernel in $K$.
quotient-transitive subgroup property	Yes	endomorphism kernel is quotient-transitive	Suppose $H\leq K\leq G$ are groups such that $H$ is an endomorphism kernel in $G$ and $K/H$ is an endomorphism kernel in $G/H$. Then, $K$ is an endomorphism kernel in $G$.
trim subgroup property	Yes	obvious	in any group $G$, the trivial subgroup and the whole group $G$ are endomorphism kernels.

Relation with other properties

Stronger properties

Property	Meaning	Proof of implication	Proof of strictness (reverse implication failure)	Intermediate notions
complemented normal subgroup		complemented normal implies endomorphism kernel	endomorphism kernel not implies complemented normal	|FULL LIST, MORE INFO
direct factor				|FULL LIST, MORE INFO
subgroup of finite abelian group		follows from subgroup lattice and quotient lattice of finite abelian group are isomorphic	(trivial subgroup, whole group are endomorphism kernels even in non-abelian groups)	|FULL LIST, MORE INFO

Weaker properties

Property	Meaning	Proof of implication	Proof of strictness (reverse implication failure)	Intermediate notions
quotient-powering-invariant subgroup		endomorphism kernel implies quotient-powering-invariant	any normal subgroup of a finite group that is not an endomorphism kernel works.	|FULL LIST, MORE INFO
powering-invariant normal subgroup		(via quotient-powering-invariant)	(via quotient-powering-invariant)	|FULL LIST, MORE INFO
powering-invariant subgroup		(via quotient-powering-invariant)	(via quotient-powering-invariant)	|FULL LIST, MORE INFO
normal subgroup		(by definition)	normal not implies endomorphism kernel	|FULL LIST, MORE INFO

Effect of property operators

BEWARE! This section of the article uses terminology local to the wiki, possibly without giving a full explanation of the terminology used (though efforts have been made to clarify terminology as much as possible within the particular context)

Operator	Meaning	Result of application	Proof
potentially operator	endomorphism kernel in some larger group	normal subgroup	normal implies potential endomorphism kernel
intermediately operator	endomorphism kernel in every intermediate subgroup	intermediately endomorphism kernel	(by definition)