Endomorphism kernel is quotient-transitive

This article gives the statement, and possibly proof, of a subgroup property (i.e., endomorphism kernel) satisfying a subgroup metaproperty (i.e., quotient-transitive subgroup property)
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Statement

Suppose $H\leq K\leq G$ are groups such that $H$ is an endomorphism kernel in $G$ and the quotient group $K/H$ is an endomorphism kernel in $G/H$. (Note that it makes sense to take quotients because endomorphism kernel implies normal). Then, $K$ is an endomorphism kernel in $G$.

Related facts

Similar facts about similar properties

• Normality is quotient-transitive
• Characteristicity is quotient-transitive
• Complemented normal is quotient-transitive

Opposite facts about endomorphism kernel

• Endomorphism kernel is not finite-intersection-closed
• Endomorphism kernel is not transitive

Definitions used

We use the following definition of endomorphism kernel: a normal subgroup $A$ of a group $B$ is an endomorphism kernel if there exists a subgroup of $B$ isomorphic to the quotient group $B/A$.

Facts used

1. Third isomorphism theorem: This basically tells us that $(G/H)/(K/H)\cong G/K$.

Proof

Given: Groups $H\leq K\leq G$ such that $H$ is an endomorphism kernel in $G$ and $K/H$ is an endomorphism kernel in $G/H$.

To prove: $K$ is an endomorphism kernel in $G$.

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	Let $L$ be a subgroup of $G$ isomorphic to $G/H$ and let $\alpha :G/H\to L$ be an isomorphism between them.		$H$ is an endomorphism kernel in $G$		Definition-direct
2	Let $M$ be a subgroup of $G/H$ isomorphic to $(G/H)/(K/H)$.		$K/H$ is an endomorphism kernel in $G/H$.		Definition-direct
3	$(G/H)/(K/H)$ is isomorphic to $G/K$.	Fact (1)	$H\leq K\leq G$, $H$ normal in $G$, $K/H$ normal in $G/H$ (normality follows from being endomorphism kernels)		Fact-direct
4	$M$ is a subgroup of $G/H$ isomorphic to $G/K$.			Steps (2), (3)	step-combination direct
5	$\alpha (M)$ is a subgroup of $L$ (and hence of $G$) isomorphic to $M$			Steps (1), (2)	$\alpha :G/H\to L$ is an isomorphism, hence it maps subgroups to subgroups isomorphic to them. Apply this to $M$.
6	$\alpha (M)$ is a subgroup of $G$ isomorphic to $G/K$.			Steps (4), (5)	step-combination direct.
7	$K$ is an endomorphism kernel in $G$.			Step (6)	definition-direct.