Endomorphism kernel is quotient-transitive

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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 \le K \le 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

Opposite facts about endomorphism kernel

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 \le K \le 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 \le K \le 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.