Homomorph-containment is not transitive

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This article gives the statement, and possibly proof, of a subgroup property (i.e., homomorph-containing subgroup) not satisfying a subgroup metaproperty (i.e., transitive subgroup property).
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Statement

Statement with symbols

It is possible to have groups H \le K \le G such that H is a homomorph-containing subgroup of K and K is a homomorph-containing subgroup of G, but H is not a homomorph-containing subgroup of G.

Related facts

Proof

An example

Consider the group:

G = SL(2,5) \times \mathbb{Z}/2\mathbb{Z}.

Let K be the first direct factor SL(2,5) and H be the center of K, which is a cyclic subgroup of order two in K. Then:

  • K is homomorph-containing in G: For any homomorphism from K to G the projection to the second direct factor is trivial since K is a perfect group and therefore has no quotient of order two. Thus, any homomorphism from K to G has image in K.
  • H is homomorph-containing in K: In fact, the only element of order two in K is the non-identity element of H.
  • H is not homomorph-containing in G: There is a a homomorphism from H to G mapping H isomorphically to the second direct factor.