Left-transitively homomorph-containing not implies subhomomorph-containing

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This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., left-transitively homomorph-containing subgroup) need not satisfy the second subgroup property (i.e., subhomomorph-containing subgroup)
View a complete list of subgroup property non-implications | View a complete list of subgroup property implications
Get more facts about left-transitively homomorph-containing subgroup|Get more facts about subhomomorph-containing subgroup
EXPLORE EXAMPLES YOURSELF: View examples of subgroups satisfying property left-transitively homomorph-containing subgroup but not subhomomorph-containing subgroup|View examples of subgroups satisfying property left-transitively homomorph-containing subgroup and subhomomorph-containing subgroup

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Proof

Let G be the direct product:

G := A_5 \times C_2

where A_5 is the alternating group of degree five and C_2 is the cyclic group of order two. Let H be the first direct factor.

Then:

  • For any group K containing G as a homomorph-containing subgroup, K also contains H as a homomorph-containing subgroup: For any homomorphism \alpha:H \to K, we can extend it to a homomorphism \beta:G \to K, since H is a direct factor of G. Since G is homomorph-containing in K, \beta(G) \le G, so \alpha(H) \le G. Now, since H is homomorph-containing in G, \alpha(H) is contained in H.
  • H is not a subhomomorph-containing subgroup of G: H has cyclic subgroups of order two, that are isomorphic to cyclic subgroups of order two in G and outside H.