Right-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., right-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 right-transitively homomorph-containing subgroup|Get more facts about subhomomorph-containing subgroup
EXPLORE EXAMPLES YOURSELF: View examples of subgroups satisfying property right-transitively homomorph-containing subgroup but not subhomomorph-containing subgroup|View examples of subgroups satisfying property right-transitively homomorph-containing subgroup and subhomomorph-containing subgroup

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:

  • Every homomorph-containing subgroup of H is a homomorph-containing subgroup of G: First, note that H itself is homomorph-containing in G, because, since H is simple, no homomorphic image of H can project nontrivially onto the second direct factor. Since H is simple, the only homomorph-containing subgroups of H are H and the trivial subgroup. Both of these are homomorph-containing in <amth>G</math>.
  • H is not subhomomorph-containing in G: H has cyclic subgroups of order two, that are isomorphic to cyclic subgroups of order two outside H, so H is not subhomomorph-containing in G.