Subisomorph-containing not implies homomorph-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., subisomorph-containing subgroup) need not satisfy the second subgroup property (i.e., homomorph-containing subgroup)
View a complete list of subgroup property non-implications | View a complete list of subgroup property implications
Get more facts about subisomorph-containing subgroup|Get more facts about homomorph-containing subgroup
EXPLORE EXAMPLES YOURSELF: View examples of subgroups satisfying property subisomorph-containing subgroup but not homomorph-containing subgroup|View examples of subgroups satisfying property subisomorph-containing subgroup and homomorph-containing subgroup

Statement

It is possible to have a group G and a subgroup H of G such that H is a subisomorph-containing subgroup of G (i.e., it contains any subgroup of G isomorphic to a subgroup of H), but H is not a homomorph-containing subgroup: there is a homomorphic image of H in G that is not contained in H.

In particular, this also shows that H is not a Subhomomorph-containing subgroup (?) of G.

Related facts

Proof

Further information: infinite dihedral group

Let G be the infinite dihedral group and H be the cyclic part:

G = \langle a,x \mid x^2 = e, xax = a^{-1} \rangle, \qquad H = \langle a \rangle.

Then:

  • H is a subisomorph-containing subgroup of G: Every subgroup of G contained in H is either trivial or infinite cyclic. On the other hand, every element in G and outside H has order two. Thus, no subgroup of G outside H is isomorphic to a subgroup of H.
  • H is not a homomorph-containing subgroup of G: H is infinite cyclic and the group \langle x \rangle, which is cyclic of order two in G and not contained in H, is a homomorphic image of H.