Descendance does not satisfy image condition

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This article gives the statement, and possibly proof, of a subgroup property (i.e., descendant subgroup) not satisfying a subgroup metaproperty (i.e., image condition).
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It is possible to have groups G and K, a descendant subgroup H of G and a surjective homomorphism \varphi:G \to K such that \varphi(H) is not a descendant subgroup of K.


Example of the infinite dihedral group

We take:

  • G to be the infinite dihedral group. We can take the presentation G = \langle a,x \mid xax = a^{-1}, x^2 = e \rangle.
  • H to be a subgroup of order two complementary to the cyclic maximal subgroup, i.e., H is a subgroup generated by a reflection. In the above presentation, we can take H = \langle x \rangle.
  • K to be the quotient of G by the subgroup generated by multiples of 3 in the cyclic maximal subgroup, with the natural quotient map. In symbols, K = G/\langle a^3 \rangle. K is isomorphic to symmetric group:S3, and the image of H under the quotient map corresponds to S2 in S3.

We can check that:

  • H is descendant in G: It is the intersection of the following descending chain of subgroups, each of index two in its predecessor, hence each normal in its predecessor:

G = \langle a,x \rangle \ge \langle a^2,x \rangle \ge \langle a^4,x \rangle \ge \dots

  • The image of H in K is not descendant in K: The image looks like S2 in S3, which is a subgroup of a finite group, and is in fact a contranormal subgroup, so it cannot be descendant.