# Descendance does not satisfy image condition

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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## Statement

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 $K$.

## Proof

### 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.