Artinian implies co-Hopfian

Statement
Any Artinian group (i.e., a group satisfying the descending chain condition on subgroups) is co-Hopfian: it is not isomorphic to any proper subgroup of itself.

Artinian group
A group $$G$$ is termed Artinian if it satisfies the following equivalent conditions:


 * If $$H_0 \ge H_1 \ge H_2 \ge \dots \ge H_n \ge \dots$$ is a descending chain of subgroups, there is a $$n$$ such that $$H_n = H_m$$ for all $$m \ge n$$.
 * Any nonempty collection of subgroups of $$G$$ has a minimal element: a subgroup not containing any other member of that collection.

co-Hopfian group
A group $$G$$ is termed co-Hopfian if there is no proper subgroup of $$G$$ isomorphic to $$G$$.

Similar facts

 * Slender implies Hopfian: An ascending chain condition on subgroups implies that the group is not isomorphic to any proper quotient.
 * Ascending chain condition on normal subgroups implies Hopfian: In fact, an ascending chain condition on normal subgroups implies that the group is not isomorphic to any proper quotient.

Proof
We prove the contrapositive here: if a group is not co-Hopfian, it is not Artinian.

Given: A group $$G$$ that is not co-Hopfian.

To prove: $$G$$ is not Artinian.

Proof: Suppose $$H \le G$$ is a subgroup and $$\alpha:G \to H$$ is an isomorphism (such a subgroup exists because $$G$$ is not co-Hopfian). Define:

$$H_0 = G, H_{i+1} = \alpha(H_i)$$.

We prove by induction that $$H_{i+1}$$ is a proper subgroup of $$H_i$$ for each $$i$$. The base case is direct, since $$H_1 = \alpha(H_0) = H < H_0 = G$$.

For the induction, suppose $$H_i < H_{i-1}$$. Since $$\alpha$$ is an isomorphism, it preserves strictness of inclusions, and we thus have:

$$\alpha(H_i) < \alpha(H_{i-1}) \qquad \implies H_{i+1} < H_i$$.

Thus, we have a strictly descending chain of subgroups of $$G$$ that does not stabilize at any finite stage. Thus, $$G$$ is not Artinian.