Ascending chain condition on normal subgroups implies Hopfian

Statement
If a group satisfies the ascending chain condition on normal subgroups, then it is Hopfian: it is not isomorphic to any of its proper quotients.

Ascending chain condition on normal subgroups
A group $$G$$ satisfies the ascending chain condition on normal subgroups if, whenever $$N_i$$ is a chain of normal subgroups of $$G$$:

$$N_1 \le N_2 \le \dots $$

then there exists a finite $$r$$ such that $$N_r = N_n$$ for all $$n \ge r$$. In other words, the series stabilizes after a finite length.

Hopfian group
A group $$G$$ is termed Hopfian if $$G$$ is not isomorphic to $$G/N$$ for any nontrivial normal subgroup $$N$$ of $$G$$.

Applications

 * Slender implies Hopfian

Similar facts

 * Artinian implies co-Hopfian: Artinianness is a descending chain condition on subgroups, and co-Hopfian is the condition of not being isomorphic to a proper subgroup.

Proof
We prove the contrapositive here: any group that is not Hopfian does not satisfy the ascending chain condition on normal subgroups.

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

To prove: $$G$$ does not satisfy the ascending chain condition on normal subgroups.

Proof: Suppose $$N$$ is a nontrivial normal subgroup of $$G$$ such that $$G \cong G/N$$ (the existence of $$N$$ follows from the assumption of not being Hopfian). Suppose $$\varphi:G \to G/N$$ is the quotient map and $$\alpha:G \to G/N$$ is an isomorphism (NOTE: $$\alpha$$ is not the quotient map).

Now, define inductively:

$$N_0 = 1, N_1 = N, \qquad N_{i+1} = \varphi^{-1}(\alpha(N_i))$$.

Note that we have:

$$N_{i+1}/N_i = \varphi^{-1}(\alpha(N_i))/\varphi^{-1}(\alpha(N_{i-1})) \cong \alpha(N_i)/\alpha(N_{i-1}) \cong N_i/N_{i-1}$$.

(the second equality here is by the fourth isomorphism theorem).

By the inductive assumption, we see that for all $$i$$:

$$N_{i+1}/N_i \cong N$$.

In particular, the $$N_i$$s are a strictly ascending chain of subgroups.

Further, induction on $$i$$ shows that, since $$N_i$$ is normal in $$G$$, $$\alpha(N_i)$$ is normal in $$G/N$$, and thus, since normality satisfies inverse image condition, $$N_{i+1}$$ is normal in $$G$$.

Thus, the $$N_i$$s form a strictly ascending chain of normal subgroups, and hence, $$G$$ does not satisfy the ascending chain condition on normal subgroups.

Alternative way of writing the above proof
If we define $$\sigma = \alpha^{-1} \circ \varphi$$, $$\sigma$$ is a surjective endomorphism from $$G$$ to itself. We can then alternatively defin $$N_i$$ as the kernel of $$\sigma^i$$, the $$i$$-fold composite of $$\sigma$$ with itself.