Noetherian implies Hopfian Can someone provide me with a reference for the following 2 statements, or indicate a proof?


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*A finitely generated nilpotent group is Noetherian

*A Noetherian group is Hopfian.

 A: The second statement can be found here.
In particular, it uses two facts:


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*Noetherian implies ascending chain condition on normal subgroups.

*Ascending chain condition on normal subgroups implies Hopfian.


More generally, see:

Wehrfritz, B. A. F. (2009). The Basic Theory of Polycyclic Groups. In Group and Ring Theoretic Properties of Polycyclic Groups (pp. 13-28). Springer London.

Theorem 2.13 (p. 20) proves that every finitely generated nilpotent group is polycyclic (i.e., Noetherian and solvable). The proof is attributed to Hirsch (1938).
A: The first result can be found in Chapter 2.3 of Avinoam Mann's book How groups grow, LMS Lecture note series 395. The idea is to prove that in a finitely generated Nilpotent group there exists a central series $1=G_0<G_1<G_2<\cdots<G_i<\cdots<G_n=G$ where each factor $G_{i+1}/G_i$ is cyclic, and then use this to prove Noetherian.
The second result is, perhaps, more interesting. It is more interesting to me anyway! So:
Recall that a group is Noetherian if every subgroup is finitely generated. Equivalently, if every ascending chain of subgroups stabilises in finite length, that is, for every chain of subgroup $$H_1\leq H_2\leq H_3\leq\cdot$$ there exists some $k\in\mathbb{N}$ such that $H_i=H_k$ for all $i>k$.
A group is Hopfian if it is not isomorphic to any of its proper quotients, so $G/N\not\cong G$ for $N\neq 1$.
Theorem: Noetherian $\Rightarrow$ Hopfian.
Proof: We shall prove that not-Hopfian implies not-Noetherian, which is sufficient. So, suppose that $G$ is not-Hopfian, so there exists a map $\phi: G\rightarrow G$ such that $\ker\phi\neq 1$. Let $N$ denote the kernel of $\phi$, $\ker\phi=N$, and more generally define $N_i:=\ker\phi^i$. Note that $\phi(N_{i+1})=N_i$. (This map is a very useful map when you are thinking about non-Hopfian groups. For another example of it, see Mal'cev's proof that residually finite groups are Hopfian.)
The idea is to prove that the following chain does not stabilise after finitely many steps.
$$1=N_0<N_1<N_2<\cdots<N_i<\cdots$$
We prove that this does not stabilise by proving that $N_{i+1}/N_i\cong N\neq 1$ for all $i$. So, consider $N_{i+1}/N_i$. As $\ker\phi=N\leq N_i$ we can quotient out $N$ from both the top line and the bottom line and apply the third$^{[1]}$ isomorphism theorem to get the following isomorphism.
$$N_{i+1}/N_i\cong (N_{i+1}/N)/(N_i/N)$$
Now, by definition, $\phi(N_{i+1})=N_i$, and so $N_{i+1}/N=N_i$. Therefore, the above isomorphism implies that $N_{i+1}/N_i\cong N_i/N_{i-1}$. By induction, we thus have the following.
$$N_{i+1}/N_i\cong N$$
As $N$ is non-trivial, this implies that $N_{i}\lneq N_{i+1}$ for all $i$, and so this chain never stabilises. This proves the theorem.
$^{[1]}$Wikipedia claims it as the third, as does my favorite text. I thought it was the second. I will believe them though.
