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I know how to prove this problem, but I just need a small check towards the end of the proof on how to use induction on the length of chains to show that $G_{1}$ is Noetherian.(below) I would appreciate your feedback!

Question:Every Polycyclic group is Noetherian and Solvable.

Proof: Since $G$ is polycyclic, there is a finite series of subgroups $${e}=G_{k}\le G_{k-1}\leq…\le G_{1}\le G_{o}=G$$ such that $G_{j+1}$ is normal in $G_{j}$ (for $0\le j\le k-1$) and $G_{j}/G_{j+1}$ is cyclic(which implies abelian) and so, $G$ is Solvable by definition. It remains to show that $G$ is Noetherian. Since $G_{1}$ is normal in $G_{0}=G$, we only need to show that $G_{1}$ and $G_{o}/G_{1}$ are Noetherian. Now, since $G_{o}/G_{1}$ is cyclic, it is Noetherian. We use induction on the length of $k$ to show that $G_{1}$ is Noetherian. If $k=1$, then we have ${e}=G_{1}$ is the identity which is Noetherian. At this point here on, this is where I am asking for a check(help).

Here's my argument: Suppose that $G_{1}$ is Noetherian for a chain of lenth $k=n$ and let $k=n+1$. Then we have the chain ${e}=G_{n+1}\le G_{n}\le...\le G_{2}\le G_{1}\le G_{0}=G$. By the inductive hypothesis, $G_{2}$ is Noetherian. Since then $G_{2}$ is normal in $G_{1}$, and $G_{1}/G_{2}$ is Noetherian(because it is cyclic), we conclude that $G_{1}$ is Noetherian.

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  • $\begingroup$ You have to prove that if $N$ is normal in $G$ and both $N$ and $G/N$ are Noetherian, then $G$ is Noetherian. $\endgroup$ – egreg Aug 1 '18 at 16:43
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    $\begingroup$ This is fine, but you've phrased it weirdly. Use the induction at the beginning: "assuming all polycyclic groups of series length $<k$ are Noetherian, consider $G$ of length $k$." Now you get $G_1$ is Noetherian by induction right away. $\endgroup$ – Steve D Aug 1 '18 at 17:54
  • $\begingroup$ Thanks @Steve D. I go it! $\endgroup$ – Djinola Aug 1 '18 at 19:56
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More is true: A solvable group is Noetherian if and only if it is polycyclic, see Lemma 2 of the paper On linear Noetherian groups by Zassenhaus.

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