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I need help to prove the following statement:

“A finite group $G$ is nilpotent iff $[N,G]<N$ for every nontrivial normal subgroup of $G$.”

I’ve found a proof for the forward direction on the comments of another question (I’ve changed the notation):

Proof: Assume by way of contradiction that $[N,G] = N$. Then $[G,[G,N]] = [G,N] = N$, so $N = [G,[G,....,[G,N]...]] ≤ [G,[G,....,[G,G]..]] = 1$ (if there are enough $[G, ...)$.

...but I don’t understand the last sentence (“if there are enough $[G,...)”$). I don’t know how to attempt the other direction.

P.S.: I’m using the following definition of “nilpotent group”: $G$ has a lower central series terminating in the trivial subgroup after finitely many steps.

P.S. 2: The forward direction is now clear, thanks to the comments.

dahemar
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  • For the other direction you can use the hypothesis to construct a lower central series which never stabilizes (until $N$ is trivial) – vujazzman Jan 17 '21 at 00:56
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    For the forward direction, the author is using the fact that if $G$ is nilpotent, then a big enough expression of the form $[G,[G,...,[G,G]]...]$ is the trivial subgroup. This is just the statement that the lower central series ends in the trivial subgroup. – vujazzman Jan 17 '21 at 01:02
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    There are a number of equivalent ways to define “nilpotent” in the case of finite groups. Could you please include explicitly the definition that you are using? – Arturo Magidin Jan 17 '21 at 01:31
  • @ArturoMagidin Sure, I’ve added that definition to the question. – dahemar Jan 17 '21 at 10:00
  • @vujazzman Oh okay, I see it now. Thanks. – dahemar Jan 17 '21 at 10:01

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