Apparently that if we have $N \lhd G, G/N$ are Nilpotent $ \not \Rightarrow G$ is Nilpotent.

I am trying to find a simple example of this.

It currently seems unintuitive since if we have $N, G/N$ nilpotent, then we can 'mash together' elements in $G/N$ by elements in $G$ to get them into $N$ in a finite number of steps, and we can also 'mash' the elements in $N$ down to the identity in a finite number of steps, so initially I'd think the opposite statement is true.

Could someone please enlighten me on how to think about this? Any obvious counterexamples?

  • 1
    $\begingroup$ You can "mash" (whatever you understand under this notion) elements of $N$ down to $1$ by assumption only when using elements of $N$, but if you are in $G$, you have to consider all elements of $G$ when trying to go further down from $N$ to $1$. See Justin's example. $\endgroup$
    – j.p.
    May 7, 2017 at 9:19

2 Answers 2


A classic example is that $S_3$ is not nilpotent, but $A_3\cong\Bbb{Z}/3\Bbb{Z}$ and $S_3/A_3 \cong \Bbb{Z}/2\Bbb{Z}$ are both nilpotent, and $A_3\lhd S_3.$ Thus, $S_3$ is an example of such a group.


$S_3$ is not very enlightening from my point of view, because it's in a sense too clumped; one can consider fundamental group of Klein bottle as more picturesque example. It's a semidirect product $\mathbb Z \rtimes \mathbb Z$, where action is by negation. (there's a presentation: $\langle a, b \, | \, [a, b] = b^2\rangle$ if you are more combinatorial-minded) Computing its lower central series is an interesting exercise.

Also reading some proof of Hall's theorem may be helpful (moreso proving it by yourself): if for $N \triangleleft G$ both $G/[N, N]$ and $N$ are nilpotent, then $G$ is.


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