Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I would like to find an exact sequence $1\rightarrow N\rightarrow G\rightarrow Q\rightarrow 1$ of groups where $N,Q$ are supersolvable and G is not. I found it but in my example $G$ is finite, I was wondering if you have a nice example with $G$ an infinite group, I tried with a lot of groups without finding it. Do you have any idea?

EDIT: Suppose that $G$ is a finite not supersolvable group and we have an exact sequence $1\rightarrow K\rightarrow G\rightarrow G/K\rightarrow 1$ with $K$ and $G/K$ supersolvable. Define $G^\prime=G\times\mathbb{Z}$, then $G^\prime$ is not supersolvable because if I have a normal-cyclic series the I project it on $G$ and I obtain a normal-cyclic series for $G$ and this is a contradiction. So I can take $1\rightarrow K\times\{0\}\rightarrow G\rightarrow G/(K\times\{0\})\rightarrow1$. Now $K\times\{0\}$ is supersolvable and $G/(K\times\{0\})\cong (G/K)\times\mathbb{Z}$ and it's supersolvable, am I right?

share|cite|improve this question
Couldn't you enlarge your class of examples by taking a direct sum with any finitely generated abelian group? – jspecter Feb 24 '12 at 3:06
up vote 4 down vote accepted

If you want a more essentially infinite example, you could take a semidirect product of ${\mathbb Z}^2$ by ${\mathbb Z}$ with an irreducible action. For example $\langle x,y,z \mid xy=yx, z^{-1}xz=y, z^{-1}yz=xy \rangle$.

share|cite|improve this answer
how did you create this example? – John Feb 28 '12 at 2:48

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.