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This is an exercise, 7.1 of the book "An Introduction to the group theory by J.J.Rotman":

Which of the following properties, when enjoyed by both $K$ and $Q$, is also enjoyed by every extension of $K$ by $Q$?

i. solvable

ii. nilpotent

iii. periodic

iv. torsion-free

I know that if $H$ and $G/H$ is solvable so is $G$, then i. is correct. ii. is not correct because $D_{\infty}$ is an extension of $\mathbb Z$ by $ \mathbb Z_2$ but it is not nilpotent. Diving in the groups makes my mind distracted and couldn't find a proper counter-example about iii. and iv. Thanks for you help.

2 Answers 2

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(iii) Suppose $\,H\triangleleft G\,\,,\,H\,,\,G/H\,\text{are periodic (torsion)}\,$ , then

$$\forall\,x\in G\,\,\exists\,n\in\Bbb N\,\,s.t.\,\,ord(xH)=n<\infty\Longrightarrow \,\exists\,n\in\Bbb N\,\,s.t.\,\,x^n\in H$$

But also $\,H\,$ is periodic, so

$$x^n\in H\Longrightarrow \exists\,m\in\Bbb N\,\,s.t.\,,\,1=(x^n)^m=x^{nm}\Longrightarrow \,x$$

has finite order and thus $\,G\,$ is torsion (periodic)

I think (iv) is trivial...

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    I see a proof for **iv** now, Don. I assume $x\in G$ such that there is a $n$ with $x^n=1$. So $x^n=1\in H$ and so $x^{n}H=H$ and so $|xH|<\infty$. A contradiction! Right?2012-11-03
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    Exactly, @BabakSorouh...with $\,x\in G-H\,$, of course.2012-11-04
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    Thanks so much. You did it as usual.2012-11-04
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Another example for (ii) is $S_3$, which is not nilpotent since it has trivial center, but is an extension of $\mathbb{Z}_3$ by $\mathbb{Z}_2$.

There are no counterexamples for (iii) or (iv).

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    Indeed, I felt there are some counterexample for iii and iv and the book didn't say that at all. Thanks for ii.2012-11-03