Let $G$ represent the group of linear functions under composition of the form $x \mapsto ax+b$ where $a,b \in \mathbb{Q}$ and $a\neq0$. Is $G$ a metabelian group?


Yes. Let $A$ be a set of functions $x\mapsto x+t$, $t\in\mathbb Q$, $B$ is a set of functions of the form $x\mapsto ax$, $0\neq a\in\mathbb Q$. Then $A,B\leq G$, $A\cong\mathbb{Q}^+$, $B\cong\mathbb{Q}^*$ and $G=A\rtimes B$, i.e. $G=AB$, $A\unlhd G$, $A\cap B=1$. So $A$ - normal abelian subgroup and $G/A\cong B$ - abelian group.

  • $\begingroup$ Thank you, Alex. I'm not sure but I think you meant that $A\cong\mathbb{Q}$ considering the addition for rational numbers. Now is it necessarily to stabilish the isomorphism for $A$? Also we can show directly that $B$ is abelian by considering $(a_1 x)\circ (a_2 x)=(a_2 x)\circ (a_1 x)$. $\endgroup$ – NumBee May 8 '15 at 20:12
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    $\begingroup$ @user129118 Yes of course, for Your goals it is not necessarily to proof that $A\cong\mathbb{Q}^+$, and $B\cong\mathbb{Q}^*$. It is enough to show, that $A,B$ abelian. Yes, $\mathbb{Q}^+$ - rationals by addition, $\mathbb{Q}^*$ - group of nonzero rationals by multiplication. $\endgroup$ – Alex W May 8 '15 at 21:13

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