I'm having a hard time understanding soluble groups.
If $G$ is a soluble group and $$G>G^{(1)}>\dots>G^{(t-1)}>G^{(t)}=\{1\}$$ is a derived series, then is it true that $G/G^{(m)}$ is finite for all $m$?
If $G$ is finite, then it is clear to me. What if $G$ is infinite?
It need not be true for infinite solvable groups. Let $G$ be the Baumslag-Solitar group $BS(1,n)=\langle a,t \mid tat^{-1}=a^n\rangle$. The group is $2$-step solvable, non-nilpotent for $n\ge 2$ and we have $G^{(1)}\cong \Bbb Z[\frac{1}{n}]$ and $G\cong \langle t\rangle \rtimes \Bbb Z[\frac{1}{n}]$, so that $$ G/G^{(1)}\cong \Bbb Z $$ is infinite. So the derived series is $$ G \triangleright G^{(1)}\triangleright G^{(2)}=1, $$ and all quotients $G/G^{(i)}$ for $i=1,2$ are infinite.