I know that finite abelian groups are solvable, because it means $ G^{(1)}=1.$ I also know that a group is solvable if you have a tower of $1=G_{s} \ \triangleleft \ G_{s-1} \ \triangleleft \ \dots \ \triangleleft \ G_1 \triangleleft \ G_0 = G$ with $G_i/G_{i+1}$ cyclic.
My intuition for the tower would be to write $1=G^{(1)} \ \triangleleft \ G$ but I don't see how $G/G^{(1)}$ is cyclic.
thanks