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
$G/1$ is not necessarily cyclic.
However, the classification theorem for finite abelian groups tells us that we can write $$ G = C_1 \times C_2 \times C_3 \times \dots \times C_n,$$ where $C_1, C_2, C_3 , \dots , C_n$ are all cyclic.
So we can take $$ G_n = 1, \ \ \ \ \ G_{n-1} = C_1, \ \ \ \ \ G_{n-2} = C_1 \times C_2, \ \ \ \ \ G_{n-3} = C_1 \times C_2 \times C_3 , \ \ \ \ \ \dots \ \ \ \ \ G_0 = G.$$
Then, for each $i$, we have $$ G_{i}/G_{i+1} = C_{n-i},$$ which is cyclic.
Unless I didn't understand your question correctly, I think you got your definitions mixed up.
A solvable group is a group which has a subnormal "tower" with abelian quotients, not necessarily cyclic.
In that case, it is clear that for a finite abelian group $G$ the tower you suggested is good, as the quotient is $G$ which is abelian.
