solvable group and abelian

Question

From wiki, a group $G$ is solvable if there are subgroups $1 = G_0 \trianglelefteq G_1$⋅⋅⋅$\trianglelefteq G_k = G$ such that $G_{j−1}$ is normal in $G_j$, and $G_j /G_{j−1}$ is an abelian group, for $j = 1, 2, …, k$.

I have the problem that we just choose $1\trianglelefteq$G and we get $G$ is sovable,so any group is solvable,am I wrong?

Answer 1

Concerning your problem: it only says that if a group $G$ is abelian, then it is solvable. Not the other way around: $S_3$ is solvable, but not abelian.

Answer 2

In the notation of the definition, for the sequence $1\trianglelefteq G$ we have $G_0 = 1$ and $G_1 = G$, and we require that $G_1/G_0$ is an abelian group. Since $G_1/G_0 = G/1 \cong G$, this sequence only shows that $G$ is solvable if $G$ is abelian.