In Serre's Linear Representations of Finite Groups, Section 8.2, the following claim is given, as equivalent to a definition of solvable groups.
Solvable groups. One says that G is solvable if there exists a sequence
$$\{1\}=G_0\subset G_1\subset\cdots\subset G_n=G$$
of subgroups of $G$, with $G_{i-1}$ normal in $G_i$ and $G_i/G_{i-1}$ abelian for $1\le i\le n$.Equivalent definition: $G$ is obtained from the group $\{1\}$ by a finite number of extensions with abelian kernels.
In the same section, the following claim is given, as equivalent to a definition of nilpotent groups to show that finite nilpotent $\Rightarrow$supersolvable.
Nilpotent groups. As Same above, except that one requires that all the $G_i$ be normal in G and $G_i/G_{i-1}$ be in the center of $G/G_i$ for $1\le i\le n$.
Equivalent definition: $G$ is obtained from the group $\{1\}$ by a finite number of central extensions.
I showed the equivalence of the definition of nilpotent groups and
the equivalent definition$\Rightarrow$ solvable. However, the converse could not be shown…
Therefore, is it equivalent to an equivalence definition in the case of solvable groups? I would like to know if there are any counterexamples.
Best Answer
Recall the definition of the commutator subgroup and derived series:
If $x,y\in G$, then $[x,y]=x^{-1}y^{-1}xy$. If $A$ and $B$ are subsets of $G$, then $[A,B]=\langle [a,b]\mid a\in A, b\in B\rangle$.
The commutator sugbroup of $G$ is $[G,G]$. This is a normal (in fact, characteristic) subgroup of $G$, and for any subgroup $N$ of $G$, $[G,G]\leq N$ if and only if $n\triangleleft G$ and $G/N$ is abelian.
The derived series of $G$ is defined inductively by $G^{(1)}=[G,G]$ and $G^{(k+1)}=[G^{(k)},G^{(k)}]$. They are all normal subgroups.
Assume that $G$ has a subnormal series $$1=G_0\lt G_1\lt\cdots \lt G_n=G$$ with $G_i\triangleleft G_{i+1}$ and $G_{i+1}/G_i$ abelian.
Claim. $G^{(i)}\leq G_{m-i}$.
We proceed inductively: Since $G/G_{m-1}$ is abelian, then $G^{(1)}\leq G_{m-1}$.
Now assume that $G^{(k)}\leq G_{m-k}$. Since $G_{m-k}/G_{m-k-1}$ is abelian, then $[G_{m-k},G_{m-k}]\leq G_{m-k-q}$. Since $G^{(k)}\leq G_{m-k}$, we have $$ G^{(k+1)} = [G^{(k)},G^{(k)}] \leq [G_{m-k},G_{m-k}]\leq G_{m-k-1}$$ we have $G^{(k+1)}\leq G_{m-(k+1)}$, as claimed.
That means that $G^{(m)}=1$. Now consider the normal series $$1 = G^{(m)}\leq G^{(m+1)} \leq\cdots\leq G^{(1)} \leq G$$ which exhibits $G$ as a series of extensions with abelian kernel starting from $1$.