I am trying to understand this proof: https://math.stackexchange.com/a/1889305/74378
To summarize what is going on there, $G$ is a solvable group with composition series
$$ 1\triangleleft G_1 \triangleleft \cdots \triangleleft G_n = G$$
At one point this answer selects a $G_{i+1}$ which is abelian and is in the series. From this fact, the answerer infers $G_{i+1}\trianglelefteq G$. Maybe I'm misunderstanding this step, but I can't think of a reason why a subgroup in the series should always be normal in $G$. Or if it has to do with the subgroup being abelian, I'm still not sure why that would entail that it is normal in $G$. If we select $g\in G$ then we would need to show $gG_{i+1}g^{-1} = G_{i+1}$ but I can't think of how this relates to two elements in $G_{i+1}$ commuting.
Best Answer
You can't conclude that. There are many normal series in which not all subgroups (including abelian subgroups) are normal in $G$. However, if $G$ is solvable then there exists some normal series with abelian quotients in which every subgroup is normal in $G$. For example, the derived series of $G$. (which indeed comes to the trivial group because $G$ is solvable) So I believe in the answer it is assumed you choose such a series from the beginning.