Is $G_i$ normal in $G_k$?

A group G is called solvable if it has a subnormal series whose factor groups (quotient groups) are all abelian, that is, if there are subgroups {1} = $G_0 < G_1 < ⋅⋅⋅ < G_k = G$ such that $G_{j − 1}$ is normal in $G_j$, and $G_j/G_{j − 1}$ is an abelian group, for $j = 1, 2, …, k$.

Question : Is $G_i$ normal in $G_k$ for $i \ge 2$?

• not necessarily. – Lord Shark the Unknown Dec 29 '17 at 16:46
• But every solvable group has such a series in which all of the subgroups are normal in $G$, namely the derived series. – Derek Holt Dec 29 '17 at 16:58
• In derived series, subgroup $G_k$ normal in $G_i$ if $G_k$ is a characteristic subgroup of $G_i$ i.e., If $f(G_k)=G_k$ where $f\in Aut(G_i)$. – 1ENİGMA1 Dec 29 '17 at 17:32

Not necessarily. Consider $$\{e\}\trianglelefteq\{e,(12)(34)\}\trianglelefteq \{e,(12)(34),(13)(24),(14)(23)\}\trianglelefteq A_4$$ $G_1$ is not normal in $A_4$. You can extend this sequence on the left by taking the direct product with an appropriate abelian group $H$: namely, $$\{0\}\times G_0\trianglelefteq 3\Bbb Z/6\Bbb Z\times G_0\trianglelefteq \Bbb Z/6\Bbb Z\times G_0\trianglelefteq\Bbb Z/6\Bbb Z\times G_1 \trianglelefteq\ \Bbb Z/6\Bbb Z\times G_2\trianglelefteq \Bbb Z/6\Bbb Z\times A_4$$ In the new sequence, $\widetilde G_3\not\trianglelefteq \widetilde G_5$.