Let $G$ be a finite solvable group of Fitting length $n$ with upper Fitting series
$1= U\_0G \leq U\_1G \leq \cdots \leq U\_nG = G$.


Is it true that every normal subgroup of Fitting length $i$ is contained in $U\_iG$? (for $i=1$ it is true because the Fitting subgroup contains all nilpotent normal subgroups)



Normal subgroup of Fitting length i contained in i-th term of upper Fitting series?
You have already done the case, when $i = 1$.


Now suppose it i true for $i$. Let's prove it for $i+1$. Suppose $H \triangleleft G$ has Fitting length $i+1$. That means $\frac{H}{U\_iH}$ is nilpotent. Now let's define $\phi\_i$ as a natural homomorphism between $G$ and $\frac{G}{U\_iG}$. Then $\phi\_i(H)$ is a nilpotent normal subgroup of $\frac{G}{U\_iG}$ because $U\_iH$ lies in $U\_iG$ as a normal subgroup (characteristic subgroup of a normal subgroup is always normal) of Fitting length $i$ by induction step. That means, that $\phi\_i(H) \triangleleft U\_1\frac{G}{U\_iG}$, which results in $H \triangleleft \phi\_i^{-1}(U\_1\frac{G}{U\_iG})=U\_{i+1}G$.


Thus, $\forall i \in \mathbb{N}$ every normal subgroup of $G$ of fitting length $i$ is contained in $U\_iG$.


