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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)

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    $\begingroup$ Yes, it is true. Use induction on the Fitting length of $N$: ,you have done the case of Fitting length $1$. Now note that $N/F(N)$ is isomorphic to a normal subgroup of $G/F(G)$. $\endgroup$
    – Geoff Robinson
    Commented Jan 16, 2020 at 12:14
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    $\begingroup$ I see. Thanks! Do you know a good standard reference for these kind of things? $\endgroup$
    – Armin Weiß
    Commented Jan 16, 2020 at 12:42
  • $\begingroup$ Almost any book which treats finite solvable groups in any depth would contain such results (though the proofs might be relegated to the exercises in some cases). $\endgroup$
    – Geoff Robinson
    Commented Jan 19, 2020 at 13:56

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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 $U_iG$.

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