# Normal subgroup of Fitting length i contained in i-th term of upper Fitting series?

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)

• 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 $G/F(G)$. Jan 16, 2020 at 12:14
• I see. Thanks! Do you know a good standard reference for these kind of things? Jan 16, 2020 at 12:42
• 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). Jan 19, 2020 at 13:56

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 $$U_iG$$.