Why Fitting subgroup of finite solvable group is self-centralizing

Question

I read the proof of the proposition that the Fitting subgroup of a finite solvable group is self-centralizing.

But I do not understand why $B$ in the proof is characteristic. The proof says that

1. $F (G)$ is characteristic in $G$;
2. $C$ is characteristic in $G$;
3. $H$ is characteristic in $G$;
4. $B/H$ is characteristic in $C/H$;
5. $B$ is characteristic in $C$;
6. $B$ is characteristic in $G$.

The steps 1-4 look fine, but I do not know why the 5th is correct. I know the quotient-transitivity of characteristicity: if $H$ is characteristic in $G$ and $K/H$ is characteristic in $G/H$ then $K$ is also characteristic in $G$, where $H \le K \le G$. To apply this transitivity to get the 5th from the 4th, I have to show that $H$ is characteristic in $C$, but I only know it so only in $G$ (the 3rd).

So... why $B$ is characteristic in $C$? Can I show that $H$ is characteristic in $C$, or are there another method to prove characteristicity of $B$ in $G$?

Answer 1

Correct the proof with the following:

5. Since $C$ is characteristic (hence normal) in $G$, $C/H$ is normal in $G/H$.
6. Since $B/H$ is characteristic in $C/H$ and since $C/H$ is normal in $G/H$, $B/H$ is also normal in $G/H$.
7. Since $B/H$ is normal in $G/H$, $B$ is normal in $G$, as being the inverse image of a normal subgroup $B/H$.