Metanilpotent groups and saturated formation Class of all metanilpotent groups is a saturated formation ?
How do I prove
 A: Some ideas (too long for a comment):
$\;G\;$ is metanilpotent iff there exists $\;N\lhd G\;$ s.t. that both $\,N\,,\,G/N\;$ are nilpotent .
If $\;|G|<\infty\;\;$  is $\;G/\Phi(G)\;$ metanilpotent ? Well, this is so iff
$$\exists\,M/\Phi(G)\lhd G/\Phi(G)\;\;\text{s.t. both}\;\;M/\Phi(G)\;,\;\left(G/\Phi(G)\right)\left(M/\Phi(G)\right)$$
are nilpotent. But 
$$\left(G/\Phi(G)\right)\left(M/\Phi(G)\right)\cong G/M$$
So $\,G/\Phi(G)\;$ is metanilpotent iff
$$\exists\;M\lhd G\;\;s.t.\,\,\Phi(G)\le M\;\;\text{and s.t. both}\;\,M/\Phi(G)\;,\;G/M\;\;\text{are nilpotent}$$
From the above is clear that we cannot have Fit$(G)\le M\;$ , as then $\,G/M\;$ isn't nilpotent , so it all boils down to check whether  there exists 
$$\Phi(G)\le M\lneq \text{Fit}(G)\;,\;\;M\lhd G\;,\;\;M/\Phi(G)\;,\;G/M\;\;\text{nilpotent}$$
I don't know the final answer, but it looks as if the last conditions don't necessarily fulfill in every case...
Also note that since $\,M\lhd G\,$ then $\,\Phi(M)\le\Phi(G)\;$ , so
$$M/\Phi(G)\cong\left(M/\Phi(M)\right)/\left(\Phi(G)/\Phi(M)\right)$$
so if $\,M\,$ is nilpotent then we already have that $\,M/\Phi(G)\;$ is nilpotent... and perhaps someone else can see something further from here.
A: Here is how to do a residual proof. You can find the general version of this in Doerk-Hawkes. This is called the “formation product”, described in IV.1.7 page 337, and happens to equal the “Fitting product” as long as the quotient groups come from a Fitting formation.
Definition: $\newcommand{\nilres}[1]{\gamma_\infty\left({#1}\right)}$ $G$ is metanilpotent iff $\nilres{G}$ is nilpotent, where $\nilres{G}$ is the intersection of the lower central series of $G$.
In particular, $\nilres{H} \leq \nilres{G}$ and $\nilres{G/N} =\nilres{G}N/N$ and $\nilres{M}\nilres{N}=\nilres{MN}$ for $H \leq G$ and $M,N \unlhd G$. Since the class of nilpotent groups is closed under subgroups, quotients, normal products, and Frattini extensions, we get that metanilpotent subgroups are as well. That is, metanilpotent groups form a subgroup closed saturated Fitting formation.
Metanilpotent is closed under subgroups since $H \leq G$ implies $\nilres{H} \leq \nilres{G}$ and $\nilres{G}$ is nilpotent by definition of metanilpotent, and $\nilres{H}$ is nilpotent since the class of nilpotents are closed under subgroups. The others follow similarly.
Metanilpotent is closed under quotient groups since $N \unlhd G$ implies $\nilres{G/N} = \nilres{G}N/N \cong \nilres{G} / (\nilres{G} \cap N)$ is a quotient of a nilpotent group ($\nilres{G}$ which is nilpotent since $G$ is metanilpotent) and so $\nilres{G/N}$ is nilpotent since the class of nilpotent groups is closed under quotients.
Metanilpotent is closed under normal products: If $M,N \unlhd G$ and $M,N \in \mathcal{F}$, then $\nilres{M}, \nilres{N}$ are nilpotent by definition of metanilpotent. Also $\nilres{M},\nilres{N} \unlhd G$ as since $\nilres{M}$ is a characteristic subgroup of $M$. Since nilpotent groups are closed under normal products, $\nilres{M}\nilres{N}$ is nilpotent. Since $\nilres{MN}=\nilres{M}\nilres{N}$, we get that $MN$ is metanilpotent. Hence metanilpotent is closed under normal products.
Meta nilpotents is closed under Frattini extensions: If $G/\Phi(G)$ is metanilpotent, then $\nilres{G/\Phi(G)} = \nilres{G}\Phi(G)/\Phi(G)$ is nilpotent by definition of metanilpotent, but then $\nilres{G}\Phi(G)$ is nilpotent by Gaschütz's result, and $\nilres{G}\Phi(G)$ is nilpotent since nilpotents are closed under normal products.
