# Is meta-$p$-nilpotent for all $p$ the same as meta-nilpotent?

I've been working on understanding solvable groups one prime at a time, and it has been very helpful. Perhaps this question is enjoyable to more than just me:

Definition: A finite group is said to be $p$-nilpotent if it has a normal Sylow $p$-complement. A finite group is said to be meta-$p$-nilpotent if it has a normal subgroup such that both the normal subgroup and the quotient group are $p$-nilpotent.

If we omit the “meta” then things work nicely:

Proposition: A finite group is nilpotent if and only if it is $p$-nilpotent for all primes $p$.

Also if we go crazy with the “meta” things work nicely:

Proposition: A finite group is $p$-solvable if and only if has a normal series with each factor $p$-nilpotent. (So $p$-solvable = meta-meta-meta-...-meta-$p$-nilpotent.) A finite group is solvable if and only if it has a normal series with factor nilpotent. (So solvable = meta-meta-meta-...-meta-nilpotent.) A finite group is solvable if and only if it is $p$-solvable for all primes $p$.

Question: Is a finite group meta-nilpotent if and only if is meta-$p$-nilpotent for all primes $p$?

I have not yet tried to prove or disprove this, but it sounds like an interesting question either way.

Clearly meta-nilpotent implies meta-$p$-nilpotent for all $p$, but I don't think the converse is true.
For a counterexample, let's take $G$ to be a soluble group with three elementary abelian layers all with different primes, and with all layers acting nontrivially on all lower layers. For example, choose $G = {\rm A \Gamma L}(1,8)$ of order 168, which has a unique chief series $1 < N_1 < N_2 < G$ with $|N_1|=8$, $|N_2|=56$.
Now neither $N_2$ nor $G/N_1$ is nilpotent, so $G$ cannot be meta-nilpotent. But $N_1$ and $G/N_1$ are both 2-nilpotent, $N_2$ and $G/N_2$ are both 7-nilpotent, and $G$ itself if 3-nilpotent.