Example of a group $G$, and prime $p$, such that every $p$-nilpotent subquotient of $G$ has normal Sylow-$p$-subgroup, but $G$ does not A few years ago I proved a theorem (invariant-theoretic) which applies to all groups $G$ and primes $p$, such that $G$ contains a subquotient which is $p$-nilpotent with non-normal Sylow-$p$-subgroup.
The theorem is not true if $G$ contains a normal Sylow-$p$-subgroup.
I wondered if anyone could give me an example of a group such that neither of these holds?
 A: Going by the problem as stated in the title, there is no $p$-solvable example of what you want. This is because if $G$ is a $p$-solvable group in which the automizer of every $p^{\prime}$-subgroup of $G$ is a $p^{\prime}$-group, then $G$ has a normal Sylow $p$-subgroup ( sketch proof: by induction, we may suppose that $O_{p}(G) = 1$, and then the Hall-Higman centralizer lemma implies that $O_{p^{\prime}}(G)$ contains its centralizer, but the hypotheses imply that it is centralized by a Sylow $p$-subgroup of $G$, which must then be trivial).
However, I think an example of what you want (if I am reading it correctly) is provided by the alternating group $A_{5}= G$ for $p = 5.$ Then every proper section of $G$ has a normal Sylow $5$-subgroup, but $G$ clearly does not have a normal Sylow $5$-subgroup ( recall that a section of a group $G$ is a group of the form $X/Y$ where $Y \lhd X$ and $Y,X$ are subgroups of $G$. The section is said to be proper if either $X \neq G$ or $Y \neq 1$).
Later edit: In fact, for every prime $p>3,$ the simple group $G = {\rm PSL}(2,p)$ has the property that every proper section of $G$ has a normal Sylow $p$-subgroup, but $G$ itself does not have a normal Sylow $p$-subgroup.
Even later edit: We have now established that there examples of the type you seek for every prime $p >3.$ Now we note that there is no example of the type you want when $p=2.$ For if $G$ is a finite group such that all $2$-nilpotent sections of $G$ are $2$-closed, then every odd order subgroup of $G$ has automizer of odd order. We claim that this implies that $G$ has a normal Sylow $2$-subgroup. As above, by induction, we may suppose that $O_{2}(G) = 1.$ But now, if $G$ has even order, $G$ contains a involution $t$ and $t$ can invert no non-identity element of odd order. Hence $\langle t,t^{g} \rangle$ is a $2$-group for each $g \in G$  and hence (by the Baer-Suzuki theorem) $t \in O_{2}(G),$ a contradiction.
To complete the picture, set $G = {\rm SL}(2,8).$ Then all proper sections of $G$ have a normal Sylow $3$-subgroup, but $G$ itself does not.
Hence there is an example of the kind you want for every odd prime $p,$ but there is no such example for $p=2.$
