About normal minimal subgroups not in the Frattini

The first claim is false: We will give an elementary proof using GAP. Consider the following code:

U := SmallGroup(720, 765);
N := Group([ (1,2,9,3)(4,6,10,8), (3,4,9,10)(5,8,6,7) ]);
P := SylowSubgroup(U, 2);

Running isSimple(N) verifies that N is a simple group, and running IsNormal(U,N) verifies that N is normal in U. We claim that U does not split over N: The subgroup N has index 2 and therefore if it had a complement, that would mean there is a cyclic subgroup of size 2 that is disjoint from N. This is impossible. Indeed, Zuppos(P) gives a list of all cyclic subgroups of P and every cyclic subgroup of size 2 is in N. Here is some quick code to check this (prints all true):

list := Zuppos(P);
for i in list do 
if Order(i) = 2 then Print(IsSubgroup(N,Group([i])));
fi; od;

Hence U does not split over N, as N is normal and all 2-Sylow subgroups are conjugate. (In fact, one can run the above code over all of U and use almost no theory whatsoever.)

Finally, N is not contained in the Frattini since it is not inside the maximal subgroup of U containing P, since P and N collectively generated U by index considerations.

PS: U never splits, as N is the only non-trivial normal subgroup of U (just run NormalSubgroups(U)). The group U is the Mathieu group of degree 10 and N is the alternating group on 6 letters. Moreover, P is the Semidihedral group SD16.

I think it is the claim from exercise 2 of Neukirch et al which is incorrect. I presume that all groups are meant to be finite in this question. The claim from exercise 2 is OK when $N$ is solvable (for then $N$ is an elementary Abelian $p$-group for some prime $p$ and there is a maximal subgroup $H$ of $G$ which does not contain $N$. Then $H \cap N$ is normal in $N$ ( as $N$ is Abelian), and also normal in $H$, so $H \cap N$ is normal in $\langle H,N \rangle= G.$ By minimality of $N$ among (non-trivial) normal subgroups, we have either $H \cap N = N$ or $H \cap N =1.$ But we can't have $H \cap N = N$ as $H$ does not contain $N$). On the other hand, there are finite non-Abelian simple groups which are not complemented in their automorphism groups, and the example of Lucchini et al is one of these. Any such example contradicts the claim of Neukirch et al ( if indeed they do not have the restriction that $N$ is assumed solvable).