9
$\begingroup$

In Neukirch--Schmidt--Wingberg, "Cohomology of Number Fields", Second edition, page 624, Exercise 2, it is stated the following fact.

$\textbf{Claim}$: If $N$ is a normal subgroup, minimal among normal subgroups, of a group $G$ not contained in the Frattini's subgroup of $G$, then $G$ is a semi-direct product over $N$, i.e. there exists $H$ subgroup of $G$ with $H \cap N=\{id\}$ and $HN=G$.

Now, if $N$ is a simple non-abelian normal subgroup of $G$, the above Claim implies that $G$ is a semi-direct product over $N$. Indeed the Frattini of $G$ is nilpotent, and since $N$ is non-abelian, it cannot be contained in a nilpotent group, otherwise $N$ itself would be nilpotent, and nilpotent and simple implies having prime order and hence abelian. Moreover $N$ is clearly minimal, since $N$ itself does not have non-trivial proper normal subgroup. But then I see a clear contradiction with Lemma 4.3 of this paper of Lucchini, Menegazzo and Morigi https://projecteuclid.org/download/pdf_1/euclid.ijm/1258488162.

Could somebody explain (in case I did not perform a mistake in reasoning) which of the two publications has a mistake?

$\endgroup$
9
$\begingroup$

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).

$\endgroup$
  • 1
    $\begingroup$ Dear Geoff, many thanks for your reply. I was also able to show the case you discuss, e.g. when N is solvable, and I got terribly stuck on the general case, until I found the paper of Lucchini and realized that after all it was a good thing I was unable to prove something that you confirm to be false. I checked indeed again up and Neukirch et alii do not assume N to be solvable. thanks for cofirming that this is wrong! $\endgroup$ – Tom1909 Oct 8 '18 at 13:41
  • $\begingroup$ Yes,if $N$ is non-Abelian simple, the argument which deals with the case $N$ solvable just does not generalize. $\endgroup$ – Geoff Robinson Oct 8 '18 at 14:22
12
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.