σ-complete groups

Consider a finite group G. In general it is interesting to know when, given that N is a maximal normal subgroup of G, we have G ≅ N x G/N. Clearly this does not always happen. For example, take a prime p and G=Zp². The maximal proper normal subgroup is Zp and G/N=Zp but clearly G is not isomorphic to Zp × Zp. So in general it cannot be done. However, if something special could be done, it happens: G is a direct product of N with a simple group (since the quotient G/N would be simple since N is maximal). So I started looking for a condition that G had to satisfy for this to happen.

A sufficient condition is to require that N be complete as we will now see (Recall that N is complete if Z(N)={e} and Inn(N)=Aut(N)). In particular, we prove the following fact:

Theorem:(1)
Let G be a finite non-simple group. Then:

1)G admits a maximal normal subgroup N

2)If N is complete then G≅N × S where S is a simple group.

Proof:
1) Let H be a non-trivial normal subgroup of G. If it is maximal we are done. Otherwise there exists H1 a normal subgroup such that H ◅ H1 ◅ G . If H1 is maximal we are done, otherwise we repeat. We find a chain of subgroups one normal inside the other, and since |G| is finite, the chain must end sooner or later. And so we have a maximal normal subgroup.

2) Since N is complete we have that the homeomorphism φ:G—>Aut(N) , g |—>φ_g is a surjective homomorphism and therefore, saying C_G(N) the centralizer of N in G we have that G/C_G(N)=G/kerψ≅ Aut(N)≅N from which G/N ≅G/G/C_G(N)≅C_G(N) therefore C_G(N) is simple (since G/N is simple). Now:

i)N,C_G(N) are subgroups of G

ii)NC_G(N)=G (since |NC_G(N)|=|N||C_G(N)|=|N||G/N|=|G|)

iii)N is normal in G

—> G≅N ⋊φ C_G(N)

where φ: C_G(N)—>Aut(N) , g|—>φ_g is the conjugation which is therefore trivial because the domain is C_G(N). It follows that:

G≅N × S

where S is a simple group. □

So this theorem already tells us something very interesting. Finite groups whose maxima is complete are the product of the maximal normal subgroup and a simple one. The maximal subgroup, however, is not easy to handle. So let’s give a further definition:

Def:

A non-simple finite group G such that every non-trivial normal subgroup is complete is called σ-complete.

So we have the following:

Theorem:
Let G be a finite group. Then:

G is σ-complete ⇔ G≅S1 × ··· × Sn where S1,…,Sn are complete simple groups.

Proof:

If G is σ-complete then from theorem (1) called N1 a maximal normal subgroup we have that G≅N1 × S1 where S1 is simple. But N1 is complete because G is σ-complete so N1≅N2 × S2 where S2 is simple and N2 the maximal normal of N1 so G≅N2 × S1 × S2 . Reasoning inductively we obtain that G is the product of simple groups S1 × ×Sn. Note that the Si are complete because the Si are normal subgroups of S1 x … x Sn. Conversely if G≅S1 × ×Sn where the Si are simple and complete then G is not simple because the Si are normal subgroups and furthermore if L is a normal subgroup of G then L≅Si for some i and therefore L is complete from which G is σ-complete. □

Example :

The Mathieu group M23 is a simple group (it falls among the sporadic groups) and is complete. It follows that M23 × M23 (repeated n-times) is σ-complete for every natural n.