Largest normal subgroup with self-centralizing Fitting subgroups

Let $$G$$ be a finite group. Let $$N$$ be the normal subgroup generated by all normal subgroups $$M$$ such that $$C_M(F(M)) \le F(M)$$ where $$F(M)$$ denotes the Fitting subgroup of $$M$$. Then $$N$$ has itself the mentioned property, i.e., we have $$C_N(F(N)) \le F(N)$$.

I tried to prove that for two normal subgroups $$A,B \unlhd G$$ with the above property, their product has this property two. By Fitting Theorem the product $$F(A)F(B)$$ is also nilpotent, hence $$F(A)F(B) \le F(AB)$$. If $$(|A|, |B|) = 1$$, as a nilpotent group is the direct product of its $$p$$-cores, we have $$F(A \times B) = F(A) \times F(B)$$. Hence $$C_{A\times B}(F(A\times B)) = C_{A\times B}(F(A) \times F(B)) = C_A(F(A)) \times C_B(F(B)) \le F(A) \times F(B).$$ But this is just one special case for the subgroups $$A$$ and $$B$$. I have no idea how to show it in general.

So how to approach this problem in the general case? How to show in general (i.e. when the subgroups do not have coprime order) that for two normal subgroups $$A,B$$ fulfilling the property, their product also fulfills it?

• You have not asked a question. Please do so! – Derek Holt Nov 25 '19 at 19:00
• @DerekHolt Added it! – StefanH Nov 25 '19 at 19:40
• In the question you mean "two normal subgroups fulfilling the property" (being contained in their centralizer), their product also satisfies the property? – YCor Nov 25 '19 at 21:42
• @YCor Thank you. Fixed that. – StefanH Nov 25 '19 at 21:55

Suppose not, and let $$N/F(AB)$$ be a minimal normal subgroup of $$AB/F(AB)$$ with $$N/F(AB) \le C_{AB}(F(AB))F(AB)/F(AB)$$. Let $$M = N \cap C_{AB}(F(AB))$$, so $$N = MF(AB)$$.
If $$N$$ was solvable, then we would have $$F(N) = F(AB)$$ and so $$C_N(F(AB)) \le F(AB)$$, contrary to assumption.
So $$N$$ is not solvable, and hence neither is $$M$$ or $$[M,M]$$. So $$[M,AB] \le [M,A][M,B]$$ is not solvable, and so at least one of $$[M,A]$$ and $$[M,B]$$, say $$[M,A]$$ is not solvable.
But then by minimality of $$N/F(AB)$$ we have $$N = [M,A]F(AB)$$, and $$[M,A] \le A \cap M \le C_A(F(AB)) \le C_A(F(A)) \le F(A)$$, so $$N=F(AB)$$, contrary to assumption.
• Thank you. But the equality $F(N) = F(AB)$ is not implied by solvability of $N$, right? I mean $F(AB) \le F(N)$ as $F(AB) \le N$ and by minimality of $N$ either $N = F(N)$ or $F(N) \le F(AB)$, the first case being excluded by the centralizer-inclusion, hence the second case must hold. And solvability then enters in the next step by the centralizer-inclusion you mention. – StefanH Nov 25 '19 at 22:31
• That's right. $F(N)$ is a normal nilpotent subgroup of $AB$, and therefore contained in $F(AB)$. – Derek Holt Nov 26 '19 at 7:22