The minimal normal subgroups of a maximal subgroup $L$ if two minimal normal subgroups of $G$ are not in $L$

Question

If a finite group $G$ contains a maximal subgroup $L$ and two minimal normal subgroups not in $L$, then every minimal normal subgroup of $L$ is contained in the subgroup generated by the minimal normal subgroups of $G$.

I have no idea so far, in what sense does it helps that two minimal normal subgroups are not in $L$... maybe I should consider the product of all minimal normal subgroups in $L$, of which $L\cap N$ and $L\cap M$ are parts if they are not trivial, and show somehow that this product is contained in the product of all minimal normal subgroups of $G$, but as I said I do not see how to use the fact that at least two minimal normal subgroups are not in $L$...

Answer 1

Here is an outline of a solution. Let $N_1$ and $N_2$ be minimal normal subgroups of $G$ that are not contained in $L$, and let $U$ be a minimal normal subgroup of $L$.

So $LN_1 = G$ and hence $UN_1 \unlhd G$. So $[U,N_2] \le UN_1 \cap N_2$.

If $[U,N_2] \le N_1$, then $[U,N_2] \le N_1 \cap N_2 = 1$ and, since $G=LN_2$, we have $U \unlhd G$, so $U$ is minimal normal in $G$ and we are done.

So assume that $[U,N_2] \not\le N_1$. Now, since $G/N_1 = LN_1/N_1 \cong L/L \cap N_1$ is isomorphic to a quotient group of $L$, $UN_1/N_1$ is a minimal normal subgroup of $G/N_1$, and since $[U,N_2]N_1/N_1$ is a nontrivial normal subgroup of $G/N_1$ contained in $UN_1/N_1$, we must have $[U,N_2]N_1/N_1 = UN_1/N_1$, so $[U,N_2]N_1 = UN_1$. Hence $U \le [U,N_2]N_1 \le N_2N_1$, and we are done.