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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 $L$...

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  • $\begingroup$ Is $G$ solvable or arbitrary ? $\endgroup$
    – mesel
    Apr 4, 2015 at 13:26
  • $\begingroup$ Arbitrary, just finite (but maybe finite is not mandatory). $\endgroup$
    – StefanH
    Apr 4, 2015 at 13:41
  • $\begingroup$ As far as I remember, minimal normal subgroup of arbitrary group is either elemantary abelain or semi-simple groups. $\endgroup$
    – mesel
    Apr 4, 2015 at 13:57
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    $\begingroup$ What is the source of question ? $\endgroup$
    – mesel
    Apr 4, 2015 at 17:33
  • $\begingroup$ @mesel It is an exercise from the book Finite Group Theory by Stellmacher, Kurzweil, from Chapter 1.7 Minimal normal subgroups. $\endgroup$
    – StefanH
    Apr 5, 2015 at 16:14

1 Answer 1

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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.

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    $\begingroup$ @DerekHolt , hello, dear professor, thanks for you nice answer! It seems kind of “magical” to apply commutator groups here. If I didn’t read your answer, I may not think it’s the key to this problem. Would you be so kind as to tell me how that idea might come to mind (although this is a post more than 2 years)? Any help would be appreciated! $\endgroup$
    – user542899
    Mar 21, 2018 at 11:46
  • $\begingroup$ Sorry but I do not think that I can help with questions like that. It is just based on experience. $\endgroup$
    – Derek Holt
    Mar 21, 2018 at 11:52
  • $\begingroup$ @DerekHolt That’s OK, thank you all the same! $\endgroup$
    – user542899
    Mar 21, 2018 at 11:58

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