# On the number of minimal normal subgroups w.r.t. to a maximal subgroup which is simple and non-abelian

Let $G$ be a finite group, and $L$ a maximal subgroup of $G$. If $L$ is non-abelian and simple, then in $G$ there exists at most two minimal normal subgroups.

What I got: Suppose we have three minimal normal subgroups $P,Q,R$. Then as $P\cap R = Q \cap R = 1$ we have $P \in C_G(R)$ and $Q \in C_G(R)$, therefore $PQ \in C_G(R)$. So I know about the centraliser of $R$ that it is i) normal in $G$, because $R$ is normal, and ii) by the above, that it is non-trivial and not a minimal normal subgroup, because $P\cap Q = 1$. So I guess I have to show somehow that the centraliser indeed has these properties for a maximal subgroup $L$ with these properties to get a contradiction, but here I have no idea how to proceed?

Let $P$ and $Q$ be minimal normal subgroups of $G$. At least one of them, say $P$, is not contained in $L$ (because $L$ is simple), so $PL=G$, Now, since $Q \cap P=1$, so $$Q \cong \frac{Q}{Q \cap P} \cong \frac{PQ}{P} \unlhd \frac{G}{P} \cong L$$ so $Q \cong L$ is nonabelian simple and hence $G = PQ \cong P \times Q$. Then $P$ must also be simple, or it wouldn't be a minimal normal subgroup, and $P$ and $Q$ are the only minimal normal subgroups of $G$.
• Thanks for your answer! ''$P$ and $Q$ are the only minimal normal subgroups'' - why that? By another theorem I know this just follows if $P$ and $Q$ are both nonabelian and simple, but I do not see why $P$ should be non-abelian... Mar 13, 2015 at 18:10
• $P$ could be abelian or nonabelian, but $P$ and $Q$ would still be the only minimal normal subgroups. Mar 13, 2015 at 19:15
• But why, for example $C_2 \times C_2$ has three minimal normal subgroups, so the factors are not the only ones... Mar 13, 2015 at 19:25
• But $Q$ is definitely nonabelian! Mar 13, 2015 at 20:10
• Okay, guess I found a proof of that fact, is this to complicated or is there any easier way to see this? My proof: Suppose $G = P\times Q$ as above and let $N\unlhd G$. If there exists no $(n_1, n_2) \in N$ with $n_2\ne 1$, then $N \le P\times \{1\}$, and by simplicity we have $N = \{1\}$ or $N = P\times\{1\}$. Otherwise there exists some $(n_1, n_2) \in N$ with $n_2 \ne 1$. Then there exists some $g_2 \in P$ with $g_2 n_2 \ne n_2 g_2$ as otherwise $Z(Q)$ would be nontrivial (using that $Q$ is nonabelian), set $g = (1,g_2)$, then $[g,n] \in N\cap Q$ and so $Q\le N$ by using simplicity... Mar 14, 2015 at 15:14