About $G=P\times Q$ with $P$ and $Q$ simple

Question

Let $G=P\times Q$, where $P$ and $Q$ are simple and minimal normal in $G$. Then $P$ and $Q$ are the only minimal normal subgroups of $G$.

My question actually came from the excellent answer of this question. The OP asked in the comments that why $P$ and $Q$ are the only minimal normal subgroups of $G$ (this question). I thought it should not have been so complicated.

Suppose $H$ to be another minimal normal subgroup of $G$, if not contained in $Q$, it must follow $H\cap Q=1$, giving $H\subseteq P$ (since $G=PQ$). Hence $H=P$. But I know it’s wrong now, how to do it otherwise?

By the way, $Q$ is non-Abelian might seem useless to me, the simplicity of $Q$ could already be sufficient. Am I right?

Would you be so kind as to answer my questions? Any help would be sincerely appreciated!

Answer 1

Derek Holt has provided the correct statement:

Let $G = P \times Q$, where $P$ and $Q$ are both simple and not both abelian of the same (finite, prime) order. Then $P$ and $Q$ are the only minimal normal subgroups of $G$

Also the elements of essentially the same proof can be found in the comments to the linked question and in the answer of Robert Chamberlain. Nevertheless, I think there is still some room for an efficient exposition.

The case that $P$ and $Q$ are abelian of different prime orders is easy and I omit it. Suppose then that $Q$ is non-abelian.

Note that $C_G(Q) = P$. In fact, if $(p,q) \in P \times Q$ centralizes $Q$, then $q \in Z(Q)$. But $Z(Q)$ is trivial since $Q$ is simple and non-abelian.

Let $N$ be a minimal normal subgroup of $G$. Note that $N \cap P \lhd G$. If $N \cap P \ne \{1\}$, then $N \cap P = P$ by simplicity of $P$ and $N \cap P = N$ by minimality of $N$, so $N = P$. Similarly if $N \cap Q \ne \{1\}$, then $N = Q$.

Suppose that $N \ne P$ and thus $N \cap P = \{1\}$. Then $N \cap C_G(Q) = N \cap P = \{1\}$. Hence if $n \in N$ is not the identity element, then there exists a $q \in Q$ such that $\def\inv{^{-1}}$ $[n,q] = nqn^{-1} q^{-1} \ne 1$. But since both $N$ and $Q$ are normal in $G$, $[n,q] \in N \cap Q$. Thus $N \cap Q \ne \{1\}$. As observed above, this implies $N = Q$.

Answer 2

You need the condition that if $P\cong Q$ then $P$ (and therefore $Q$) is non-abelian, otherwise $P\cong Q\cong C_p\times C_p$ for some prime $p$ (where $C_p$ is the cyclic group of order $p$) which has minimal normal subgroups $\langle (0,1)\rangle$, $\langle (1,i)\rangle$ for $i=1,\ldots,p-1$. The proof would then have two cases.

First case, $P\cong C_p$, $Q\cong C_q$ for some primes $p\ne q$. Then $G\cong C_{pq}$ only has subgroups of orders $1,p,q,pq$ and these subgroups are unique so $P,Q$ must be the only minimal normal subgroups.

The second (harder) case is where one of $P,Q$, lets say $Q$, is non-abelian. Suppose $H$ is a minimal normal subgroup of $G$ and, identifying $G$ with $P\times Q$, define the projection $\phi_Q:H\to Q$ by $\phi_Q(x,y)=y$. It is easy to see that this is a group homomorphism and that $\mathrm{ker}(\phi_Q)\le P$.

If $\mathrm{ker}(\phi_Q)$ is non-trivial then it is a non-trivial normal subgroup of $P$ so, by simplicity of $P$, $P=\mathrm{ker}(\phi_Q)\le H$ and, by minimality of $H$, $P=H$.

If $\mathrm{ker}(\phi_Q)$ is trivial, let $(x,y)\in\mathrm{ker}(\phi_Q)$. As $Q$ is simple and non-abelian, it has trivial center, so there is some $z\in Q$ with $w=y^{-1}z^{-1}yz\ne 1$. As $H$ is normal in $G$, $(1,z)^{-1}(x,y)(1,z)\in H$, so $(1,1)\ne(1,w)=(x,y)^{-1}(1,z)^{-1}(x,y)(1,z)\in H$. As $(1,w)\in Q\cap H$ the smallest normal subgroup $\langle (1,w)^G\rangle$ containing $(1,w)$ is then contained in $H$ and $Q$. By minimality of $H$ and $Q$, $H=\langle (1,w)^G\rangle=Q$.