3
$\begingroup$

I'd appreciate it if you consider this question and together with its hint:

Let $G=AB$ be a finite group which is the internal direct product of $A$ and $B$ which are non-abelian simple groups. Show that the only proper, nontrivial normal subgroups of $G$ are $A$ and $B$.

Hint: If $N$ is another such subgroup, we would have to consider the commutator $[N,A]$.

In fact I am not sure how to use this hint!

|cite|improve this question
$\endgroup$
  • 1
    $\begingroup$ $[N,A]$ is a normal subgroup, and is contained in both $A$ and $N$. $\endgroup$ – user641 Aug 12 '13 at 0:02
7
$\begingroup$

Suppose $N \leq G$ is normal, and $A$ is not contained in $N$, nor is $B$. By normality of $N$ and $A$, $[N,A] \leq A \cap N$, and $A \cap N$ is normal in $A$. By simplicity, $[N,A]=1,$ so for all $n \in N, a \in A,$ $nan^{-1}a^{-1}=1,$ and therefore $(n^{-1})^{a}=n^{-1}$, which implies that $A \subset C_G(N)$. Similarly, $B \subset C_G(N)$. It follows that $N \leq Z(G).$

Assuming that $A,B$ are nonabelian, we get that $Z(G)=Z(A)\times Z(B)=1.$ Therefore, $N$=1. The case of $A \lneq N$ should contradict the simplicity of $B$.

|cite|improve this answer
$\endgroup$
  • 1
    $\begingroup$ The normal subgroups of $G$ above $A$ corresponds to the normal subgroups of $G/A \cong B$. $\endgroup$ – Dune Aug 12 '13 at 0:37
  • 1
    $\begingroup$ It's worth noting that nowhere in this argument is the assumption that $G$ is finite used. $\endgroup$ – B. Mackey Aug 12 '13 at 1:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.