Let $G$ be a nontrivial finite group. Then $H$ the intersection of all nontrivial normal subgroups has the property that if $K$ is a normal subgroup of $G$ such that $K \leq H$, then $K = H$ or $K$ is trivial. What is an example of a $G$ such that $H$ is not simple?

I suspect that $H$ is not necessarily simple because being a normal subgroup is not transitive.

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    $\begingroup$ One example is $G=S_4$. $\endgroup$ – Pavel Čoupek Oct 25 '14 at 12:47

Expanding PavelC's comment into an answer:

$S_4$ has 2 normal subgroups: $A_4$, and $V$, the group of double transpositions. $V\subset A_4$, so if you take $G$ to be $S_4$, your $H$ is $V$. $V$ is isomorphic to the Klein 4-group, so it is not simple.

What's going on in this case (the substance of your observation that normality is not transitive) is that subgroups of $V$ are not normal in $S_4$ even though they are normal in $V$. More specifically, the subgroups of $V$ cannot be conjugated to each other inside $V$ since it is abelian; therefore they are all normal in $V$. However, they are all conjugate inside $G$. For example, one of the subgroups of $V$ is the order 2 cyclic group generated by $(12)(34)$ (call it $J$). This subgroup is normal in $V$ since $V$ is abelian, but we have

$$(13)(12)(34)(13) = (14)(23)$$

and therefore $(13)J(13)$ is the subgroup generated by $(14)(23)$. Similarly,

$$(23)(12)(34)(23) = (13)(24)$$

and therefore $(23)J(23)$ is the subgroup generated by $(13)(24)$. Thus all three cyclic subgroups of $V$ are conjugate in $S_4$.


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