4

How do I calculate the Frattini subgroup of a simple non-abelian group $G$?

My efforts:

Well, it's a well know fact that $\Phi(G) \trianglelefteq G$ because $\Phi(G)$ is a characteristic subgroup. Thus $\Phi(G) = 1$ or $\Phi(G) = G$ due to the simplicity of $G$.

I'm wondering if the hypothesis '$G$ is non-abelian' helps me decide which of the two possibilities above actually occurs.

Since $G$ is a simple non-abelian group, we obtain $Z(G)=1$ (consequently $G$ is not nilpotent) and $G' = G$. How do I use this to prove what I want?

Remark: Note that the group $G$ is arbitrary, so it can be finite or infinite.

Shaun
  • 47,747
nom
  • 579
  • 2
    If the group is finite, then it needs to be the trivial subgroup, since $\Phi(G)=G$ if and only if $G$ has no maximal subgroups. And if you are working with an infinite group, then "simple non-abelian" is a bit redundant. – Arturo Magidin Aug 09 '22 at 22:51
  • 1
    So the question reduces to "do there exist infinite simple groups with no maximal subgroups?" Good question! – Derek Holt Aug 10 '22 at 09:53

0 Answers0