Let $R$ a ring and $J$ a ideal of $R$. Show that $M$ is a maximal ideal of $R$ if, and only if, $\overline{M}=M/J$ is a maximal ideal of $\overline{R}$.
Proof: Lets suppose $M$ maximal. So, if $\overline{I}=I/J$ is a ideal of $\overline{R}$ that contains properly $\overline{M}$, i.e., $$\overline{M}\subsetneq \overline{I}$$ We need to show that $\overline{I}=\overline{R}$. In fact, since $$\overline{M}\subsetneq \overline{I}$$ we have $$M\subsetneq I$$ But, since M is a maximal ideal of $R$, this implies that $I=R$. Then, $\overline{I}=\overline{R}$. Therefore, $\overline{M}$ is a maximal ideal of $\overline{R}$.
Conversely, lets suppose $\overline{M}=M/J$ is a maximal ideal of $\overline{R}$. So, if $I$ is a ideal of $R$ that contains properly $M$, i.e., $$M\subsetneq I$$ We need to show that $I=R$. In fact, since $J\subset M$ (by correspondence theorem), we have $J\subset I$. So, we have $$\overline{M}=M/J \subsetneq I/J=\overline{I}$$ But, since $\overline{M}$ is maximal ideal of $\overline{R}$, we have $\overline{I}=\overline{R}$. This implies that, $I=R$. Therefore, $M$ is a maximal ideal of $R$.
My question is: How to generalizes this proof for simple groups? Thank you.