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Let $M$ be a maximal normal subgroup of $G$. Consider the canonical homomorphism $\phi : G \to G/M$ Now $\phi^{-1}$ of any nontrivial proper normal subgroup of $G/M$ is a proper normal subgroup of $G$ properly containing $M$. But M is maximal, so this can not happen, Thus $G/M$ is simple.

I couldn't understand the part in bold. How did we conclude that? , $\phi$ is an homomorphism and we are sure that homomorphisms preserve normal subgroups. A $\phi^{-1}$ of a normal subgroup of $G/M$ will be normal, but I can't conclude that it will properly contain $M$.

I understood the converse.

Any help?

2 Answers2

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This is The Correspondence Theorem : the canonical homomorphism (the projection) $\;\phi\;$ determines a $\;1-1\;$ correspondence between subgroup of $\;G/M\;$ and subgroups of $\;G\;$ containing the subgroup $\;M\;$ . This correspondence preserves indexes and normality.

DonAntonio
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The key word here is nontrivial. Let $H$ be a normal subgroup of $G/M$, and $K=\phi^{-1}(H)$. Since $H$ is a group, it must have the identity, so $K$ must contain $M$, since $\phi^{-1}(e) = M$ by the construction of $\phi$. If $H$ is nontrivial, then the pullback of $H$ is a normal subgroup which must contain $M$.

Santana Afton
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