Minimal normal subgroups.

**ENRIQUESTEFANINI** · March 19th 2012, 06:14 AM

Hi:

In a certain book the following definition is given: Let G be a group. A normal subgroup N $\neq$ 1 of G is a minimal normal subgroup of G if 1 and N are the only normal subgroups of G that are contained in N.

Then the following proposition is stated: Let $N$ be a minimal normal subgroup of $G$ . If $\varphi$ is an epimorphism from $G$ to a group $H$ , then $N^\varphi = 1$ or $N^\varphi$ is a minimal normal subgroup of $H$ .

The author gives this proof: Let $A \neq 1$ be a normal subgroup of $H$ that is contained in $N^\varphi$ . Then $A^{\varphi^{-1}}\cap N$ is a normal subgroup of $G$ , and $A^{\varphi^{-1}}\cap N \neq 1$ since $A \neq 1$ . Hence $A^{\varphi^{-1}}\cap N = N$ and $N^\varphi = A$ .

OK. But why $A \neq 1 \Rightarrow A^{\varphi^{-1}}\cap N \neq 1$ ? Since $\varphi$ is homomorphism and $A$ normal in $H$ , $A^{\varphi^{-1}}$ is normal in $G$ . And the intersection of two normal subgroups is a normal subgroup, that is $A^{\varphi^{-1}}\cap N$ is normal in $G$ . But $A^{\varphi^{-1}}\cap N \subseteq N$ and N is minimal normal in $G$ . Hence either $A^{\varphi^{-1}}\cap N = 1$ or $A^{\varphi^{-1}}\cap N = N$ . Hence, if it were true that $A^{\varphi^{-1}}\cap N \neq 1$ , we would have $A^{\varphi^{-1}}\cap N = N$ . Assuming the latter is true, we have the following: $N \subseteq A^{\varphi^{-1}}$ , $N^\varphi \subseteq (A^{\varphi^{-1}})^\varphi$ . But since $\varphi$ is onto, $(A^{\varphi^{-1}})^\varphi = A$ . Hence $N^\varphi \subseteq A$ . But $A \subseteq N^\varphi$ by assumption. Hence $N^\varphi = A$ .

As you see, I can follow the steps in the proof, except for the inference " $A \neq 1 \Rightarrow A^{\varphi^{-1}}\cap N \neq 1$ ". Any suggestion?

The book is The Theory of Finite Groups, An Introduction, by Kurzweil and Stellmacher, and the proposition is proposition 1.7.1, p. 36.

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