# Minimal normal subgroups.

• Mar 19th 2012, 06:14 AM
ENRIQUESTEFANINI
Minimal normal subgroups.
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.