# 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 $\displaystyle \neq$ 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 $\displaystyle N$ be a minimal normal subgroup of $\displaystyle G$. If $\displaystyle \varphi$ is an epimorphism from $\displaystyle G$ to a group $\displaystyle H$, then $\displaystyle N^\varphi = 1$ or $\displaystyle N^\varphi$ is a minimal normal subgroup of $\displaystyle H$.

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

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

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