## Frattini argument, a special variant.

Hi: In the following, G is a group that acts on the set $\displaystyle \Omega$. For $\displaystyle \alpha \in \Omega, G_\alpha := \{x \in G | \alpha x = \alpha\}$.

Proposition: Suppose that G contains a normal subgroup N, which acts transitively on $\displaystyle \Omega$. Then $\displaystyle G = G_\alpha N$ for every $\displaystyle \alpha \in \Omega$. In particular, $\displaystyle G_\alpha$ is a complement of N in G if $\displaystyle N_\alpha= 1$.

This proposition is in a book, Kurzweil - Stellmacher, The Theory of Finite Groups, An Introduction, Springer, 2004. What I do not understand is why N has to be normal. In the proof given by the author, no use is made of that fact. The proof runs like this:

Let $\displaystyle \alpha \in \Omega$ and $\displaystyle y \in G$. The transitivity of $\displaystyle N$ on $\displaystyle \Omega$ gives an element $\displaystyle x \in N$ such that $\displaystyle \alpha y = \alpha x$. Hence $\displaystyle \alpha yx^{-1} = \alpha$ and thus $\displaystyle yx^{-1} \in G_\alpha$. This shows that $\displaystyle y \in G_\alpha x \subseteq G_\alpha N$.