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Frattini's argument, a variant.

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Frattini's argument, a variant.

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

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

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

Any hint will be most welcome.

Thank you very much, csguy.

N doesn't need to be normal. Rotman's book has the exact same problem without the normality assumption.

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