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I would like to show that,

If $G$ acts transitively on a set $X$, and $K$ is regular normal subgroup of $G$. Then $G = K \operatorname{Stab}(a)$. ($K \operatorname{Stab}(a)$. w.r.t G) for any $a \in X$.

I see why this is true if $G$ acts primitively, but I can't work out why this is true as it is currently stated. (Indeed it may not be!)

Note: I write $K \operatorname{Stab}(a)$ to be the (group) $\{k*g : \text{where $k\in K$ and $g\in \operatorname{Stab}(a)$.}\}$.

Also both $G$ and $X$ are finite.

Def of regular normal subgroup is : $K$ is a normal subgroup of $G$ which acts trans on $X$ and $\operatorname{Stab}(a) = 1$ for all $a\in X$ ($\operatorname{Stab}(a)$ w.r.t $K$)

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1 Answer 1

up vote 1 down vote accepted

You just need $K$ to be regular, and then you get the stronger result that $G = K\operatorname{Stab}(a)$ with $K \cap \operatorname{Stab}(a) = 1$.

This is because if $a \in X$ is fixed, then for each $g \in G$ there is $k \in K$ such that $(a^g)^k = a$, so that $g k \in \operatorname{Stab}(a)$, and $g \in \operatorname{Stab}(a) K$, so that $G = \operatorname{Stab}(a) K$, and then also $G = K \operatorname{Stab}(a)$.

By regularity $1 = \operatorname{Stab}_{K}(a) = K \cap \operatorname{Stab}(a)$.

If in addition $K$ is normal, then this is a semidirect product.

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