Regular Normal Subgroups and Semidirect Products

Let $G$ act transitively on $\Omega$ throughout. We shall be concerned with the case where $G$ has a regular normal subgroup $K$, meaning that $K \unlhd G$ acts transitively on $\Omega$ and the stabilizer $K_\alpha$ is trivial for every $\alpha$ (this is the regular action, it's equivalent to the action on the cosets $1 g$).

Lemma If $G$ acts transitively on $\Omega$ with regular normal subgroup $K$ choose $\alpha \in \Omega$ the action of $G_\alpha$ on $K$ by conjugation is equivalent to its action on $\Omega$ with $1 \in K$ corresponding to $\alpha \in \Omega$.
proof: Define $\theta : K \to \Omega$ by $k \mapsto \alpha k$. This is a bijection since $K$ is regular. We just need to show that the two actions are equivalent: $$(k \theta) g = \alpha k g = \alpha g k^g = \alpha k^g = k^g \theta.$$

If $G$ is multiply transitive and has a regular normal subgroup $K$ the subgroup $G_\alpha$ is transitive on $\Omega \setminus \{\alpha\}$ and hence $K^\# = K \setminus \{1\}$ as conjugation is an automorphism of a group, this means in particular that the group $K$ ???? automorphisms ?????? ff

Lemma An automorphism preserves the order of a group element.
proof: Let $\theta : G \to G$ be an automorphism and $g$ have order $n$, then $(\theta g)^i \not = 1$ for any $0 < i < n$ since if it did $\theta(g^i) = 1$ implies that $g^i = 1$.

Proposition Let $K$ be a group then

1. If $\operatorname{Aut} K$ acts transitively on $K^\#$ then $K$ is elementary abelian
2. If $\operatorname{Aut} K$ is 2-transitive on $K^\#$ then either $K \simeq C_2^s$ or $C_3$
3. If $\operatorname{Aut} K$ is 3-transitive on $K^\#$ then $K \simeq C_2^2$
4. $\operatorname{Aut} K$ is not 4-transitive.

 proof: (1) Since automorphisms preserve the order of elements all elements must have the same order, and that order must be a prime otherwise there would be elements with smaller order, by Cauchy's theorem we can say this is a $p$-group and so $Z(K)$ is non-trivial. In fact we know $Z(K) \operatorname{char} G$ (because the center is preserved by conjugation) now any automorphism of $K$ fixes a characteristic subgroup like $Z(K)$ so for transitivity to be possible $Z(K)=K$. This implies $K$ is abelian and therefore a direct product of cyclic groups, which we know must all be $C_p$!
(2) Since 2-transitivity implies primitivity of the action we define a relation on $K^\#$ that is an $\operatorname{Aut} K$-congruence: $k R k'$ if $k=k'$ or $k^{-1} = k'$. Clearly this is an equivalence relation preserved by the action hence by primitivity it must be the equality relation (implying $k^{-1} = k$ so the group is $C_2^s$) or the entire relation (implying every two non-identity elements are equal or inverse so the group will be $C_3$).
(3) If the group is further 3-transitive, then it's impossible that it's $C_3$: That's too small! Take some $k \in K^\#$ and consider the stabilizer $(\operatorname{Aut} K)_k$ which is 2-transitive hence primitive so again let's define a congruence on it $k_1 R k_2$ if $k_2 = k_1$ or $k_2 = k k_1$ - meaning that $k_1$,$k_2$ "differ by $k$" - this can't be the equality relation since that would imply $k=1$ so it's the universal relation implying $K \simeq C_2^2$.
(3)  This is impossible again because the group is too small.

Corollary Suppose $G$ acts $t$-transitively on $\Omega$ with a regular normal subgroup $K$ so $|\Omega| = |K|$ then

1. If $t=2$ then $K \simeq C_p^s$
2. If $t=3$ then $K \simeq C_2^s$ or $C_3$
3. If $t=4$ then $K \simeq C_2^2$
4. $t < 5$

Proposition If $(G,\Omega)$ is a primitive permutation group such that some $G_\alpha$ is simple (therefore all $G_\alpha$ are simple) then either $G$ is simple or has a regular normal subgroup.
proof: Suppose $G$ is not simple, then there's a $K$ such that $1 \not = K \lhd G$. In that case $K \cap G_\alpha \unlhd G_\alpha$ is simple so $K \cap G_\alpha = 1$ or $K \cap G_\alpha = G_\alpha$ but $K$ is transitive by a corollary on primitive permutation groups and $G_\alpha$ is a maximal subgroup by another corollary so we cannot have $G_\alpha \le K$ (certainly $G_\alpha < K$ can't happen, and if $G = G_\alpha$ then \alpha $g G_\alpha = \alpha g$ so $G_\alpha = G_\beta = \ldots = G$ contradiction). Then $K \cap G_\alpha = 1$ so $K$ is regular because it's a normal subgroup which we showed acts transitively and also it has the regular action because we saw that $K_\alpha = 1$.

Definition Let $H,K$ be groups with a homomorphism $\theta : H \to \operatorname Aut K$. For $k \in K$, $h \in H$ write $k^h$ short for $k^{h \theta}$. Then the semidirect product $K \rtimes H$ is defined as the group $K \times H$ with binary operation:
$$(k_1,h_1)(k_2,h_2) = (k_1 k_2^{h_1^{-1}},h_1h_2).$$
proof: prove this is a group

Define subgroups of the semidirect product $\tilde{K} = \{(k,1)\}$ and $\tilde{H} = \{(1,h)\}$ then $\tilde{K} \le K \rtimes H$, $\tilde{H} \le K \rtimes H$, $\tilde K \cap \tilde H = 1$,,$\tilde K \tilde H = K \rtimes H$ and $\tilde K \unlhd K \rtimes H$ but we don't know that $\tilde H \unlhd K \rtimes H$. This is an "external" construction of the semidirect product. We also have an internal one. An extension $G$ of $K$ by $H$ is such that $K \lhd G$, $G/K \simeq H$ i.e. the short exact sequence $$1 \longrightarrow K \longrightarrow G \longrightarrow H \longrightarrow 1$$ in which case we write $G = K . H$, the $.$ is "neutral" meaning that it doesn't really tell you how the group has been extended.

Definition if we have a s.e.s. like above with $K \unlhd G$, $K \cap H = 1$ and $KH=G$ we call $G$ a split extension of $K$ by $H$ and write $G = K:H$.

Theorem If $G$ is a split ext then define $\theta : H \to \operatorname{Aut} K$ by $k^{h \theta} = k^h$. Then $G \simeq K \rtimes H$.

Corollary If $G$ is transitive on $\Omega$ with a regular normal subgroup then $G = K : G_\alpha$.