I'm trying to understand the proof of this Lemma.
Lemma
Suppose that $G$ acts transitively on $Ω$. If $N$ is a normal subgroup of $G$, then the orbits of $N$ on $Ω$ form a block system. In particular, if $G$ is primitive on $Ω$ and $N$ is not in the kernel of the action, then $N$ acts transitively on $Ω$.
Proof
The orbits of $N$ surely partition $Ω$. We just need to show that $G$ leaves this partition invariant. So let $Λ$ be an $N$-orbit. Then, for $g ∈ G$ we need to show that $Λg$ is an $N$-orbit. Let $n ∈ N$. Then $$(Λg)n = Λ(gn) = Λ(gng^{−1}g) = Λn^{g^{−1}} g = Λg.$$ Hence $Λg$ is a union of $N$-orbits. Let $λ_1g,λ_2g ∈ Λg$. Then there exists $n ∈ N$ such that $λ_2 = λ_1n$. Thus $$λ_2g= λ_1ng = λ_1gg^{−1}ng =λ_1gn^g. $$ and so, as $n^g ∈ N^g = N$, $N$ acts transitively on $Λg$. The rest of the claim now follows.
These are my questions.
- "The orbits of $N$ surely partition $Ω$." Yes, straightforward from the definition. So, they mention this because the block system forms a partition of $\Omega $ right?
- "We just need to show that $G$ leaves this partition invariant." Why do we have to show this? Is it because this is one of the properties of a block system?
- "The rest of the claim now follows." How come the rest of the claim follows? All we did is prove that $N$ acts transitively on $Λg$.
Remark: From what I understand: If we can prove the first part of the lemma, the "in particular" part follows because $Λ$ is a block so, as $G$ is primitive and $N$ is not in the kernel of the action, $Λ=\Omega$. From the proof we have that $N$ acts transitively on $Λg$. Hence, this means $N$ acts transitively on $\Omega.$
As for your first two questions, remember what you are trying to prove: the orbits of $N$ on $\Omega$ form a block system. By definition, a block system for the action of $G$ on $\Omega$ is a partition of $\Omega$ that is left invariant by $G$. So in order to show that the orbits of $N$ on $\Omega$ form a block system, you need to show that these partition $\Omega$ and are $G$-invariant. It's just by the definition of what you are trying to prove really.
As for the third question, we want to show that the orbits of $N$ are $G$-invariant. That means that if $\Lambda$ is an orbit of $N$ and $g \in G$, then $\Lambda g$ is also an orbit of $N$. You don't ask about this so I suppose that the proof that $\Lambda g$ is a union of orbits of $N$ is clear. Now we need to show that $\Lambda g$ is actually just one orbit. This follows from the fact that $N$ acts transitively on $\Lambda g$. Indeed, suppose that $\Lambda_1, \Lambda_2$ are two orbits contained in $\Lambda g$. Since $N$ acts transitively on $\Lambda g$, given $x_1 \in \Lambda_1$ and $x_2 \in \Lambda_2$, there exists $n \in N$ such that $x_1 n = x_2$. Therefore $x_1$ and $x_2$ are in the same $N$-orbit. So $\Lambda_1 = \Lambda_2$ (recall that two orbits are either equal or disjoint). We conclude that $\Lambda g$ is a union of orbits any two of which are equal, so it is just an orbit.