Normal subgroup has at most $|G:N|$ orbits

Question

$G$ is a transitive group action. Normal subgroup of transitive group $G$ has at most $|G:N|$ orbits, and if $|G:N|$ is finite, then the number of orbits of $N$ divides $|G:N|$.

My attempt: I have to use orbit-stabilizer theorem to solve second part. I am confused because we have action $G/N$ on the set of orbits $\{gNx, g \in G\}$. I know the action $G/N$ is transitive, but why is set of orbits less than $|G:N|$?

Answer 1

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The inequality you seek to prove can be established as follows: let us first assume $G$ is a group acting transitively on set $A$ and that $H \trianglelefteq G$.

1. Consider first the natural action of $G$ on $\mathscr{Part}(A)$, the set of all partitions of $A$, given by $\lambda \mathscr{P}=\{\lambda X\}_{X \in \mathscr{P}}$.

2. Consider next the partition $A/H$ given by the orbits under the action of $H$. The normality of $H$ is what makes this partition to be a fixed point under the action of $G$ introduced above, since $\lambda (Hx)=H(\lambda x)$ for any $\lambda \in G$ and $x \in A$.

3. By virtue of the previous observation, one can consider the natural action of $G$ on $A/H$, given by $\lambda X=\{\lambda x\}_{x \in X}$ for any $\lambda \in G$ and $X \in A/H$. The transitivity of the original action of $G$ on $A$ will induce the transitivity of this new action on $A/H$: as $A$ is nonempty (by definition), we can fix a certain $B \in A/H$, in its own turn nonempty (since it is an orbit), hence we can furthermore fix an arbitrary $a \in B$; consider also an arbitrary $X \in A/H$; again, since $X$ is nonempty, there must exist a certain $t \in X$; since $G$ acts transitively on $A$, there will exist $\lambda \in G$ such that $\lambda a=t$; then we have $\lambda B, X \in A/H$ and $t \in \lambda B \cap X$, hence $X=\lambda B$.

4. Keeping the notations of 3), by the orbit-stabilizer theorem we have that $\left|A/H\right|=\left|G:\mathrm{Stab}_{G}(B)\right|$; as $B$ is an orbit under $H$, it is immediate that $H \leqslant \mathrm{Stab}_{G}(B)$ and therefore that $\left|A/H\right| \leqslant |G:H|$.

As a side-note, it can actually be shown that $\mathrm{Stab}_{G}(B)=H \mathrm{Stab}_{G}(a)$.

Hope this helps.

Answer 2

Every $g\in G$ lays in some right coset of $N$ in $G$. So, denoted with $R\subseteq G$ a complete set of coset representatives, for $x\in X$ we get: \begin{alignat}{1} \operatorname{Orb}_G(x) &= \{g\cdot x, g\in G\} \\ &= \{(ng_i)\cdot x, g_i\in R \text{ and } n\in N\} \\ &= \{n\cdot(g_i\cdot x), g_i\in R \text{ and } n\in N\} \\ &= \{n\cdot y_i, y_i\in Y(x) \text{ and } n\in N\} \\ &= \bigcup_{y_i\in Y(x)}\{n\cdot y_i, n\in N\} \\ &= \bigcup_{y_i\in Y(x)}\operatorname{Orb}_N(y_i) \\ \end{alignat} where $Y(x):=\{g_i\cdot x, g_i\in R\}$. Now, $|Y(x)|\le |R|=|G:N|$, so if the $G$-action is transitive, then the $N$-action has at most $|G:N|$ orbits.

So, the normality of $N$ is not necessary up to this point (though it enters in the second part of the claim).