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

abstract-algebragroup-actionsgroup-theory

$G$ 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 $|G:N|$?

Best Answer

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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.

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