orbits of a normal subgroup are equal in size when the full group acts transitively

The following theorem proves that if a group acts transitively on a finite set, then any of the orbits of any normal subgroup are equal in size and the group acts transitively on them. We also derive an explicit formula for the size of each orbit and the number of orbits.

Theorem 1.

Let $H$ be a normal subgroup of $G$ , and assume $G$ acts transitively on the finite set $A$ . Let $\\mathcal{O}_{1},\\ldots,\\mathcal{O}_{r}$ be the orbits of $H$ on $A$ . Then

1.
$G$ permutes the $\\mathcal{O}_{i}$ transitively (i.e. for each $g\\in G,1\\leq j\\leq r$ , there is $1\\leq k\\leq r$ such that $g\\mathcal{O}_{j}=\\mathcal{O}_{k}$ , and for each $1\\leq j,k\\leq r$ , there is $g\\in G$ such that $g\\mathcal{O}_{j}=\\mathcal{O}_{k}$ ), and the $\\mathcal{O}_{i}$ all have the same cardinality.
2.
If $a\\in\\mathcal{O}_{i}$ , then $\\lvert\\mathcal{O}_{i}\\rvert=\\lvert H:H\\cap G_{a}\\rvert$ and $r=\\lvert G:HG_{a}\\rvert$ .

Proof.

Note first that if $g\\in G,a\\in\\mathcal{O}_{i}$ , and $g\\cdot a\\in\\mathcal{O}_{j}$ , then $g\\mathcal{O}_{i}\\subset\\mathcal{O}_{j}$ . For suppose also $b\\in\\mathcal{O}_{i}$ . Then since $a, b$ are in the same $H$ -orbit, we can choose $h\\in H$ such that $b=h\\cdot a$ . Then

g\\cdot b=g\\cdot h\\cdot a=g\\cdot h\\cdot g^{-1}\\cdot g\\cdot a=h^{\\prime}\\cdot g%\\cdot a\\in h^{\\prime}\\mathcal{O}_{j}\\subset\\mathcal{O}_{j}

since $H$ is normal in $G$ . Thus for each $g\\in G,1\\leq j\\leq r$ , there is $1\\leq k\\leq r$ such that $g\\mathcal{O}_{j}\\subset\\mathcal{O}_{k}$ .

Given $j, k$ , choose $a_{j}\\in\\mathcal{O}_{j},a_{k}\\in\\mathcal{O}_{k}$ . Since $G$ is transitive on $A$ , we may choose $g\\in G$ such that $g\\cdot a_{j}=a_{k}$ . It follows from the above that $g\\mathcal{O}_{j}\\subset\\mathcal{O}_{k}$ .

To prove 1), given $j, k$ , choose $g$ such that $g\\mathcal{O}_{j}\\subset\\mathcal{O}_{k}$ and $g^{\\prime}$ such that $g^{\\prime}\\mathcal{O}_{k}\\subset\\mathcal{O}_{j}$ . But then $\\lvert\\mathcal{O}_{j}\\rvert\\leq\\lvert\\mathcal{O}_{k}\\rvert\\leq\\lvert\\mathcal{O%}_{j}\\rvert$ so that $\\lvert\\mathcal{O}_{j}\\rvert=\\lvert\\mathcal{O}_{k}\\rvert$ and the subset relationships in the previous two paragraphs are actually set equality.

To prove 2), consider the following diagram:

\\xymatrix{&G\\ar@{-}[d]\\inner@par&HG_{a}\\ar@{-}[ld]\\ar@{-}[rd]\\inner@par H\\ar@{%-}[rd]&&G_{a}\\ar@{-}[ld]\\inner@par&H\\cap G_{a}}

Clearly $H\\cap G_{a}=H_{a}$ , and $\\lvert H:H_{a}\\rvert=\\lvert\\mathcal{O}_{i}\\rvert$ by the orbit-stabilizer theorem. Using the second isomorphism theorem for groups, we then have

\\lvert\\mathcal{O}_{i}\\rvert=\\lvert H:H_{a}\\rvert=\\lvert H:H\\cap G_{a}\\rvert=%\\lvert HG_{a}:G_{a}\\rvert

But $\\lvert G\\rvert=r\\lvert\\mathcal{O}_{i}\\rvert$ by the above, so

r\\lvert\\mathcal{O}_{i}\\rvert=\\lvert G\\rvert=\\lvert G:HG_{a}\\rvert\\cdot\\lvert HG%_{a}:G_{a}\\rvert=\\lvert G:HG_{a}\\rvert\\cdot\\lvert\\mathcal{O}_{i}\\rvert

and the result follows.∎