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

###### 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. 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. 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. ∎

Title orbits of a normal subgroup are equal in size when the full group acts transitively OrbitsOfANormalSubgroupAreEqualInSizeWhenTheFullGroupActsTransitively 2013-03-22 17:17:56 2013-03-22 17:17:56 rm50 (10146) rm50 (10146) 5 rm50 (10146) Theorem msc 20M30