Exercise 4.1.9 (Action of a normal subgroup)

Proof. We assume that $G$ acts transitively on the finite set $A$. Let $H$ be a normal subgroup of $G$, and let ${𝒪}_1,{𝒪}_2,\dots ,{𝒪}_r$ be the distinct orbits of $H$ on $A$, i.e., if $H=\{{\sigma }_1,{\sigma }_2,\dots ,{\sigma }_k\}$, then the orbit of $a\in A$ for the action of $H$ on $A$ is

Then ${𝒪}_i,1\le i\le r$, are distinct orbits, and form a partition of $A$:

Consider a complete system of representatives $\{a_1,a_2,\dots ,a_r\}$ of the orbits, so that $a_i\in {𝒪}_i$ and ${𝒪}_i=𝒪(a_i)$.

(a)

The orbits ${𝒪}_1,{𝒪}_2,\dots ,{𝒪}_r$ are blocks for the action of $G$ on $A$. Indeed, consider the orbit ${𝒪}_i$: $${𝒪}_i=\{\tau \cdot a_i\mid \tau \in H\}.$$

Since $H⊴G$, then for all $\sigma \in G$, the map

$$f_{\sigma }\{\begin{array}{ccc}\hfill H\hfill & \hfill \to \hfill & H\hfill \\ \hfill \tau \hfill & \hfill \mapsto \hfill & \mathit{𝜎𝜏}{\sigma }^{-1}\hfill \end{array}$$

is bijective, therefore

$$\begin{array}{llllll}\hfill \sigma {𝒪}_i& =\{\sigma \cdot (\tau \cdot a_i)\mid \tau \in H\}& \hfill & & \hfill & \\ \hfill & =\{(\mathit{𝜎𝜏})\cdot a_i\mid \tau \in H\}& \hfill & & \hfill & \\ \hfill & =\{(\mathit{𝜎𝜏}{\sigma }^{-1})\cdot (\sigma \cdot a_i)\mid \tau \in H\}& \hfill & & \hfill & \\ \hfill & =\{\lambda \cdot (\sigma \cdot a_i)\mid \lambda \in H\}& \hfill (\lambda =\mathit{𝜎𝜏}{\sigma }^{-1})& & \hfill & & \hfill \\ \hfill & & \hfill & & \hfill \end{array}$$

So $\sigma {𝒪}_i$ is the orbit of $\sigma \cdot a_i$ under the action of $H$ on $A$, thus

$$\sigma {𝒪}_i={𝒪}_j\text{for some }j\in [[1,k]].$$

Note that since ${({𝒪}_i)}_{1\le i\le r}$ is a partition of $A$, then $B={𝒪}_i$ satisfies $\sigma (B)={𝒪}_j$, so $B=\sigma (B)$ (if $i=j$) or $B\cap \sigma (B)=\varnothing$ (if $i\ne j$), for all $\sigma \in G$: this shows that each orbit ${𝒪}_i$ is a block for the action of $G$ on $A$.

In other words, if $R=\{{𝒪}_1,{𝒪}_2,\dots ,{𝒪}_r\}$, the map $f:G\times R\to R$ defined by $(\sigma ,{𝒪}_i)\mapsto \sigma {𝒪}_i$ is an action of $G$ on $R$, so $G$ permutes the sets ${𝒪}_1,{𝒪}_2,\dots ,{𝒪}_r$: ${\phi }_{\sigma }:R\to R$ defined by ${𝒪}_i\mapsto \sigma {𝒪}_i$ is a permutation of $R$, i.e., ${\phi }_{\sigma }$ is a bijection for every $g\in G$.

We show that the action of $G$ on $R$ is transitive. Consider ${𝒪}_i,{𝒪}_j\in R$. Since these orbits are non empty, pick $a\in {𝒪}_i$ and $b\in {𝒪}_j$. Since the action of $G$ on $A$ is transitive, there is some $\sigma \in G$ such that $\sigma \cdot a=b$. Then $b\in {𝒪}_j\cap \sigma {𝒪}_i$, therefore these two orbits are non disjoint, hence they are equal: ${𝒪}_j=\sigma {𝒪}_i$. This shows that the action of $G$ on $R$ is transitive.

Since the map $f:{𝒪}_i\to {𝒪}_j$ defined by $\sigma \mapsto \sigma \cdot a$ for all $a\in {𝒪}_i$ is bijective, $|{𝒪}_j|=|\sigma {𝒪}_i|=|{𝒪}_i|$. This is true for every $i,j$, so all orbits of $H$ on $A$ have the same cardinality.

(b)

Pick some element $a\in A$, and let ${𝒪}_1$ be the orbit of $a$. Note that for all $\sigma \in G$, and for all $a\in A$, $$\begin{array}{llll}\hfill \sigma \in H_a& ⟺\sigma \in H\text{ and }\sigma \cdot a=a& \hfill & \\ \hfill & ⟺\sigma \in H\cap G_a.& \hfill & \end{array}$$

So $H_a=H\cap G_a$.

Since $H_a$ is the stabilizer of $a$ for the action of $H$ on $A$, the orbit-stabilizer formula (Proposition 2) shows that

$$\begin{array}{lll}\hfill |{𝒪}_1|=|H:H_a|=|H:H\cap G_a|.& & \hfill \text{(1)}\end{array}$$

Since $G$ acts transitively on $A$, then $A$ is the orbit of $a$ for the action of $G$ on $A$, so the orbit-stabilizer formula gives

$$\begin{array}{lll}\hfill |G:G_a|=|A|.& & \hfill \text{(2)}\end{array}$$

By part (a), all the orbits ${𝒪}_i$ have the same cardinality, and there are $r$ orbits which form a partition of $A$, therefore

$$\begin{array}{lll}\hfill |A|=r|{𝒪}_1|.& & \hfill \text{(3)}\end{array}$$

Finally, the Second Isomorphism Theorem for the subgroups $H$ and $G_a$ of $G$ (see figure) shows that $H_a⊴G_a$ and $(H{G}_a)∕H\simeq G_a∕H_a$, so

$$\begin{array}{lll}\hfill |H{G}_a:H|=|G_a:H_a|.& & \hfill \text{(4)}\end{array}$$

Put $x=|H{G}_a:H|$. The equality (4) shows that

$$\begin{array}{llll}\hfill |H{G}_a:H_a|& =|H{G}_a:H|\cdot |H:H_a|=x|H:H_a|,& \hfill & \\ \hfill |H{G}_a:H_a|& =|H{G}_a:G_a|\cdot |G_a:H_a|=|H{G}_a:G_a|x,& \hfill & \end{array}$$

so

$$\begin{array}{lll}\hfill |H{G}_a:G_a|=|H:H_a|.& & \hfill \text{(5)}\end{array}$$

Putting all together, the equalities (1), (2) and (3) give

$$\begin{array}{lll}\hfill |G:G_a|& =r|H:H_a|,& \hfill \text{(6)}\end{array}$$

and $|G:G_a|=|G:H{G}_a|\cdot |H{G}_a:G_a|$, so by (5)

$$\begin{array}{lll}\hfill |G:G_a|& =|G:H{G}_a|\cdot |H:H_a|.& \hfill \text{(7)}\end{array}$$

The comparison of (6) and (7) shows that

$$r=|G:H{G}_a|.$$