Exercise 4.1.10 (Double cosets)

Proof. Let $H$ and $K$ be subgroups of the group $G$. (The statement assumes implicitly that $G$ is a finite group.)

(a)

We know that $G$ acts on the set $S$ of left cosets of $K$ by $g\cdot (g^{\prime }K)=g{g}^{\prime }K$. This action restricted to the subgroup $H$ gives an action of $H$ on the set of left cosets of $K$. We write ${𝒪}_{\mathit{𝑥𝐾}}$ the orbit containing $\mathit{𝑥𝐾}$ for this action of $H$ on $S$. Explicitly, if $H=\{h_1,\dots ,h_r\}$, then $$\begin{array}{lll}\hfill {𝒪}_{\mathit{𝑥𝐾}}=\{h_1(\mathit{𝑥𝐾}),h_2(\mathit{𝑥𝐾}),\dots ,h_r(\mathit{𝑥𝐾})\}=\{x_{1}K,\dots ,x_{n}K\},& & \hfill \text{(1)}\end{array}$$

where the $x_{i}K$ are distinct after removing the duplicates in the list $h_1(\mathit{𝑥𝐾}),h_2(\mathit{𝑥𝐾}),\dots ,h_r(\mathit{𝑥𝐾})$.

Then, using (1),

$$\begin{array}{llll}\hfill \mathit{𝐻𝑥𝐾}& =\underset{h\in H}{\bigcup }\mathit{h𝑥𝐾}& \hfill & \\ \hfill & =\underset{1\le i\le r}{\bigcup }{h}_i(\mathit{𝑥𝐾})& \hfill & \\ \hfill & =\underset{1\le j\le n}{\bigcup }{x}_{j}K.& \hfill & \end{array}$$

So $\mathit{𝐻𝑥𝐾}={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{j=1}^n{x}_{j}K$ is the union of the left cosets $x_{1}K,\dots ,x_{n}K$ where $\{x_{1}K,\dots ,x_{n}K\}$ is the orbit containing $\mathit{𝑥𝐾}$ of $H$ acting by left multiplication on the set of left cosets of $K$.

(b)

Using the right action of $K$ on the right cosets of $H$, we obtain similarly $$\mathit{𝐻𝑥𝐾}=\underset{1\le i\le m}{\bigcup }H{y}_i,$$

where $\{H{y}_1,\dots ,H{y}_m\}$ is the orbit containing $\mathit{𝐻𝑥}$ for this right action.

(c)

We can use the parts (a) and (b), but alternatively, we may use the action of $\mathrm{\Gamma }=H\times K$ on $G$ defined by $$(h,k)\cdot x=\mathit{h𝑥}{k}^{-1},(h,k)\in H\times K,x\in G.$$

We verify that this defines an action.

• For all $x\in G$,

  $$(1,1)\cdot x=1x{1}^{-1}=x.$$

• For all $(h,k)\in \mathrm{\Gamma }$ and all $(h^{\prime },k^{\prime })\in \mathrm{\Gamma }$,

  $$\begin{array}{llll}\hfill (h,k)\cdot ((h^{\prime },k^{\prime })\cdot x)& =h(h^{\prime }x{k{}^{\prime }}^{-1})k^{-1}& \hfill & \\ \hfill & =(h{h}^{\prime })x{(k{k}^{\prime })}^{-1}& \hfill & \\ \hfill & =(h{h}^{\prime },k^{\prime }k)\cdot x& \hfill & \\ \hfill & =((h,k)(h^{\prime },k^{\prime }))\cdot x.& \hfill & \end{array}$$

  Let’s characterize the orbits of this action. For all $y\in G$,

  $$\begin{array}{llll}\hfill y\in \mathrm{\Gamma }\cdot x& ⟺\exists <!--\; FUNCTION\; APPLICATION\; -->h\in H,\exists <!--\; FUNCTION\; APPLICATION\; -->k\in K,y=\mathit{h𝑥}{k}^{-1}& \hfill & \\ \hfill & ⟺y\in \mathit{𝐻𝑥𝐾}.& \hfill & \end{array}$$

  Indeed, $\mathit{h𝑥}{k}^{-1}\in \mathit{𝐻𝑥𝐾}$ if $h\in H,k\in K$, and conversely, an element $y\in \mathit{𝐻𝑥𝐾}$ is of the form $h=\mathit{h𝑥𝑙},h\in H,l\in K$, and since $k=l^{-1}\in K$, $y=\mathit{h𝑥}{k}^{-1}$, where $h\in H,k\in K$.

  We have proven

  $$\mathrm{\Gamma }\cdot x=\mathit{𝐻𝑥𝐾}.$$

  The double cosets are therefore the orbits of the action of $\mathrm{\Gamma }=H\times K$ on $G$, and these orbits thus form a partition of $G$.

  Let us choose a complete system $S$ of representatives of the orbits, i.e., $|\mathit{𝐻𝑥𝐾}\cap S|=1$ for all $x\in G$. Then $G$ is a disjoint union of the orbits $\mathrm{\Gamma }\cdot x=\mathit{𝐻𝑥𝐾},x\in S$:

  $$G=\underset{x\in S}{\bigcup }\mathit{𝐻𝑥𝐾}\text{(disjoint union)}.$$

(d)

First proof. Let us calculate the cardinality of the orbit $\mathrm{\Gamma }\cdot x=\mathit{𝐻𝑥𝐾}$, where $\mathrm{\Gamma }=H\times K$.

For every $(h,k)\in \mathrm{\Gamma }$,

$$\begin{array}{llll}\hfill (h,k)\in {\mathrm{\Gamma }}_x& ⟺(h,k)\cdot x=x& \hfill & \\ \hfill & ⟺\mathit{h𝑥}{k}^{-1}=x& \hfill & \\ \hfill & ⟺k=\mathit{𝑥h}{x}^{-1},& \hfill & \end{array}$$

so the stabilizer of $x$ is

$$\begin{array}{lll}\hfill {\mathrm{\Gamma }}_x& =\{(h,k)\in H\times K\mid k=\mathit{𝑥h}{x}^{-1}\}.& \hfill \text{(2)}\end{array}$$

Consider the map

$$\phi \{\begin{array}{ccc}\hfill H\cap x^{-1}\mathit{𝐾𝑥}\hfill & \hfill \to \hfill & {\mathrm{\Gamma }}_x\hfill \\ \hfill h\hfill & \hfill \mapsto \hfill & (h,\mathit{𝑥h}{x}^{-1}).\hfill \end{array}$$

By (2), $\phi$ is surjective, and

$$\phi (h)=\phi (h^{\prime })\Rightarrow (h,\mathit{𝑥h}{x}^{-1})=(h^{\prime },x{h}^{\prime }{x}^{-1})\Rightarrow h=h^{\prime },$$

so $\phi$ is injective. Therefore $\phi$ is a bijection, hence

$$|{\mathrm{\Gamma }}_x|=|H\cap x^{-1}\mathit{𝐾𝑥}|.$$

The orbit-stabilizer formula shows that

$$\begin{array}{llll}\hfill |\mathit{𝐻𝑥𝐾}|& =|\mathrm{\Gamma }\cdot x|& \hfill & \\ \hfill & =|\mathrm{\Gamma }:{\mathrm{\Gamma }}_x|& \hfill & \\ \hfill & =|H||K|∕|H\cap x^{-1}\mathit{𝐾𝑥}|,& \hfill & \end{array}$$

so, since $H\cap \mathit{𝑥𝐾}{x}^{-1}$ is a subgroup of $H$,

$$|\mathit{𝐻𝑥𝐾}|=|K|\cdot |H:H\cap \mathit{𝑥𝐾}{x}^{-1}|.$$

(In particular, for $x=1$, we obtain a new proof of Proposition 13, p. 93:

$$|\mathit{𝐻𝐾}|=\frac{|H||K|}{|H\cap K|}.)$$

Second proof. (Using Proposition 13, p. 93.)

The map $G\to G$ defined by $g\mapsto g{x}^{-1}$ is bijective, with inverse $g\mapsto \mathit{𝑔𝑥}$. The image of $\mathit{𝐻𝑥𝐾}$ under this bijection is $\mathit{𝐻𝑥𝐾}{x}^{-1}$, therefore

$$|\mathit{𝐻𝑥𝐾}|=|\mathit{𝐻𝑥𝐾}{x}^{-1}|.$$

Since $H$ and $\mathit{𝑥𝐾}{x}^{-1}$ are two subgroups, the Proposition 13 (p. 93) gives

$$|\mathit{𝐻𝑥𝐾}|=|H(\mathit{𝑥𝐾}{x}^{-1})|=\frac{|H||\mathit{𝑥𝐾}{x}^{-1}|}{|H\cap \mathit{𝑥𝐾}{x}^{-1}|}=\frac{|H||K|}{|H\cap \mathit{𝑥𝐾}{x}^{-1}|}.$$

So

$$|\mathit{𝐻𝑥𝐾}|=|K|\cdot |H:H\cap \mathit{𝑥𝐾}{x}^{-1}|.$$

(e)

Similarly, the equalities $$|\mathit{𝐻𝑥𝐾}|=|(x^{-1}\mathit{𝐻𝑥})K|=\frac{|x^{-1}\mathit{𝐻𝑥}||K|}{|x^{-1}\mathit{𝐻𝑥}\cap K|}=\frac{|H||K|}{|x^{-1}\mathit{𝐻𝑥}\cap K|}$$

give

$$|\mathit{𝐻𝑥𝐾}|=|\mathit{𝐻𝑥𝐾}|=|H|\cdot |K:K\cap x^{-1}\mathit{𝐻𝑥}|$$