Double cosets

We turn now to another important decomposition of an arbitrary group G into disjoint complexes. Such decompositions will play an important role in our later considerations of the Sylow Theorems.

Let $G$ be an arbitrary group and let $H_1$ and $H_2$ be subgroups of $G$. For $a,b \in G$, we define

\begin{displaymath} a \sim b \ \ \ \ {\rm if\ and\ only\ if}\ \ \ h_1ah_2 = b \end{displaymath} (8.3)

where $h_1 \in H_1$ and $h_2 \in H_2$. We contend first of all that the relation given in (8.3) is an equivalence relation on $G$. To see this, we note

  1. (Reflexivity): $a \sim a$ since $eae = a$, where $e$ is the identity element and $e \in H_1$ and $e \in H_2$ since $H_1\leq G$, $H_2 \leq G$.

  2. (Symmetry): $a \sim b$ implies there exist $h_1 \in H_1$ and $h_2 \in H_2$ such that $h_1ah_2 = b$, but then $a=h_1^{-1}bh_2^{-1}$, so $b \sim a$ since $h_1^{-1} \in H_1$ and $h_2^{-1} \in H_2$.

  3. (Transitivity): If $a \sim b$ and $b \sim c$, then there exist elements $h_1,h_1'\in H_1$ and $h_2,h_2'\in H_2$ such that

    \begin{displaymath} h_1ah_2=b\ \ \ \ {\rm and}\ \ \ h_1'ah_2'=c, \end{displaymath}


    \begin{displaymath} h_1'h_1ah_2'h_2=c. \end{displaymath}

    Since $h_1h_1'\in H_1$ and $h_2h_2'\in H_2$, we have that $a \sim c$.

Thus $\sim$ given by (8.3) is indeed an equivalence relation on $G$. We next take a look at the equivalence classes.

Definition 8.2.1   Let $G$ be a group with subgroups $H_1$ and $H_2$ (not necessarily distinct). If $a \in G$, the complex $H_1aH_2$ is called a double coset with respect to $H_1$ and $H_2$. By definition,

\begin{displaymath} H_1aH_2 = \{h_1ah_2 \in G \ \vert\ h_1 \in H_1, h_2 \in H_2\}. \end{displaymath}

For $a \in G$, the equivalence class $[a]$ of $a$ as we recall, contains all $b \in G$ with $b \sim a$. By the definition given in (8.3), this means $b = h_1ah_2$, where $h_1 \in H_1$ and $h_2 \in H_2$. Thus $b \in H_1aH_2$. As the above statements are all ``if and only if'', we see that $[a] = H_1aH_2$, the double coset given in Definition 8.2.1. By our general theorem on equivalence relations, Theorem 1.1.4, we know that either

\begin{displaymath} H_1aH_2 = H_1bH_2\ \ \ {\rm or}\ \ \ H_1aH_2 \cap H_1bH_2 = \emptyset \end{displaymath}


\begin{displaymath} G =\coprod_a H_1aH_2 \end{displaymath}

where the union is taken over certain $a \in G$. The identity element $e$ belongs to the complex $H_1H_2$.

If $H_2 = \{e\}$, then we simply get the right coset decomposition of $G$ with respect to $H_1$. If $H_1 = \{e\}$, then we have the left coset decomposition of $G$ with respect to $H_2$. Thus the double coset decomposition of a group may be viewed as a generalization of the coset (right or left) decomposition of a group. However, the reader should be careful not to generalize all facts related to coset decompositions to the case of double coset decompositions. For example, we saw that any two cosets of a finite group have the same number of elements. We shall presently see that this is not the case with double cosets.

Let us consider the double coset $H_1aH_2$. Clearly $H_1aH_2$ contains all right cosets of the form $H_1ah_2$, where $h_2 \in H_2$ and $H_1aH_2$ contains all left cosets of the form $h_1aH_2$ where $h_1 \in H_1$. We claim, as a matter of fact, that $H_1aH_2$ is a union of right or left cosets of the above form. For suppose that

\begin{displaymath} gH_2 \cap H_1aH_2\not= \emptyset. \end{displaymath}

Then there exist elements $h_2,h_2'\in H_2$ and $h_1 \in H_1$ such that

\begin{displaymath} gh_2'=h_1ah_2 \end{displaymath}

or $g=h_1ah_2(h_2')^{-1}$. This implies that

\begin{displaymath} gH_2=h_1aH_2, \end{displaymath}

and so $gH_2 \subset H_1aH_2$. Since this shows that any left coset which has anything at all in common with $H_1aH_2$, must be totally contained in $H_1aH_2$, we have
\begin{displaymath} H_1aH_2 = \cup_{h_1} h_1aH_2. \end{displaymath} (8.4)

Similarly, it can be shown that
\begin{displaymath} H_1aH_2 = \cup_{h_2} H_1ah_2 . \end{displaymath} (8.5)

Next, we wish to ascertain the number of left and right cosets in the double coset. Even though this number can be finite for an infinite double coset, we assume $\vert G\vert < \infty$. This is contained in

Theorem 8.2.2   Let $G$ be a finite group, let $H_1\leq G$, $H_2 \leq G$, and let $a \in G$. Then
The number of right cosets of $H_1$ in $H_1aH_2$ is $[H_2 : H_2 \cap a^{-1}H_1a]$.

The number of left cosets of $H_2$ in $H_1aH_2$ is $[a^{-1}H_1a : H_2 \cap a^{-1}H_1a]$.

Proof: We first note that $a^{-1}H_1a$ is a subgroup by Proposition 7.2.1. Consider the mapping of the double coset $H_1aH_2$ onto the complex $a^{-1}H_1aH_2$ given by $h_1ah_2\longmapsto a^{-1}h_1ah_2$. It is easy to show that this map is well-defined, 1-1, and onto (see exercise 2 for this section). Thus $\vert H_1aH_2\vert = \vert a^{-1}H_1aH_2\vert$. But $a^{-1}H_1aH_2$ is the product of two subgroups $a^{-1}H_1a$ and $H_2$ so by the product theorem (Theorem 4.3.6),

\begin{displaymath} \vert a^{-1}H_1a\cdot H_2 \vert = \frac{\vert a^{-1}H_1a\vert\vert H_2 \vert}{\vert a^{-1}H_1a\cap H_2 \vert}. \end{displaymath}

Now according to (8.4), the number of left cosets of $H_2$ in $H_1aH_2$ is $\vert H_1aH_2\vert/ \vert H_2\vert$. Thus the number of left cosets of $H_2$ in $H_1aH_2$ is

\begin{displaymath} \frac{\vert a^{-1}H_1a\vert}{\vert a^{-1}H_1a\cap H_2 \vert} =[a^{-1}H_1a:a^{-1}H_1a\cdot H_2 ]. \end{displaymath}

This establishes part (b) of the theorem. A similar argument establishes part (a) (this is left as an exercise). $\Box$

Under the same hypotheses as in Theorem 8.2.2, we use the notation $\char93 (H_1aH_2)$ to be the number of right cosets of $H_1$ in $H_1aH_2$ times $\vert H_1\vert$ (note from (8.5) that $\char93 (H_1aH_2) = \vert H_1aH_2\vert$). Thus Theorem 8.2.2 implies that

\begin{displaymath} \char93 (H_1aH_2)= \frac{\vert H_2\vert}{\vert a^{-1}H_1a\cap H_2 \vert}\cdot \vert H_1\vert. \end{displaymath}

This proves the following result.

Corollary 8.2.3   Let $G$ be a finite group and let $H_1\leq G$, $H_2 \leq G$. If $G=\coprod_a H_1aH_2$ (disjoint), then
\begin{displaymath} \vert G\vert=\sum_{j=1}^n \frac{\vert H_1\vert\vert H_2\vert}{d_j} \end{displaymath} (8.6)

where $d_j=\vert a_j^{-1}H_1a_j\cap H_2 \vert$.


David Joyner 2007-08-06