# Abstract Nonsense

## Double Cosets

Point of Post: In this post we discuss the notion of double cosets in a group $G$.

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Motivation

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There is a natural extension of the notion of cosets which are double cosets, namely cosets on ‘both’ sides. While of algebraic interest in their own right we shall see that they are mostly of interest in other fields such as ring theory and representation theory.

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Double Cosets

Let $G$ be a group and $H,K\leqslant G$. Consider the mapping of the direct product $H\times K$ times $G$ to $G$ given by left multiplication by $H$ and right inverse multiplication by $K$.  In other words $\cdot:\left(H\times K\right)\times G\to G$ given by $(h,k)\cdot g=hgk^{-1}$. We claim that $\cdot$ is an action of $H\times K$ on $G$. Indeed, it’s clear that $(e,e)\cdot g=ege=g$ for every $g\in G$ and for every $(h_1,k_1),(h_2,k_2)\in H\times K$ and $g\in G$ we have

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\displaystyle \begin{aligned}(h_1,k_1)\cdot((h_2,k_2)\cdot g) &=(h_1,k_1)\cdot h_2gk_2^{-1}\\ &=h_1h_2gk_2^{-1}k_1^{-1}\\ &=(h_1h_2,k_1k_2)\cdot g\\ &=\left((h_1,k_1)(h_2,k_2)\right)\cdot g\end{aligned}

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Thus, $\cdot$ is a $H\times K$-action on $G$ as claimed. Let $\mathcal{O}_g$ be an orbit of $g\in G$. It’s evident that $\mathcal{O}_g=HgK=\left\{hgk:h\in H\text{ and }k\in K\right\}$. We call such an orbit a double coset of $H$ and $K$. We denote the set of all double cosets of $H$ and $K$ in $G$ as $H\backslash G/K$

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We can think about double cosets in another way. Namely, let $K$ act on $G$ by the usual right inverse multiplication (or not inverse if you dont’ care about right actions). Consider then the action of $H$ on $G/K$ given by left multiplication. Clearly the double cosets and this action are very related. Indeed:

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Theorem: Let $G$ be a group and $H,K\leqslant G$. Then, if $HgK$ is a double coset of $H$ and $K$ then

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$\displaystyle HgK=\bigsqcup_{xK\in\mathcal{O}_{gK}}xK$

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It’s evident that $\displaystyle \bigcup_{xK\in\mathcal{O}_{gK}}xK$ is contained in $HgK$. To see the converse we note that if $hgk\in HgK$ then $hgk\in hgK$ but by definition we have that $hgK=h\cdot gK$ and so $hgk\in hgK\in\mathcal{O}_{gK}$ and so evidently $gk$ is in $\displaystyle \bigcup_{xK\in\mathcal{O}_{gK}}xK$. To see that they are disjoint we merely recall that the set of right cosets partition $G$.

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Of course there is an analogous s result by considering the obvious $K$-action on the set of all right cosets of $H$ in $G$.

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We can now use to find the cardinality of $HgK$. Namely:

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Theorem: Let $G$ be a finite group and $H,K\leqslant G$. Then, for any double coset $HgK\in H\backslash G/K$.

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$\displaystyle \#\left(HgK\right)=|K|\left|H:H\cap gKg^{-1}\right|$

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Proof: We have by the previous theorem that

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$\displaystyle \#\left(HgK\right)=\#\left(\bigsqcup_{xK\in\mathcal{O}_{gK}}xK\right)=\sum_{xK\in\mathcal{O}_{gK}}\#\left(xK\right)=|K|\#\left(\mathcal{O}_{gK}\right)$

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But, by the orbit-stabilizer theorem we have that $\#\left(\mathcal{O}_{gK}\right)=\left|H:\text{stab}\left(gK\right)\right|$. What we claim though is that $\text{stab}\left(gK\right)=H\cap gKg^{-1}$. Indeed, if $h\in H\cap gKg^{-1}$ then $h=gkg^{-1}$ for some $k\in K$ and so $h\cdot gK=gkg^{-1}gK=gkK=gK$ and thus $h\in\text{stab}(gK)$. Conversely, if $h\in\text{stab}(gK)$ then $h\in H$ and $hgK=gK$ so that there exists some $gk\in gK$ so that $hg=gk$ and so $h=gkg^{-1}$ so that $h\in H\cap gKg^{-1}$. Thus, $\text{stab}(gK)=H\cap gKg^{-1}$ as claimed and so the theorem follows. $\blacksquare$

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Applying the exact same technique considering instead the $K$-action on $G/H$ we see that

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Theorem: Let $G$ be a finite group and $H,K\leqslant G$. Then, for any double coset $HgK\in H\backslash G/K$ one has

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$\displaystyle \#\left(HgK\right)=|H|\left|K:K\cap g^{-1}Hg\right|$

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References:

1. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print

May 2, 2011 -