Abstract Nonsense

Crushing one theorem at a time

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

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May 2, 2011 - Posted by | Algebra, Group Theory | , , , , , , ,

3 Comments »

  1. […] Let and let be a transveral of the set of double cosets  then for any one has […]

    Pingback by Composition of the Restriction Map and the Induction Map « Abstract Nonsense | May 4, 2011 | Reply

  2. […] Reciprocity and the our knowledge of we know that, picking some transversal for the set of double cosets  we see […]

    Pingback by Mackey Irreducibility Criterion « Abstract Nonsense | May 6, 2011 | Reply

  3. […] It is a commonly used theorem in finite group theory that if is a finite group and such that is the smallest prime dividing then . We have already seen a proof of this fact by considering the homomorphism which is the induced map from acting on by left multiplication, and proving that . We now give an even shorter (and the just mentioned proof is already short) proof of this fact using double cosets. […]

    Pingback by A Clever Proof of a Common Fact « Abstract Nonsense | September 16, 2011 | Reply


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