Facts about double cosets

Let G be a group and H,K \leq G subgroups. For each x \in G, define the (H,K) double coset of x by HxK = \{ hxk \ |\ h \in H, k \in K \}.

  1. Prove that HxK is the union of the left cosets x_iK where x_iK \in H \cdot xK.
  2. Prove (as above) that HxK is a union of right cosets of H.
  3. Show that the set of (H,K) double cosets partitions G.
  4. Prove that |HxK| = |K| \cdot [H : H \cap xKx^{-1}].
  5. Prove that |HxK| = |H| \cdot [K : K \cap x^{-1}Hx].

  1. (\subseteq) Let hxk \in HxK. Now hxK = h \cdot xK \in H \cdot xK and hxk \in hxK. Thus hxk \in \bigcup_{yK \in H \cdot xK} yK.

    (\supseteq) Let g \in \bigcup_{yK \in H \cdot xK} yK. Then g \in yK for some yK \in H \cdot xK, so that yK = h \cdot xK = hxK for some h \in H. Then g = hxk for some k \in K. So g \in HxK.

  2. (\subseteq) Let hxk \in HxK. Now Hxk = Hx \cdot k \in Hx \cdot K and hxk \in Hxk. Thus hxk \in \bigcup_{Hy \in Hx \cdot K} Hy.

    (\supseteq) Let g \in \bigcup_{Hy \in Hx \cdot K} Hy. Then g \in Hy for some Hy \in Hx \cdot K, so that Hy = Hx \cdot k = Hxk for some k \in K. Then g = hxk for some h \in H. So g \in HxK.

  3. Note that every element is in some double coset- in particular, x \in HxK for all x \in G. So G = \bigcup_{x \in G} HxK.

    Note that if y \in HxK, then HyK \subseteq HxK.

    Now suppose x,y \in G such that HxK \cap HyK \neq \emptyset. Then we have h_1xk_1 = h_2yk_2 for some h_i \in H and k_i \in K. Then x = h_1^{-1}h_2yk_2k_1^{-1} \in HyK, so that HxK \subseteq HyK. Similarly HyK \subseteq HxK. Thus two double cosets are either disjoint or equal. Thus the (H,K) double cosets form a partition of G.

  4. First we prove a lemma.

    Lemma 1: \mathsf{stab}_H(xK) = H \cap xKx^{-1}. Proof: (\subseteq) Suppose h \in \mathsf{stab}_H(xK). Then hxK = h \cdot xK = xK, and we have x^{-1}hx \in K. So h \in xKx^{-1}, hence h \in H \cap xKx^{-1}. (\supseteq) Suppose h \in H \cap xKx^{-1}. Then x^{-1}hx \in K, so that h \cdot xK = hxK = xK, thus h \in \mathsf{stab}_H(xK). \square

    We saw in part 1 above that HxK = \bigcup_{yK \in H \cdot xK} yK; moreover, this union is disjoint because the yK are distinct left cosets of K, each of order |K|. Thus |HxK| = |K| \cdot |H \cdot xK| = |K| \cdot [H : \mathsf{stab}_H(xK)] = |K| \cdot [H : H \cap xKx^{-1}], using Lemma 1.

  5. First we prove a lemma.

    Lemma 2: \mathsf{stab}_K(Hx) = K \cap x^{-1}Hx. Proof: (\subseteq) Suppose k \in \mathsf{stab}_K(Hx). Then Hxk = Hx \cdot k = Hx, and we have xkx^{-1} \in H. So k \in x^{-1}Hx, hence k \in K \cap x^{-1}Hx. (\supseteq) Suppose k \in K \cap x^{-1}Hx. Then xkx^{-1} \in H, so that Hx \cdot k = Hxk = Hx, thus k \in \mathsf{stab}_K(Hx). \square

    We saw in part 2 above that HxK = \bigcup_{Hy \in Hx \cdot K} Hy; moreover, this union is disjoint. Thus |HxK| = |H| \cdot |Hx \cdot K| = |H| \cdot [K : \mathsf{stab}_K(Hx)] = |K| \cdot [K : K \cap x^{-1}Hx].

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  • Gobi Ree  On December 5, 2011 at 3:17 am

    The notations of 1,2 are very delicate but a little strict. A (tough) simplication of 1: HxK=\bigcup_{h \in H} hxK= \bigcup_{i=1}^n x_i K

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