**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 4.1 Exercise 4.1.10**

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

Solution:

(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 $$|HxK| = |H| \cdot |Hx \cdot K| = |H| \cdot [K : \mathsf{stab}_K(Hx)] = |K| \cdot [K : K \cap x^{-1}Hx].$$