Section 4.1, problem 10.
This was a really weird problem and I have no idea if I did anything correctly.
Let $H$ and $K$ be subgroups of the group $G$. For each $x \in G$ define the $HK$ double coset of $x$ in $G$ to be the set $$HxK = \{hxk \, | \, h \in H, k \in K\}.$$ (a) Prove that $HxK$ is the union of the left cosets $x_1K, \ldots x_nK$, where $\{x_1K, \ldots x_nK\}$ is the orbit containing $xK$ of $H$ acting by left multiplication on the set of left cosets of $K$.
(b) Prove that $HxK$ is a union of right cosets of $H$.
(c) Show that $HxK$ and $HyK$ are either the same set or are disjoint for all $x, y \in G$. Show that the set of $HK$ double cosets partitions $G$.
(d) Prove that $|HxK| = |K| \cdot |H:H \cap xKx^{-1}|$.
Attempt:
(a) Let $g \in HxK$ and write $g = hxk$, for some $h \in H$ and $k \in K$. Then $g \in (hx)K = h(xK)$, which is some element $x_iK$ in the orbit of $xK$ under the action of $H$ on the set of left cosets of $K$ in $G$. Hence $g \in \displaystyle \bigcup_i x_iK$ and $HxK \subset \displaystyle \bigcup_i x_iK$.
Conversely, let $g \in \displaystyle \bigcup_i x_iK$. Then for some $i, g \in x_iK$, i.e. $g = x_ik$, for some $k \in K$. Now $x_iK$ is in the orbit of $xK$ under the action of $H$ on the set of left cosets of $K$ in $G$, so there exists $h \in H$ such that $h(xK) = x_iK$, so $g \in h(xK) = (hx)K$, i.e there exists $k \in K$ such that $g = hxk$, i.e. $g \in HxK$. Hence $\displaystyle \bigcup_i x_iK \subset HxK$.
(b) Suppose $g \in \displaystyle \bigcup_i Hx_i$. Then for some $i$, $g \in Hx_i$, which is in the orbit of $Hx$ under the (right) action of $K$ on the right cosets of $H$ in $G$, so there is a $k \in K$ such that $Hx_i = xk$. Since $g \in Hx_i$ there exists an $h \in H$ such that $g = hxk$, i.e. $g \in HxK$.
Conversely, let $g \in HxK$ and write $g = hxk$ for some $h \in H, k \in K$. Then $g \in (Hx)k = H(xk)$ so $g$ is in a right coset of $H$.
(c) We show that the relation $\sim$ on $G$ defined by $$x \sim y \quad \mathrm{if \, there \, exist} \quad h \in H \mathrm{ \, and \, } k \in K \mathrm{\, such \, that \, } hxk = y$$ is an equivalence relation.
Note that $x \sim x$ because $exe = x$ ($e \in H, K$ because $H, K$ are subgroups). If $x \sim y$, then $hxk = y$ for some $h \in H$, $k \in K$, and then $h^{-1}yk^{-1} = x$, so $y \sim x$. If $x \sim y$ and $y \sim z$, then there exist $h, h_1 \in H$ and $k, k_1 \in K$ such that $hxk = y$ and $h_1yk_1 = z$. Then $hxk = y = h_1^{-1}zk_1^{-1}$ and $h_1hxkk_1 = z$, so $x \sim z$ and $\sim$ is an equivalence relation. Now the equivalence classes (which are precisely the set defined as $HxK$) partition $G$ so the set $HK$ of double cosets partitions $G$. In particular, $HxK$ and $HyK$ are either the same double coset or they are disjoint.
(d) Recall that for two subgroups $H, K$ of $G$, $|HK| = \frac{|H| \cdot |K| }{|H \cap K|}$. Since conjugation is an isomorphism, $|K| = |xKx^{-1}|$, so $|H \cap xKx^{-1}| = |H \cap K|$. Hence $|HxK| = \frac{|K| \cdot |H|}{|H \cap K|} = |K| \cdot \frac{|H|}{|H \cap K|} = |K| \cdot \frac{|H|}{|H \cap xKx^{-1}|} = |K| \cdot |H:H \cap xKx^{-1}|$.