Question

Let H and K be subgroups of a group G, and let g be an element of G. The set $\math HgK = \{x \in G \mid x = hgk \text {for some} h \in H, k \in K\}$ is called a double coset. Do the double cosets partition G?

Answer 1

Answer:

Yes, double cosets partition G.

Step-by-step explanation:

We are going to define a relation over the elements of G.

Let $x,y\in G$. We say that $x\sim y$ if, and only if, $y\in HxK$, or, equivalently, if $y=hxk$, for some $h\in H, k\in K$.

This defines an equivalence relation over G, that is, this relation is reflexive, symmetric and transitive:

• Reflexivity: ($x\sim x$ for all $x\in G$.) Note that we can write $x=exe$, where $e$ is the identity element, so $e\in H,K$ and then $x\in HxK$. Therefore, $x\sim x$.
• Symmetry: (If $x\sim y$ then $y\sim x$.) If $x\sim y$ then $y=hxk$ for some $h\in H$ and $k\in K$. Multiplying by the inverses of h and k we get that $x=h^{-1}yk^{-1}$ and is known that $h^{-1}\in H$ and $k^{-1}\in K$. This means that $x\in HyK$ or, equivalently, $y\sim x$.

• Transitivity: (If $x\sim y$ and $y\sim z$, then $x\sim z$.) If $x\sim y$ and $y\sim z$, then there exists $h_1,h_2\in H$ and $k_1,k_2\in K$ such that $y=h_1xk_1$ and $z=h_2yk_2$. Then, $\\ z=h_2yk_2=h_2(h_1xk_1)k_2=(h_2h_1)x(k_1k_2)=h_3xk_3$ where $h_3=h_2h_1\in H$ and $k_3=k_1k_2\in K$. Consequently, $z\sim x$.

Now that we prove that the relation "$\sim$" is an equivalence over G, we use the fact that the different equivalence classes partition G. Since the equivalence classes are defined by $[x]=\{y\in G\colon x\sim y\} =\{y\in G \colon y=hgk\ \text{for some } h\in H, k\in K \}=HxK$, then we're done.