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

Yes, double cosets partition G.

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.