Coset of Subgroup of Subgroup

Theorem
Let $G$ be a group.

Let $H, K \le G$ be subgroups of $G$.

Let $K \subseteq H$.

Let $x \in G$.

Then either:
 * $x K \subseteq H$

or:
 * $x K \cap H = \O$

where $x K$ denotes the left coset of $K$ by $x$.

Proof
Suppose $x K \cap H \ne \O$.

Then:

We have that:
 * $y \in H$

and:
 * $K \subseteq H$

As $H$ is a group, it is closed.

Hence:
 * $\forall x \in K: y x \in H$

which means, by definition of subset product:
 * $y K \subseteq H$

Hence the result.