Coset Product on Non-Normal Subgroup is not Well-Defined

Theorem
Let $\struct {G, \circ}$ be a group.

Let $H$ be a subgroup of $G$ which is not normal.

Let $a, b \in G$.

Then it is not necessarily the case that the coset product:
 * $\paren {a \circ H} \circ \paren {b \circ H} = \paren {a \circ b} \circ H$

is well-defined.

Proof
Proof by Counterexample:

Let $S_3$ denote the Symmetric Group on 3 Letters, whose Cayley table is given as:

Consider the subgroups of $S_3$:

Let $H = \set {e, \tuple {12} }$.

From Normal Subgroups in Symmetric Group on 3 Letters, $H$ is not normal.

Let $A = \set {\tuple {123}, \tuple {23} }$.

We have:

and so $A$ is the left coset of $H$ by both $\tuple {123}$ and $\tuple {23}$.

Let $B = \set {\tuple {132}, \tuple {13} }$.

and so $B$ is the left coset of $H$ by both $\tuple {132}$ and $\tuple {13}$.

Now consider:

But:

So two different evaluations of the coset product give two different result.

Hence by definition the coset product is not well-defined on $H$.

The result follows.