Define a binary operation on {e,(12)}×{e,(123),(132)} so that it becomes isomorphic to $S_3$ [closed]

Since {e,(12)} isomorphic to $Z_2$ and
{e,(123),(132)} isomorphic to $Z_3$ and gcd(2,3)=1, so, {e,(12)}×{e,(123),(132)} isomorphic to $Z_6$ which is not isomorphic to $S_3$.
But can a binary operation be defined for the direct product {e,(12)}×{e,(123),(132)} so that it is isomorphic to $S_3$ ?

As markvs has noted in the comments, every set with 6 elements has an operation that gives it a group structure isomorphic to $S_{3}$.

Here's how it works: let $X$ be a set with $6$ elements. Since $S_{3}$ also has 6 elements, there exists a bijection $f \colon X \to S_{3}$. Now, we can define a binary operation $\circ \colon X \times X \to X$ on $X$ by using the bijection $f$.

Informally, given two elements $x, y$ of $X$, we define their product in the following way: first, use the bijection $f$ to find the corresponding elements $f(x), f(y) \in S_{3}$. Then, find the product of $f(x)$ and $f(y)$ in $S_{3},$ where we already have a defined group operation. Finally, use the inverse bijection $f^{-1}$ to find the element of $X$ that corresponds to the product in $S_{3}$; this will be the product of $x$ and $y$.

So, this gives us the following definition: For each $(x, y) \in X \times X,$ let $$x \circ y = f^{-1}(f(x) \cdot f(y)),$$ where $\cdot$ denotes the group operation of $S_{3}$. You should check that $\circ$ actually makes $X$ a group, by checking the following:

1. $\circ$ is well-defined.
2. $\circ$ is associative.
3. $(X, \circ)$ has an identity element $e$.
4. Each element $x \in X$ has an inverse $x^{-1}$ such that $x \circ x^{-1} = e = x^{-1} \circ x.$

All of these can be verified by using the fact that $f, f^{-1}$ are both bijections, and that $S_{3}$ already has a group structure.

Now, we claim that $(X, \circ)$ and $(S_{3}, \cdot)$ are isomorphic as groups. We basically get this for free: we already have a bijection $f \colon X \to S_{3},$ and note that the definition of $\circ$ implies that for all $x, y \in X,$ $$f( x \circ y) = f(x) \cdot f(y).$$ Clearly, $f$ is a homomorphism from $(X, \circ)$ to $(S_{3}, \cdot).$ Since it is also bijective, is is an isomorphism, and we are done.

This process is known as transport of structure, and allows us to give a set $X$ a group structure whenever there exists a bijection between $X$ and another set $G$ which already has a group structure.

Applying this to your question, you can simply take $f \colon \{e,(12)\} \times \{e,(123),(132)\} \to S_{3}$ to be your favorite bijection between the two, and then define the group structure using the above process. Of course, given the names of the elements in $\{e,(12)\} \times \{e,(123),(132)\}$, things might get a little confusing.