The Group R\{1} with the Quasi-Product

Let:

(1)

\begin{align} \quad G = \mathbb{R} \setminus \{ 1 \} \end{align}

Define a binary operation $\cdot$ on $G$ for all $g_1, g_2 \in G$ by:

(2)

\begin{align} \quad g_1 \cdot g_2 = g_1 + g_2 - g_1g_2 \end{align}

Where the operations on the righthand side of the equation are standard addition and multiplication. The operation $\cdot$ defined above is called the Quasi-Product of $a$ and $b$. We will now see that $G$ becomes a group with the operation $\cdot$.

First, let $g_1, g_2 \in G$ and suppose that $g_1 \cdot g_2 = 1$. Then:

(3)

\begin{align} \quad g_1 + g_2 - g_1g_2 = 1 \\ \end{align}

Rearranging the above equation and factoring gives us that:

(4)

\begin{align} \quad g_1 -g-1g_2 &= 1 - g_2 \\ \quad g_1(1 - g_2) &= 1 - g_2 \end{align}

Since $g_2 \in G$ we have that $g_2 \neq 1$ and so $1 - g_2 \neq 0$. Dividing both sides by $1 - g_2$ gives us that $g_1 = 1$, a contradiction, since $g_1 \in G$.

So for all $g_1, g_2 \in G$ we have that $g_1 \cdot g_2 \neq 1$, i.e., if $g_1, g_2 \in G$ then $g_1 \cdot g_2 \in G$. So $G$ is closed under the operation $\cdot$.

Now we check that this operation is associative. For all $g_1, g_2, g_3 \in G$ we have that:

(5)

\begin{align} \quad g_1 \cdot (g_2 \cdot g_3) = g_1 \cdot (g_2 + g_3 - g_2g_3) = g_1 + (g_2 + g_3 - g_2g_3) - g_1(g_2 + g_3 - g_2g_3) \end{align}

(6)

\begin{align} \quad (g_1 \cdot g_2) \cdot g_3 = (g_1 + g_2 - g_1g_2) \cdot g_3 = (g_1 + g_2 - g_1g_2) + g_3 - (g_1 + g_2 - g_1g_2)g_3 \end{align}

Comparing the two equalities above, we see that $g_1 \cdot (g_2 \cdot g_3) = (g_1 \cdot g_2) \cdot g_3$. So the operation of quasi product is associative.

We now show then existence of an identity for $G$. First suppose $e$ is such that $g \cdot e = g$ for all $g \in G$. Then:

(7)

\begin{align} \quad g + e - ge = g \quad \forall g \in G \end{align}

So $e - ge = 0$, i.e., $(1 - g)e = 0$ for all $g \in G$. Since $g \neq 1$ for all $g \in G$, we see that $1 - g \neq 0$ and so $e = 0$.

On the other hand if $e \cdot g = g$ for all $g \in G$ then:

(8)

\begin{align} \quad e + g - eg = g \quad \forall g \in G \end{align}

So $e - eg = 0$, i.e., $e(1 - g) = 0$ for all $g \in G$. Again, since $g \neq 1$ for all $g \in G$, we see that $1 - g \neq 0$ and so $e = 0$.

Thus the only candidate for an identity is $e = 0$. But $e = 0 \in G$. So indeed, $G$ has an identity element $e = 0$ and $g \cdot e = g = e \cdot g$ for all $g \in G$.

Lastly, we show that each $g \in G$ has an inverse in $G$. Let $g \in G$. Then we want to find $h \in G$ such that $g \cdot h = e = h \cdot g$. We must have that:

(9)

\begin{align} \quad g \cdot h &= e \\ \quad g + h - gh &= 0 \\ \quad h(1 - g) &= -g \\ \quad h &= \frac{-g}{1 - g} \end{align}

And also:

(10)

\begin{align} \quad h \cdot g &= e \\ \quad h + g - hg &= 0 \\ \quad h(1 - g) = -g \\ \quad h = \frac{-g}{1 - g} \end{align}

So if $g \in G$ then a candidate for $g^{-1} = h$ is $g^{-1} = \frac{-g}{1 - g}$. Since $g \in G$ we see that $1 - g \neq 0$, so $g^{-1} \in \mathbb{R}$. Also note that $g^{-1} \neq 1$ since $-g \neq 1 - g$. So $g^{-1} \in G$.

So indeed, $G$ with the quasi-product is a group.