Problem source

Let $a$ be a non-zero real number and define a binary operation
on the set of real numbers by
$$
x*y = x+y+axy \,.
$$
Show that the operation $*$ is associative.
Show that $(G,*)$ is a group, where
$G$ is the set of all real numbers except for one number which you should 
identify.
Find a subgroup of $(G,*)$ which has exactly 2 elements.

Solution source

Claim: $*$ is associative.

Proof:  Then $x*(y*z) = x*(y+z+ayz) = x + (y+z+ayz) + ax(y+z+ayz) = x + y + z + a(yz + xy + zx) + a^2xyz$ and $(x*y)*z = (x+y+axy)*z = (x+y+axy) + z+ a(x+y+axy)z = x + y + z + a(yz + xy + zx) + a^2xyz$ so $x*(y*z) = (x*y)*z$ and we are done.

Let $G = \mathbb{R} \setminus \{-\frac1{a} \}$
In order to show that $(G, *)$ is a group we need to check:

\begin{enumerate}
\item \textbf{closure} $x*y = x + y + axy$ is a real number
\item \textbf{associativity} we have already checked
\item \textbf{identity} $x*0 = x + 0 + 0 = x = 0*x$, so $0$ is an identity
\item \textbf{inverses} $x*\left ( \frac{-x}{1+ax} \right ) = x -  \frac{x}{1+ax} + ax \frac{-x}{1+ax} = \frac{x +ax^2 - x - ax^2}{1+ax} = 0$ so $x$ has an identity, assuming $1+ax \neq 0$ which is true for everything in our set 
\end{enumerate}

Consider the set $\{0, \frac{-2}{a} \}$. Then $\frac{-2}{a} * \frac{-2}{a} = \frac{-4}{a} + a \frac{4}{a^2} = 0$, so this is a group of order $2$