basic question about Group structure (answering a small exercise..) The operation * defines a binary operation in $\mathbb R\times \mathbb R$ by $(X_1,Y_1)*(X_2,Y_2) = (X_1X_2, Y_1X_2+Y_2)$


*

*defines a group structure (i found out..), but shouldn't we exclude the elements of the form $(0,Y)$ from $\mathbb R\times \mathbb R$ to retain a group structure ? , because after finding out the identity element is $(1,0)$ : $(0,Y)$ cannot have a inverse , because $(0,Y)*(a,b)$ = $(0,Y.a+b)$ which cannot be $(1,0)$ . that is for any element $Y$ of $\mathbb R$ there is no inverse of $(0,Y)$. 

 A: You are correct that $(1, 0)$ is the identity of that binary operation and elements of the form $(0, x)$ do not have an inverse.  This means that $\mathbb R \times \mathbb R$ is not a group under this operation.
If you exclude elements of the form $(0, x)$ then the remaining elements, $\mathbb R^\ast \times \mathbb R$, do indeed form a group.  In fact this group is the semidirect product $\mathbb R^\ast \ltimes \mathbb R$ you get using the homomorphism $\mathbb R^\ast \to \mathrm{Aut} \ \mathbb R$ defined by $a \mapsto [x \mapsto ax]$.
A: I believe the general argument one might want to apply here is the following.
First note that the operation is associative.
In fact
$$
((X_1,Y_1)*(X_2,Y_2))*(X_3,Y_3) = (X_1 X_2, Y_1 X_2+Y_2)*(X_3,Y_3)
=
(X_1 X_2 X_3, (Y_1 X_2+Y_2) X_3 + Y_3)
=
(X_1 X_2 X_3, Y_1 X_2 X_3+Y_2 X_3 + Y_3))
$$
and
$$
(X_1,Y_1)*((X_2,Y_2)*(X_3,Y_3)) = (X_1,Y_1)*(X_2 X_3, Y_2 X_3+Y_3)
=
(X_1 X_2 X_3, Y_1 X_2 X_3 + Y_2 X_3+Y_3).
$$
Then note that $(1,0)$ is the identity. 
So you have a monoid. If you take only the invertible elements, they will form a group, by a standard result. And in fact there is $(X_2, Y_2)$ such that
$$
(X_1,Y_1)*(X_2,Y_2) = (X_1 X_2, Y_1 X_2+Y_2) = (1, 0)
$$
iff $X_1 \ne 0$. We have then $X_2 = X_1^{-1}$ and $Y_{2} = - X_{1}^{-1} Y_{1}$, and for these values we also have  $
(X_2,Y_2) * (X_1,Y_1) = (1, 0).
$
