Is this binary operation a group?

Question

Let $Y=\left\{(a,b)\in\mathbb{R}\times\mathbb{R}\ |\ a\ne 0\right\}$. Given $(a,b),(c,d)\in Y$, define $(a,b)∗(c,d)=(ac,ad+b)$. Prove that $Y$ is a group with the operation $*$.

I already did the proof that $*$ is an operation on $Y$. To prove that $Y$ is a group with the operation $*$ I do the follow:

Associative proof: \begin{align} \left((A,B)*(C,D)\right)*(E,F) &= (A,B)*\left((C,D)*(E,F)\right) \\ (AC,AD+B)*(E,F) &= (A,B)*(CE,CF+D) \\ (ACE,ACF+AD+B) &= (ACE,ACF+AD+B) \end{align}

Y has an identity proof: Suppose the inverse is $(I,J)$. We should have $(A,B)*(I,J)=(A,B)$ and $(I,J)*(A,B)=(A,B)$.

If we have $AI=A$, then $I=1$, and if $AJ+B=B$, then $J=0$ so we get that $(1,0)$ is the identity element. Therefore, $(1,0)*(A,B)=(1A,1B+0)=(A,B)$.

Inverse proof: We should find $(X,Y)$ so that $(X,Y)*(A,B)=(1,0)=(A,B)*(X,Y)$.

We have that $XA=1$ or $X=1/A$ and $XB+Y=0$ or $Y=-XB=-B/A$.

So inverse is $(1/A,-B/A)$. Since $A$ is not zero, we have the inverse as an element of \mathbb{R}\times\mathbb{R}. $(A,B)(1/A,-B/A)=(A*1/A,((A*-B)/A)+B)=(1,0).$

Hence $*$ as a binary operation is a group. We find that $(a,b)*(c,d) = (ac,ad+b)$.

Is all this correct?

Answer 1

This basically seems OK, but there are a few things that need improving:

1. For item 1, you need to start with one side and compute that it is equal to the other side. The way you have phrased the argument, you are assuming the result and showing that you get an identity. This is not a valid proof. For example, you could write: \begin{align} \left((a,b)*(c,d)\right)*(e,f) &= (ac, ad+b)*(e,f) = (ace, acf+ad+b) \\ (a,b)*\left((c,d)*(e,f)\right) &= (a,b)*(ce, cf+d) = (ace, a(cf+d)+b)) = (ace, acf+ad+b). \end{align} Since the two computations give the same result, $*$ is associative.
2. For item 2, you have shown that a left identity must be $(1,0)$. You must also show that this is a right identity.
3. Similarly, you have found a right inverse; you should also show that this is a left inverse.

Answer 2

An alternative approach is to note that, if we identify $(a,b)$ as the $2\times 2$ matrix $\begin{pmatrix} a & b \\ 0 & 1 \end{pmatrix}$, then the operation $*$ corresponds to matrix multiplication. Because $\begin{pmatrix} a & b \\ 0 & 1 \end{pmatrix}^{-1} = \begin{pmatrix} a^{-1} & -a^{-1} b \\ 0 & 1 \end{pmatrix}$, the collection of matrices of this form is a subgroup of $GL(2)$.