how to prove this group of the binary operation

Question

Let $Y=\{(a,b)∈ \Bbb R\times \Bbb R∣ a≠0\}$. 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 do the proof of ∗ is an operation on Y. And proof is associative like this: $$\{(A,B)*(C,D)\}*(E,F)=(A,B)*\{(C,D)\}*(E,F)\}$$

$$(AC,AD+B)(E,F)=(A,B)(CE,CF+D)$$

$$(ACE,ACF+AD+B)=(ACE,ACF+AD+B)$$

but I get stock trying the identity and the inverse proof. dont know how to start

Answer 1

Hint

Prove the associativity of $*$, find the neutral element and prove that each element has an inverse.

Answer 2

There are four things you need to accomplish, here:

• You need to show that $*$ is an operation on $Y.$ In particular, you must show that if $(a,b),(c,d)\in Y,$ then $(a,b)*(c,d)\in Y.$ This should be straightforward from the definition of $Y.$
• You need to show that $*$ is an associative operation on $Y.$ That is, you need to show that if $(a,b),(c,d),(e,f)\in Y,$ then $$\bigl[(a,b)*(c,d)\bigr]*(e,f)=(a,b)*\bigl[(c,d)*(e,f)\bigr].$$
• You need to show that $Y$ has a $*$-identity. That is, you must find some $(i,j)\in Y$ such that no matter what $(a,b)\in Y$ we choose, we will always have $$(a,b)*(i,j)=(a,b)=(i,j)*(a,b).$$
• You need to show that every element of $Y$ has a $*$-inverse. That is, you must show that for every $(a,b)\in Y,$ there is some $(c,d)\in Y$ such that $$(a,b)*(c,d)=(i,j)=(c,d)*(a,b),$$ where $(i,j)$ is the $*$-identity that you should already have found.

Which of these have you accomplished so far? What have you managed to show for the others?