Prove $G$ is a group under $\ast$ given as $(a, b)*(c, d) = (ac,bc+d)$ Let $G = \{(a, b) \text{ }|\text{ } a, b \in \mathbb{R}, a \neq 0 \}$. Let $*$ be an operation on $G$ defined by $$(a, b)*(c, d) = (ac,bc+d)$$ Prove $G$ is a group.
Is this a group or not? because the commutative property does not hold: $$(a,b)*(c,d) = (ac,bc+d) \quad \text{  but }\quad (c,d)*(a,b) = (ca,da+b)$$
 A: A group is a pair $(G, \ast)$, where $G$ is a set and $\ast$ is a binary operation on $G$, satisfying three properties:


*

*$\ast$ is associative (i.e. $(g \ast h)\ast k = g\ast(h\ast k)$ for all $g, h, k \in G$),

*there is $e \in G$ such that $e\ast g = g\ast e = g$ for all $g \in G$,

*for every $g \in G$, there is $g^{-1} \in G$ such that $g\ast g^{-1} = g^{-1}\ast g = e$.


If in addition, $\ast$ is commutative (i.e. $g\ast h = h\ast g$ for every $g, h \in G$), then $(G, \ast)$ is called an abelian group.
What you have deduced is that $(G, \ast)$ is not an abelian group, but it may still be a group (you need to check the three properties above to see if it is).
A: You need to check closure: make sure you maintain a isn't zero! Check that for all sets of valid numbers you can apply the given multiplication to get another valid set.
Identity: see above comments.
Inverses: once you have found the identity as discussed above try to find an element for each g in G that will multiply with g to give your identity. Confirm this is in G and confirm it's a two sided inverse.
If you need help with any step then comment.
A: It is worth mentioning that this operation has a name and can be generalized. The operation is called the semi-direct product and the group you have formed is $(\mathbb{R}^*,\times)\ltimes (\mathbb{R},+)$. Here, by $(\mathbb{R}^*,\times)$ I mean the group of nonzero real numbers with multiplication as the operation and by $(\mathbb{R},+)$ I mean the real numbers with addition as the operation. 
This construction arises naturally when looking at extensions of groups. Let $G$ be a group (not necessarily abelian) and call an extension of $G$ by $N$ the short exact sequence of groups $$0\longrightarrow N \longrightarrow E \longrightarrow G \longrightarrow 1 .$$
Classifying these extensions amounts to finding ways of constructing an $E$ with $N$ as a normal subgroup and $E/N\cong G$. A solution to this problem of finding an $E$ is $N \rtimes G$ where the symbol indicates that $G$ acts on $N$. If this interests you, I recommend Kenneth Brown's Group Cohomology. It is accessible after some courses in abstract algebra and perhaps a course in algebraic topology.
