The groups to be determined are
$$(\mathbb{R},*) \; \mathbb{defined \; by \;} a*b=a+b+ab\;\;\;\;\;\forall \;a,b\in\mathbb{R}$$
and
$$\mathbb{Z\times Z\;defined \;by} \; (a,b)*(c,d)=(ad+bc,bd) \;\;\;\;\;\forall\;a,b,c,d\in\mathbb{Z}$$
I've made an attempt at the both, but am not sure if I am correct or even have sound arguments.
Denote our group to $G$. Here's an attempt
If $a,b \in G=(\Bbb R,*)$, then it follows that $a+b \in G$ and $ab \in G$. Thus, $a + b + ab \;\mathbb{must\;} \in G$ so we have closure. Now, suppose $c \in G$. Then,
$$a*b*c=a+(b+c+abc) = (a +b+c)+abc$$
and so $G$ satisfies the associative property. Now, we need to find an identity. Let's denote our potential identity to be $e$ $\in G$. Then, for $a\in G$
$$a=a*e=a+e+ae=a+0+a(0)=a$$
since under addition, we have that $e=0$. Lastly, we need to see if there exists an inverse for $G$. We have to satisfy $a*f=e$, so we let f be our potential inverse. Hence,
$a*f=a+f+af=a+a^{-1}+aa^{-1} \neq e = 0$.
Under addition, $a^{-1}=-a$, but $a(-a)=-a^2$. This does not give us 0, unless $a=0$. Therefore, G is not a group.
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Now, if $a,b\in G=(\Bbb Z\times \Bbb Z,*)$, then it follows that $(a,b)*(c,d)=(ad+bc,bd) \;\mathbb{must}\; \in G$ so we have closure. Now, suppose $(q,r)\in G$. Then
$$(a,b)*(c,d)*(q,r)=(ad+bc,bd)+(cr+dq,dr)$$
It is impossible to satisfy the associative property with our $*$ is defined, so $G$ is not a group.
So I'm sure there are many mistakes, as I began to get very confused with this problem. Could anyone guide me along the correct road?
Best Answer
Closures are trivial. Let's list what we have to prove:
i) Associativity;
ii) Existence of identity;
iii) Existence of inverses.
Now, let's see the other one.