Proving $G$ is a group under a specific operation Let $G = \mathbb{R} \setminus \{-1\}$ and define the binary operation on $G$ by $a*b=a+b+ab$.
Prove $G$ is a group under this operation.
So to prove $G$ is a group, I know we have to show it is associative, has an identity element, and contains inverses for all elements. 
How do I specifically show those 3 properties though?
 A: To show something is a group you need:
$1.$ Well defined binary operation on a nonempty set.
$2.$ The group operation to be associative.
$3.$ There to be an identity under the operation.
$4$. The set to be closed under the operation.
$5.$ Closure under inverses.
(This might be overkill but early on it is good to justify them all). Alright, here we go...
$1.$ The operation is binary (if you really want you can check well defined). $\mathbb{R}$ is clearly nonempty so our set should be nonempty.
$2$. Is the operation $*$ associative? You need $(a*b)*c=a*(b*c)$.

 This operation is associative. Notice that $$ \begin{align} (a*b)*c&=(a+b+ab)*c\\&=(a+b+ab)+c+(a+b+ab)c\\&=a+b+ab+c+ac+ab+abc\\ &= a+(b+c+bc)+a(b+c+bc)\\&=a*(b+c+bc)\\&=a*(b*c)\end{align}$$

$3.$ Is there an identity? That is, is there an $e$ so that $a*e=a$ for all $a$?

 Let $b=0$, then $a*0=a+0+a\cdot 0=a+0+0=a$ for any $a \in \mathbb{R}$.

$4.$ Check that $a*b \in \mathbb{R}$ and not $-1$

 $a*b=a+b+ab$, but if $a,b \in \mathbb{R}$, then $a+b+ab \in \mathbb{R}$. Done! Now if $a,b \neq -1$, can $a*b=a+b+ab=-1$? Solving for $a$ would yield $a=-1$ and similarly for $b$, but this is impossible as $a,b \neq -1$, so $a*b=a+b+ab\neq -1$ so long as $a,b \neq -1$. Therefore, the set is closed under $*$.

$5.$ That is, if $a \in \mathbb{R}$, we need to find an element $i$ so that $a*i=e$, where $e$ is the identity of the group we found before.

 The identity is $0$. We want $i$ such that $a*i=0$. Well, $a*i=a+i+ai=0$. Solving for $i$ gives, $i=-\frac{a}{a+1}$, so if $a\neq -1$, we can always find an inverse. Notice if we define $i$ as before then $$a*i=a+\frac{-a}{a+1}+a\frac{-a}{a+1}=\frac{(a^2+a)-a-a^2}{a+1}=0$$.

But then we've shown this set under $*$ is a group! Moreover, we can show this group is commutative:
$$
\begin{align}
a*b&=a+b+ab \\
&=b+a+ba \\
&=b*a
\end{align}
$$
because $a,b \in \mathbb{R}$ and addition and multiplication are commutative in $\mathbb{R}$.
A: Hint $a+b+ab=a+a(b+1)=(b+1)(a+1)-1$. We're effectively translating usual multiplication. Does this give you any idea?
A: HINT
For associativity:  $\forall a,b,c\in G\ \ a*(b*c)=(a*b)*c$.
For identity:  show that $\exists e\in G $ such that $\forall a\in G\ e*a=a*e=a$
For inverse:  show that $\forall a\in G,\ \exists a^{-1}\in G$ such that $a*a^{-1}=a^{-1}*a=e$
A: You also need to show that you have a well-defined operation on $G$, that is if $a$ and $b$ belong to $\mathbb{R} \setminus \{ -1 \}$ then so does $a*b$.  Equivalently, you can prove that if $a*b = -1$, then either $a = -1$ or $b = -1$.
To show the other three properties, you have to use the definitions.  Here is a start with proving associativity.  You need to show that $(a * b) * c = a * (b * c)$ for all $a, b, c \in G$.  Since $a * b = a + b + ab$, it follows that $$(a * b) * c = (a * b) + c * (a * b)c = (a + b + ab) + c + (a + b + ab)c = a + b + c + ab + ac + bc + abc.$$  Work out $a * (b * c)$ similarly and check that you get the same result.
