Abstract Algebra – Proving G is a Group Under a Specific Operation

abstract-algebragroup-theory

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?

Best Answer

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 $\mathbb{R}$.

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