Today we'll be discussing how to prove that something is or isn't a group. In order to prove anything, we need to start off by undertanding what it is that defines that thing.

A group is defined by a set, $G$, along with a binary operation, $\bullet$. We generally write this tuple as $\langle G, \bullet \rangle$. The set can be any well-defined set and it can be finite or infinite. Additionally, the binary operation can be anything, it need not be limited to common addition or multiplication.

Finally, the set coupled with the binary operation must satisfy four axioms:

*Closure*. For any $x \in G$ and $y \in G$, $x \bullet y \in G$. This simply means that if we are to take any two elements from $G$ and apply the operation, the result must also be an element of $G$.*Associativity*. For any $x, y, z \in G$, $(xy)z = x(yz)$. In order for something to satisfy this axiom, we should be able to move around the parentheses and still get the same answer.*Identity*. For all $x \in G$, there must exist some $e$ such that $x \bullet e = e \bullet x = x$.*Inverse*. For all $x \in G$, there must exist some $x^{-1}$, such that $x \bullet x^{-1} = x^{-1} \bullet x = e$.

Suppose we have a set $G = \mathbb{R} \backslash \{ -1 \}$ with binary operation $x\bullet y := x + y + x\cdot y$. In order to determine if $\langle G , \bullet \rangle$ is a group, we need to determine if it satisfies all four of the axioms or not.

Starting with closure, we need to know if given any $x, y \in G$, is $x \bullet y$ also an element of $G$. In this case, we know that $-1$ is not an element of $G$, so we need to know that $x \bullet y$ will never result in a value of $-1$. How is this done? Well, we can see our equation equal to $-1$ and then solve for $x$ and $y$, seeing what we get.

$\begin{aligned} -1 &\neq x + y + xy \\ &\neq x + y(1 + x) \\ \end{aligned}$ © Daniel Marvin. Last modified: December 03, 2020. Website built with Franklin.jl and the Julia programming language.