Generalized Abstraction

Proving the Axioms of a Group

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, GG, along with a binary operation, \bullet. We generally write this tuple as G,\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:

  1. Closure. For any xGx \in G and yGy \in G, xyGx \bullet y \in G. This simply means that if we are to take any two elements from GG and apply the operation, the result must also be an element of GG.

  2. Associativity. For any x,y,zGx, y, z \in G, (xy)z=x(yz)(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.

  3. Identity. For all xGx \in G, there must exist some ee such that xe=ex=xx \bullet e = e \bullet x = x.

  4. Inverse. For all xGx \in G, there must exist some x1x^{-1}, such that xx1=x1x=ex \bullet x^{-1} = x^{-1} \bullet x = e.

An Example

Suppose we have a set G=R\{1}G = \mathbb{R} \backslash \{ -1 \} with binary operation xy:=x+y+xyx\bullet y := x + y + x\cdot y. In order to determine if G,\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,yGx, y \in G, is xyx \bullet y also an element of GG. In this case, we know that 1-1 is not an element of GG, so we need to know that xyx \bullet y will never result in a value of 1-1. How is this done? Well, we can see our equation equal to 1-1 and then solve for xx and yy, seeing what we get.

1x+y+xyx+y(1+x) \begin{aligned} -1 &\neq x + y + xy \\ &\neq x + y(1 + x) \\ \end{aligned}