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 and it can be finite or infinite. Additionally, the binary operation can be anything, it needn't be limited in any way other than it being an operation applied to two elements.

We used a \(\bullet\) symbol for our operation, but that choice was arbitrary and could be anything we wanted to use, e.g. \(\langle G, \# \rangle\) and then apply it like \(x \# y\).

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. We should be clear that we're*not*talking about commutativity here, \(ab\) doesn't necessarily need to be equal to \(ba\).*Identity*. For all \(x \in G\), there must exist an \(e\) such that \(x \bullet e = e \bullet x = x\).*Inverse*. For all \(x \in G\), there must exist an \(x^{-1}\) such that \(x \bullet x^{-1} = x^{-1} \bullet x = e\).

Let \(G \subseteq \mathbb{R} \backslash \{ -1 \} \) where for all \(x, y \in G\), \(x \bullet y := x + y + xy\). We want to know if \(\langle G, \bullet \rangle\) is a valid group.

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 set our equation equal to \(-1\) and then solve for \(x\) and \(y\), seeing what we get.

\[ \begin{aligned} -1 &\neq x + y + xy \\ &\neq x + y(1 + x) \\ \end{aligned}\]The only way that we can get the result of \(-1\) is if \(x\) or \(y\) is \(-1\). Since this can never be the case, closure passes. Just to be completely explicit, suppose that \(x = -1\), then \(-1 = (-1) + y(1 - 1) = -1 + y \cdot 0 = -1 + 0 = -1\). Conversely, suppose that \(y = -1\), then \(-1 = x + (-1)\cdot ( 1 + x) = x - x - 1 = 0 - 1\).

Next, we need to check associativity. We need to make sure that for \(x, y, z \in G\), \((xy)z = x(yz)\).

\[ \begin{aligned} (x \bullet y) \bullet z &= x \bullet (y \bullet z) \\ (x + y + xy) + z + (x + y + xy)z &= x + (y + z + yz) + x(y + z + yz) \\ x + y + xy + z + xz + yz + xyz &= x + y + z + yz + xy + xz +xyz \\ x (1 + y + z + yz) + y + z &= x (1 + y + z + yz) + y + z \\ \end{aligned}\]We can see that associativity passes too. We can move around the parentheses and the result is the same.

The last two axioms which we must verify are that each element has an identity and an inverse.

For every \(x \in G\), there must be an identity element, \(e\), which yields \(x\).

\[\begin{aligned} x + e + x\cdot e &= x + 0 + x\cdot 0 = x \\ \end{aligned}\]Does the identity element need to be \(0\)? What if we chose \(1\) instead?

\[x + 1 + x\cdot 1 = 2x + 1 \neq x \]Therefore, we can see that our identity element is \(0\).

Finally, we need to verify that every element in \(G\) has an inverse which yields the identity element.

For every \(x \in G\), there is some \(y \in G\), such that \(x \bullet y = 0\).

\[\begin{aligned} x + y + xy &= 0 \\ y(1 + x) &= -x \\ y &= -\frac{x}{1+x} \\ \end{aligned}\]We can see that there is an inverse for every element. Just for kicks, let's plug in a couple of values to try it out.

Suppose \(x = 5\), then \(x^{-1} = -\frac{5}{1 + 5} = -\frac{5}{6}\). We should be able to compute \(5 \bullet -\frac{5}{6}\) and yield \(0\).

\[ 5 - \frac{5}{6} + 5\cdot -\frac{5}{6} = 5 - \frac{5}{6} - \frac{25}{6} = 5 - \frac{30}{6} = 5 - 5 = 0\]We could also try \(x = -\frac{3}{8}\) where the inverse is \(x^{-1} = \frac{3/8}{1 - \frac{3}{8}} = \frac{3}{5}\). Plugging these values in yields \(0\) as expected.

\[-\frac{3}{8} + \frac{3}{5} - \frac{3}{8}\cdot\frac{3}{5} = 0\]And there you have it, folks. We've shown that \(\langle G, \bullet \rangle\) is indeed a group by showing that it satisfies all four axioms of a group. We went a little further than necessary, showcasing a few concrete cases for the sake of the readers who may require a bit of equestrial abuse. For real though, there's no need to harm any horses in official proofs.

Alright, if you've actually read through up to this point, then you either really like math, really want to know how to prove if something is or isn't a group, or you have more time on your hands than you may be willing to admit.

In any of these cases, one of the things that people tend to miss out on is why these things matter. If you're not a *math person* or happen to define yourself in this way, you may be wondering why we care if something is a group or not and wondering how this can be applied to that less abstract thing we call "the real world."

Well, that's a pretty big topic and definitely one for another time and another post. Until that time comes, here are two parting questions and partial answers for the tragically curious.

Why did we define the operation one way instead of another way? We could've defined the binary operation as \(x - y + xy\) instead of \(x + y + xy\), or even \(x - y^2\). The operation is basically the thing that creates the underlying structure of the group. There are finite groups called

*cyclic groups*which are generated by a single element. They have a lot of really interesting properties and are used extensively in physics, cryptography, and biology. We are also interested in groups which have elements which commute, called Abelian groups. Commutativity is something we take for granted with normal addition and multiplication, but when we're talking about matrix multiplication or similar operations, commutativity becomes something a little more special.Why did we choose the set to be a subset of the reals and why did we exclude \(-1\)? The latter question is easy to answer considering that dividing by zero doesn't make any sense and we can't have some inverses blowing up to infinity. We could've constrained our set to only be integers or primes or anything else. This further impacts the structure of the group.

Groups are incredibly robust mathematical objects which help unlock many deep secrets regarding algebraic structure. The best way to get a handle on them is to play with them. Pick a set, define an operation, and check if the axioms of a group hold. Once you've found one, add some additional constraints, or restrict the set in some way. You just might be surprised what you find.

© Daniel Marvin. Last modified: July 20, 2021.