Consider an object \(X\) with some symmetries \(S\). We've seen that we can compose any of the symmetries in \(S\) and obtain another symmetry of \(X\). We've also seen that these symmetries obey certain rules. We can now, at last, define a group.

##### Definition 2.1.0: Group

A *group* is a set \(S\) with an operation \(\circ: S\times S\rightarrow S\) satisfying the following properties:

- Identity: There exists an element \(e\in S\) such that for any \(f\in S\) we have \(e\circ f = f\circ e = f\).
- Inverses: For any element \(f\in S\) there exists \(g\in S\) such that \(f\circ =e\).
- Associativity: For any \(f,g,h \in S\), we have \((f\circ g)\circ h = f\circ (g\circ h)\).

An essential notion in mathematics is *abstraction*. Note that our definition certainly applies to any collection \(S\) of symmetries of an object, but in fact there are other contexts where the definitions apply as well! The operation can be any way of combining two things in \(S\) and getting another back; \(S\) doesn't need to be a collection of functions, and the operation doesn't need to be composition. A group is defined purely by the rules that it follows! This is our first example of an algebraic structure; all the others that we meet will follow a similar template: A set with some operation(s) that follow some particular rules.

For example, consider the integers \(\mathbb{Z}\) with the operation of addition. To check that the integers form a group, we need to check four things:

- Addition takes two integers and gives another integer back. (Here we're checking the requirement that the operation is one from \(S\times S\rightarrow S\). Notice that the the output of the operation is always in \(S\)! This is called
*closure* of the operation.) - There's an identity element, \(0\), where for any integer \(n\), we have \(n+0=0+n=n\).
- Every integer \(n\) has an inverse, \(-n\), with \(n+(-n)=(-n)+n=0\).
- Addition of integers is associative.

Thus, the integers - with the operation of addition - form a group.

On the other hand, the set of integers with the operation of multiplication do not form a group. Multiplication does indeed take two integers and return another integer, and there is an identity \(1\), and multiplication is associative. But not every element has an inverse that is also an integer. For example, the multiplicative inverse of \(2\) is \(\frac{1}{2}\), but this isn't an integer! Thus, integers with multiplication do not form a group.

##### An important note about inverses: An inverse means, roughly, that we can go back to where we started after applying an operation. Algebraically, this means we can cancel elements. When we have something like \(gh=gk\), we can multiply both sides on the left by \(g^{-1}\) to get \(h=k\). We have to be careful to multiply on the same side on both sides, since groups aren't always commutative! If \(gh=kg\), it doesn't necessarily tell us that \(h=k\)!

##### Show that the symmetries of an equilateral triangle are not commutative. In other words, find two symmetries \(f, g\) of the equilateral triangle such that \(fg\neq gf\).

## Contributors and Attributions

- Tom Denton (Fields Institute/York University in Toronto)