## 4.1 Basic definitions

Let G be a set. A binary operation on G is a function that takes as input a pair of elements of G and whose output is a single element of G: that is, a function $$G \times G \to G$$. We usually write these binary operations in infix notation, so if the binary operation is called $$*$$ we write $$g*h$$ for the result of applying $$*$$ to the pair (g,h) of elements of G.

Example 4.1

• $$+$$ is a binary operation on the set of integers $$\mathbb{Z}$$.
• $$-$$ is a binary operation on the set of complex numbers $$\mathbb{C}$$.
• $$-$$ is not a binary operation on the set of strictly positive whole numbers $$\mathbb{N}$$, because it doesn’t always output an element of $$\mathbb{N}$$.
• Binary operations needn’t be some mathematical operation we’re familiar with, they can be something silly: defining $$a*b = 2$$ for all $$a,b$$ creates a perfectly good, if useless, binary operation on the real numbers $$\mathbb{R}$$.

Definition 4.1 A group $$(G, *)$$ is a set G with a binary operation $$*$$ which contains an element e such that

• (Identity axiom) For all $$g \in G$$, we have $$e*g=g*e=g$$.
• (Inverses axiom) For all $$g \in G$$ there exists $$h \in G$$ such that $$h*g=g*h=e$$.
• (Associativity axiom) For all $$g,h,k \in G$$ we have $$(g*h)*k = g*(h*k)$$.

Strictly speaking, to specify a group I have to tell you two things: the underlying set G and the binary operation $$*$$, which we call the group operation. In practise people often do not specify the group operation when there is only one standard choice of group operation. For example, “the group $$\mathbb{Z}$$” refers to $$(\mathbb{Z}, +)$$ and “the symmetric group $$S_n$$” refers to $$(S_n, \circ)$$.

Sometimes people add an extra axiom to the ones above called the closure axiom which says that if $$g, h \in G$$ then $$g*h\in G$$. There is no need for us to include this because it is part of the definition of binary operation that $$g*h \in G$$. It is important to remember that if you want to check if a certain set $$G$$ under a certain operation $$*$$ is a group, you have to check that $$*$$ is genuinely a binary operation on $$G$$ as well as checking that the three group axioms hold.

The element $$e$$ referred to in the first group axiom is called an identity element for $$G$$. We can show that in any group there is one and only one element of G which satisfies the the first axiom from Definition 4.1.

Proposition 4.1 Suppose $$(G, *)$$ is a group. Then there is a unique element e of G such that $$e*g=g*e=g$$ for all $$g \in G$$.
Proof. Suppose that $$e, e' \in G$$ and for any $$g\in G$$ we have \begin{align} \tag{4.1} e*g & =g*e=g \\ \tag{4.2} e' *g &= g*e' = g \end{align} Then putting $$g=e$$ in (4.2) we get $$e = e *e'$$ and putting $$g=e'$$ in (4.1) we get $$e*e'=e'$$. So $$e = e*e' = e'$$.

An element $$h \in G$$ is called an inverse of $$g \in G$$ if $$h*g=g*h=e$$. Such an element is guaranteed to exists for every g by the second group action, we’ll show that in fact each g has only one inverse.

Proposition 4.2 Suppose $$(G,*)$$ is a group, $$g \in G$$. Then g has only one inverse element.
Proof. Suppose h and k are inverses of g, so that in particular $$h*g=e$$ and $$g*k=e$$. Then $$(h*g)*k = e*k = k$$, but $$h*(g*k) = h*e = h$$. But the associativity law tells us $$(h*g)*k=h*(g*k)$$, which says $$k=h$$.

This proposition means that we are justified in talking about the inverse element of an element of a group. We usually write $$g^{-1}$$ for the inverse of g, so that $g*g^{-1} = g^{-1}*g = g.$

Proposition 4.3 Let $$(G,*)$$ be a group and let $$g \in G$$. Then $$(g^{-1})^{-1} = g$$, that is, the inverse of $$g^{-1}$$ is g.
Proof. To show that g is the inverse of $$g^{-1}$$ we have to check that $$g* g^{-1}=g^{-1} *g = e$$. But this is true because $$g^{-1}$$ is the inverse of g!

Remark. There are only two ways to bracket a product of three elements $$a,b,c$$ of a group: $a*(b*c) \,\,\, \text{ or } \,\,\, (a*b)*c$ and the associativity axiom tells you that these are equal, so that the product $$a*b*c$$ is unambiguous. But the associativity axiom doesn’t tell you immediately that longer products are independent of how they are bracketed, for example, is $(a*b)*(c*d) \,\,\, \text{ equal to } \,\,\, ((a*b)*c)*d?$ In fact the answer is yes, for a product of any length: any bracketing you use to work out a product like $$g_1* \cdots * g_n$$ gives the same result. You can prove this by induction: one such proof is given in the book (Green 1988) by J.A. Green in the suggested reading, or there is a proof at MSE 21581.

Proposition 4.4 Let $$(G,*)$$ be a group and $$g,h \in G$$. Then $$(g*h)^{-1} = h^{-1}* g^{-1}$$.
Proof. We are going to start by showing that $$(h^{-1}*g^{-1})*(g*h) = e$$. We are free to bracket the product on the left in whichever way we like as discussed in the previous remark, so $$$\tag{4.3} (h^{-1}*g^{-1})*(g*h) = h^{-1}*(g^{-1}*g)*h = h^{-1} * e * h = h^{-1}*h = e.$$$ If $$x*y=e$$ then multiplying on the right by $$y^{-1}$$ gives $$x*y*y^{-1}=e*y^{-1}$$ so $$x = y^{-1}$$. Applying that to (4.3) gives that $$h^{-1}*g^{-1} = (g*h)^{-1}$$ as required.

There is some special notation for what happens when you multiply a group element by itself some number of times.
Definition 4.2 Let $$n \in \mathbb{Z}$$, let $$(G,*)$$ be a group and let $$g \in G$$. Then we define $$g^n$$ as follows: $g^n = \left\{ \begin{array}{ll} g * g* \cdots * g & n>0 \\ g^{-1} * g^{-1} * \cdots * g^{-1} & n<0 \\ e & n=0 \end{array} \right.$ where in the first case there are n copies of g in the product and in the second there are $$-n$$ copies of $$g^{-1}$$, so that $$g^{n}=(g^{-1})^{-n}$$.
These exponents behave exactly how you would expect them to.

Proposition 4.5 Let $$n,m \in \mathbb{Z}$$ and let $$(G, *)$$ be a group. Then

• $$g^n * g^m = g^{n+m}.$$
• $$(g^n)^m = g^{nm}$$.

Proof.

• This result is clear if either n or m is zero, and follows directly from associativity of $$(G,*)$$ if n and m have the same sign. So suppose $$n>0$$ and $$m<0$$. The proof is by induction on n, and when $$n=1$$, $\begin{equation*} g* g^{m} = g* (\underbrace{g^{-1} * \cdots * g^{-1}}_{-m }) = (g*g^{-1}) * g^{m+1} = g^{m+1} \end{equation*}$ as required. Now for $$n>1$$ we have $\begin{equation*} g^n * g^m = g* g^{n-1} * g^m = g* g^{m+n-1} \end{equation*}$ by induction, and using either the base case (if $$m+n-1<0$$) or the comment at the start of the proof, this equals $$g^{m+n}$$ as required. The case $$n<0$$ and $$m>0$$ is similar.
• For m positive or zero this follows immediately from the definition, so suppose $$m<0$$. We do the case $$n>0$$. The first part of this proposition implies $$(g^n)^{-1}= g^{-n}$$, so $\begin{equation*} (g^n)^m = g^{-n} * \cdots * g^{-n} \end{equation*}$ with $$-m$$ copies of $$g^{-n}$$ appearing. This is the product of $$-mn$$ copies of $$g^{-1}$$, so equals $$g^{mn}$$ by definition. The case when $$n<0$$ is similar.

Very often we’ll write the product $$g*h$$ of two group elements simply as $$gh$$, especially when the group operation $$*$$ is some kind of multiplication.

Definition 4.3 Let $$(G,*)$$ be a group. If G is a finite set with exactly n different elements we write $$|G|=n$$ and say G has order n, and that G is a finite group. Otherwise we say that $$(G,*)$$ is an infinite group.

### References

Green, J.A. 1988. Sets and Groups : A First Course in Algebra. Library of Mathematics. Springer Netherlands.