## 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 \[\begin{equation} \tag{4.3} (h^{-1}*g^{-1})*(g*h) = h^{-1}*(g^{-1}*g)*h = h^{-1} * e * h = h^{-1}*h = e.\end{equation}\] 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 \(g^{n}=(g^{-1})^{-n}\).

**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.