Next: Modular arithmetic Up: MT2002 Algebra Previous: Introduction Contents

Groups - definition and basic properties

$\begin{de} A {\em group} is a set $G$\ on which a binary operation $\ast$\ is de... ... $x^{-1}\in G$such that $x\ast x^{-1}=x^{-1}\ast x=e$. \end{description}\end{de}$

$\begin{no} The symbol one uses to denote the operation is unimportant. When we w... ...by $-x$, rather than $x^{-1}$, and is called the {\em negative of $x$}. \end{no}$

$\begin{de} A group $G$\ is said to be {\em commutative}, or {\em abelian}, if th... ...vity:}] For all $x,y\in G$\ we have $x\ast y=y\ast x$. \end{description}\end{de}$

$\begin{exs} Each of the sets $\mathbb{Z}$, $\mathbb{Q}$, $\mathbb{R}$\ and $\mat... ...bb{Z}}\setminus\{0\}$are not (why?). All the above groups are abelian. \end{exs}$

Next we give two examples of finite groups. For a finite group we denote by $\vert G\vert$ the number of elements in . A finite group can be given by its multiplication table (also called the Cayley table). This is a square table of size $\vert G\vert\times\vert G\vert$ ; the rows and columns are indexed by the elements of ; the entry in the row and column is $g\ast h$ .

$\begin{ex} The table \begin{displaymath} \begin{array}{c\vert cccc} \ast&e&a&b&c... ...led the {\em Klein four group}, and is denoted by $K_4$. It is abelian. \end{ex}$

$\begin{ex} The table \begin{displaymath} \begin{array}{c\vert cccccc} \ast & e &... ...rse.) It is not abelian, since, for example, $p\ast q=t\neq s=q\ast p$. \end{ex}$

$\begin{de} For a group $G$, the number of elements of $G$\ is called the {\em order} of $G$. \end{de}$

We are now going to list some basic consequences of our defining axioms for groups. First of all, we note that associativity implies that in a product of any number of elements $x_1,\ldots , x_n$ in that order, the arrangement of brackets does not matter. For example, we have , since

$\begin{displaymath} (x_1(x_2x_3))(x_4x_5)=((x_1x_2)x_3)(x_4x_5)= (x_1x_2)(x_3(x_4x_5))=(x_1x_2)((x_3x_4)x_5). \end{displaymath}$

We can therefore omit the brackets altogether and write simply $x_1x_2\ldots x_n$ . (A rigorous proof of this is a somewhat tedious induction on

.) This, in turn means that we can use the power notation:

$\begin{displaymath} a^n=\underbrace{aa\ldots a}_n\ (n>0). \end{displaymath}$

The existence of inverses implies that we can extend this (as we do in $\mathbb{Q}$ ) to

$\begin{displaymath} a^0=e,\ a^{-n}=(a^{-1})^n. \end{displaymath}$

With this in mind, we have the following natural rules:

$\begin{eqnarray*} && a^i a^j=a^{i+j},\\ && (a^i)^j=a^{ij}. \end{eqnarray*}$

The proof follows straight from the definition, but one has to consider all possible cases, depending on the signs of

and

$\begin{thm} The following statements are true in any group $G$. \begin{itemize} ... ...(a_1a_2\ldots a_n)^{-1}=a_n^{-1}\ldots a_2^{-1}a_1^{-1}$. \end{itemize}\end{thm}$

$\begin{proof} (i) $e=xx^{-1}=(e_1x)x^{-1}=e_1(xx^{-1})=e_1e=e_1$. \par (ii) $y=e... ...dots =a_n^{-1}\ldots a_2^{-1}a_1^{-1}, \end{displaymath}as required. \end{proof}$

Next: Modular arithmetic Up: MT2002 Algebra Previous: Introduction Contents

Edmund F Robertson

11 September 2006