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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 $G$ we denote by $\vert G\vert$ the number of elements in $G$. 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 $G$; the entry in the row $g$ and column $h$ 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 $(x_1(x_2x_3))(x_4x_5)=(x_1x_2)((x_3x_4)x_5)$, 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 $n$.) 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 $i$ and $j$.


\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 up previous contents
Next: Modular arithmetic Up: MT2002 Algebra Previous: Introduction   Contents

Edmund F Robertson

11 September 2006