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Introduction

Modern algebra is a study of sets with operations defined on them. It begins with the observation that certain familiar rules hold for different operations on different sets.

Let us consider the set of natural numbers $\mathbb{N}$. The operations of addition and multiplication satisfy the following rules:

    $\displaystyle (x+y)+z=x+(y+z),$ (1)
    $\displaystyle (xy)z=x(yz),$ (2)
    $\displaystyle x+y=y+x,$ (3)
    $\displaystyle xy=yx,$ (4)

as well as
\begin{displaymath}
x(y+z)=xy+xz.
\end{displaymath} (5)

Also there is a distinguished element 1, which has the property
\begin{displaymath}
1x=x.
\end{displaymath} (6)

In the sets of integers $\mathbb{Z}$, rationals $\mathbb{Q}$, reals $\mathbb{R}$ and complex numbers $\mathbb{C}$ all the above rules remain valid. In addition, in each of them there is a distinguished element $0$ such that

    $\displaystyle 0+x=x,$ (7)
    $\displaystyle 0x=0.$ (8)

Also, for every element $x$ there is an element $-x$ (its negative) such that
\begin{displaymath}
x+(-x)=0.
\end{displaymath} (9)

Moreover, in $\mathbb{Q}$, $\mathbb{R}$ and $\mathbb{C}$, for every $x\neq 0$ there is an element $1/x$ such that
\begin{displaymath}
x(1/x)=1.
\end{displaymath} (10)

At this stage you should note that the following pairs of rules are very similar: (1) and (2); (3) and (4); (6) and (7); (9) and (10). The only difference is that they refer to different operations.

Our next example is the set $M_{n,n}$ of all $n\times n$ matrices with real number entries. Again, we can add and multiply matrices, and we have the following rules:

\begin{eqnarray*}
&&(A+B)+C=A+(B+C),\\
&&(AB)C=A(BC),\\
&&A+B=B+A,\\
&&A(B+C)=AB+AC,\ (A+B)C=AC+BC.
\end{eqnarray*}



This time, however, we do not have $AB=BA$. Also, we have distinguished matrices $0$ (the zero matrix) and $I$ (the identity matrix) such that

\begin{eqnarray*}
&&A+0=A,\\
&&AI=IA=A.
\end{eqnarray*}



For every matrix $A$ there exists a matrix $-A$ such that

\begin{displaymath}
A+(-A)=0.
\end{displaymath}

However, it is not true that for every matrix there exists a matrix $A^{-1}$ such that $AA^{-1}=A^{-1}A=I$; such a matrix exists if and only if $A$ is invertible (i.e. if $A$ has a non-zero determinant).

Let us now fix a set $X$, and consider the set $T(X)$ of all mappings $f\st X\rightarrow X$. For such a mapping $f$ and $x\in X$, we shall write the image of $x$ under $f$ as $f(x)$. We can compose two mappings $f$ and $g$ by applying first one and then the other; the resulting mapping is denoted by $g\circ f$. Thus

\begin{displaymath}
(g\circ f)(x)=g(f(x)).
\end{displaymath}

(Note that this means that we multiply mappings, slightly unnaturally, from right to left. In some books you will find mappings written to the right of their argument, i.e. $(x)f$ instead of $f(x)$; a benefit of doing this is that the composition law becomes $(x)(f\circ g)=((x)f)g$.)

Straight from the definition it follows that

\begin{displaymath}
(f\circ g)\circ h=f\circ (g\circ h),
\end{displaymath}

and that there exists a mapping $\rm id$ (the identity mapping, sending every $x$ to itself) satisfying

\begin{displaymath}
f\circ{\rm id}={\rm id}\circ f=f.
\end{displaymath}

However, $f\circ g=g\circ h$ does not hold in general, and it is not true that for every mapping $f$ there exists a mapping $g$ such that $f\circ g=g\circ f={\rm id}$.


\begin{exc}
Find examples confirming the last two claims.
\end{exc}


\begin{exc}
Let $X$\ be a set, and let ${\cal P}(X)$\ be the
set of all subsets ...
...operties
of the operations $\cup$\ (union) and $\cap$\ (intersection).
\end{exc}

In algebra, one studies an abstract set with one or more operations defined on it. These operations are assumed to satisfy some basic properties, and the aim is to study the consequences of these properties.


next up previous contents
Next: Groups - definition and Up: MT2002 Algebra Previous: About the course   Contents

Edmund F Robertson

11 September 2006