<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\usepackage{epsfig}
\begin{document}
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\begin{center}
\S 12. Generators and Relations.
\end{center}

Let S be a set of symbols, e.g. $\{a,b\}$.  Let $s^{-1}$ be the
set of symbols $\{s^{-1}|s \in S\}$.  At this stage we have no
multiplication defined, so we are thinking of $s^{-1}$ as a symbol
and not as the inverse of $s$.  We assume $S \cap
S^{-1}=\emptyset$. A word is a finite string of symbols from $S
\cup S^{-1}$ possibly with repetitions.  Thus if $S=\{a,b\}$, then
$aba^{-1}bb^{-1}ab$ is a word.  Let $W_S$ be the set of all words
in symbols from $S \cup S^{-1}$. Also we regard the empty word as
a word and denote it 1.  Multiplication is defined on $W_S$ by
defining $w_1w_2$ to be the word obtained by writing $w_1$ and
then $w_2$: e.g. if $w_1=aba^{-1}$ and $w_2=ab^{-1}b$, then
$w_1w_2=aba^{-1}ab^{-1}b$. With multiplication defined in this
way, $W_S$ is a semi-group but not a group because \vspace{2in}

Let R be a subset of  $W_S$. If $u,v \in W_S$, we write $u=v$(mod
$R$), if there is a finite sequence

$$u=w_1,w_2, ..., w_k=v$$

of words such that $w_i$ is obtained from $w_{i-1},\ i=2,...,k$ by
inserting or deleting either an element of $R$ or a word of the
form $ss^{-1}$ or $s^{-1}s$ where $s \in S$.

\bigskip

EXAMPLE.

If $S=\{a,b\}$ and $R=\{a^2,b^3,a^{-1}bab^{-2}\}$, then
$aba^{-1}bb^{-1}aba=b$ (mod $R$). (Note that we write $a^2$ to
denote the word $aa$ and $b^{-2}$ to denote the word
$b^{-1}b^{-1}$ etc.)

In this case the sequence

$\hspace{1in} aba^{-1}bb^{-1}aba,\ aba^{-1}aba,\ abba,\
abaa^{-1}ba,\ abaa^{-1}bab^{-1}b,$

$\hspace{1in} abaa^{-1}bab^{-1}b^{-1}bb,\ abab^2,\ aaa^{-1}bab^2,\
a^{-1}bab^2,\ a^{-1}bab^{-1}b^3,$

$\hspace{1in} a^{-1}bab^{-2}b^4,\ b^4,\ b$

has the required property.

\newpage

THEOREM 12.1.

If $u=v$(mod $R$) and $w \in W_S$, then $wu=wv$(mod $R$) and
$uw=vw$(mod $R$)

\bigskip
\underline{Proof}. \vspace{2in}

Using this result the example above becomes much easier. Thus
$a^{-1}bab^{-2}=1$ (mod$R$) $\Rightarrow ba=ab^2$ (mod$R$),
multiplying on the left by $a$ and on the right by $b^2$.  Hence
\begin{eqnarray*} aba^{-1}bb^{-1}aba &amp; = &amp; ab^2a\ ({\rm mod} R)\\
 &amp; = &amp; baa\ ({\rm mod} R)\\ &amp; = &amp; b\ ({\rm mod} R) \end{eqnarray*}

\bigskip
THEOREM 12.2.

Equality (mod$R$) is an equivalence relation.

\bigskip
\underline{Proof}. \vspace{3in}

\newpage
Let $&lt;w&gt;$ denote the equivalence class of $W_S$ containing $w$. It
follows from Theorem 12.1 that if $u_1=u_2$ (mod$R$) and $v_1=v_2$
(mod$R$), then $u_1v_1=u_2v_2$ (mod$R$), since

$\hspace{1in} u_1u_2=$
\medskip

i.e. if $&lt;u_1&gt;=&lt;u_2&gt;$ and $&lt;v_1&gt;=&lt;v_2&gt;$, then $&lt;u_1u_2&gt;=&lt;v_1v_2&gt;$.
This means that it is possible to define a multiplication on the
equivalence classes of $W_S$ under equality (mod$R$) by putting
$&lt;u&gt;&lt;v&gt;=&lt;uv&gt;$.

\bigskip
THEOREM 12.3.

The equivalence classes of $W_S$ determined by equality (mod$R$)
form a group with multiplication as defined above.

\bigskip
\underline{Proof}.

\newpage

The above group is called the group with the elements of $S$ as
generators and the elements of $R$ as relations.

Relations are usually written as equations, thus the group

$$(a,b|a^2=b^3=1,\ ab=b^2a)$$

is the group as described above, in which $$S=\{a,b\} \textrm{ and
} R=\{a^2,\ b^3,\ aba^{-1}b^{-2}\}.$$

If $R=\emptyset$, the $w_1=w_2$ (mod$R$) if $w_2$ can be obtained
from $w_1$ by inserting or deleting words of the form $ss^{-1}$ or
$s^{-1}s$, where $s \in S$.  The group obtained in this case is
called the free group on the set $S$.

\bigskip

\begin{center}
\underline{Exercises}
\end{center}

\begin{description}
\item[12.1.]

If $S=\{a,b\}$ and $R=\{a^{-1}bab^4,\ a^2\}$, prove that

$\hspace{2in} b^{15}=1$ (mod$R$).

\item[12.2.]
Prove that the group $$(a,\ b\ |a^{-1}ba=b^2,\ b^{-1}ab=a^2)$$

has order 1.
\end{description}

\newpage
\begin{center}
\S 13. The Word Problem.
\end{center}

Let $S$ be a set of symbols, and let $R \subseteq W_S$.

\bigskip
PROBLEM.  Is it possible to program a computer so that if the
computer is given two words $w_1,w_2$ it will be able to give the
correct answer to whether or not $w_1=w_2$ (mod$R$)?

This problem is in fact \lq\lq unsolvable".  It was shown by P. S.
Novikov (1955) that there is a finite set $S$ and a finite set $R
\subseteq W_S$ for which it is impossible to program a computer to
answer the above question.

In particular cases the question above can be solved. If the group
with $S$ as set of generators and $R$ as set of relations is
finite, then the word problem can be solved (Mendelsohn 1964),
although it has to be assumed that the computer involved has an
unbounded amount of storage space and unlimited time to work on
the problem.

The following theorem is useful in solving the word problem in
some cases.

\bigskip
THEOREM 13.1.

Let $G$ be a group with $S$ as set of generators and $R$ as set of
relations.  Let $H$ be a group and suppose $\theta:\ S
\longrightarrow H$ is a mapping such that for all $w \in R$, if
$w=s_1^{\epsilon_1}s_2^{\epsilon_2}...s_r^{\epsilon_r}$, then

$$(s_1\theta)^{\epsilon_1}(s_2\theta)^{\epsilon_=2}...(s_r\theta)^{\epsilon_r}=1_H.$$

Under these circumstances there exists a homomorphism $\theta_1:\
G \longrightarrow H$ such that $&lt;s&gt;\theta_1=s\theta$ for all $s
\in S$.

\bigskip
EXAMPLE

Let $G=(a,\ b\ |a^2=b^3=1)$.  If $S=\{a,b\}$, then $\theta:\ S
\longrightarrow S_3$, $a\theta=(12),\ b\theta=(123)$ satisfies
$(a\theta)^2=1,\ (b\theta)^3=1.$ Theorem 13.1 then states that
there is a homomorphism $\theta_1:\ G \longrightarrow S_3$ such
that $&lt;a&gt;\theta_1=(12),\ &lt;b&gt;\theta_1=(123)$.

\bigskip
\underline{Proof}. (Theorem 13.1).

We define a mapping $\theta':\ W_S \longrightarrow H$ as follows.
If $w=s_1^{\epsilon_1}s_2^{\epsilon_2}...s_r^{\epsilon_r}$, then
$w\theta'=(s_1\theta)^{\epsilon_1}(s_2\theta)^{\epsilon_=2}...(s_r\theta)^{\epsilon_r}.$

From the hypothesis of the theorem if $w \in R$, then
$w\theta'=1$. Suppose $w_1=w_2$(mod$R$).

\vspace{3in}

Hence $w_1\theta'=w_2\theta'$, i.e. if $&lt;w_1&gt;=&lt;w_2&gt;$, then
$w_1\theta'=w_2\theta'$.  It follows that we can define a mapping
$\theta_1:\ G \longrightarrow H$ by putting $&lt;w_1&gt;=w\theta'$.  The
mapping $\theta_1$ is a homomorphism because

\vspace{2in}

EXAMPLE.

Let $G=(a,\ b,\ |a^2=b^5=1,\ a^{-1}ba=b^4)$. Let $S=\{a,b\}$ and
$R=\{a^2,\ b^5,\ a^{-1}bab^{-4}\}$.  Since $ba=ab^4$ (mod$R$),
every word $w$ of $W_S$ is equal (mod$R$) to a word of the form
$a^ib^j$ and since $a^2=1$ (mod$R$) and $b^5=1$ (mod$R$) every
word of $W_S$ is equal (mod$R$) to one of the set

$$\{1,\ a,\ b,\ ab,\ b^2,\ ab^2,\ b^3,\ ab^3,\ b^4,\ ab^4\}.$$

The problem is to show that these elements lie in distinct
equivalence classes under equality (mod$R$).  If they are
distinct, then the Cayley homomorphism

$$\rho:\ G \longrightarrow A(G)$$

gives

\hspace{1in} $a\rho=$

\hspace{1in} $b\rho=(1,\ b,\ b^2,\ b^3,\ b^4)(a,\ ab,\ ab^2,\
ab^3,\ ab^4)$.

(Here we should really write $&lt;a&gt;,&lt;b&gt;$ etc.  instead of $a,\ b$.)
This gives us a clue as to how to show that $G$ has 10 elements.
Let

$\hspace {1in} \alpha=(1,6)(2,10)(3,9)(4,8)(5,7)$

$\hspace {1in} \beta=(1,2,3,4,5)(6,7,8,9,10)$\ ,

then $\alpha^2=\beta^5=1$ and $\alpha^{-1}\beta\alpha=\beta^4$.
Hence by Theorem 10.1 there is a homomorphism $\theta_1:\ G
\longrightarrow S_{10}$ such that $&lt;a&gt;\theta_1=\alpha$,\
$&lt;b&gt;\theta_1=\beta$.  It is easy to check that $\theta'$ maps each
element of the set $$\{1,\ a,\ b,\ ab,\ b^2,\ ab^2,\ b^3,\ ab^3,\
b^4,\ ab^4\}$$ to a distinct element of $S_{10}$ and so no two
elements of this set represent the same element of $G$. Thus
$o(G)=10$.

\newpage

DEFINITION.

A subset $K$ of ${\bf R}^3$ is called a knot if there is a
continuous injective mapping $\theta:\ C \longrightarrow {\bf
R}^3$ where $C \subseteq {\bf R}^2$ is the set $C=\{(x,y) |
x^2+y^2=1\}$ and $Im \theta = K$

\bigskip
EXAMPLES.


\begin{center}
$\begin{array}{cc} \textrm{The trivial knot} &amp; \textrm{The
trefoil}\\ \epsfig{file=MJD-1.eps, width=30mm} &amp;
\epsfig{file=MJD-2.eps, width=40mm}\\
 &amp; \\
\textrm{The figure-eight knot} &amp; \textrm{The granny knot}\\
\epsfig{file=MJD-3.eps, width=30mm} \ \ &amp; \ \
\epsfig{file=MJD-4.eps, width=50mm}
\end{array}$
\end{center}

\
The main problem in knot theory is to give a procedure for
deciding whether or not two knots are the same (whatever that
means). Loosely speaking if one thinks of two knots as being made
up of string, then they are the same if one can be moved to take
up the same position as the second knot without cutting the
string.

The problem above is unsolved.  However it is often possible to
show that two knots are different by computing knot groups.

In a knot diagram insert arrows giving one direction around the
curve.  Label each continuous piece of line by a symbol.  Let $S$
be the set of all labelling symbols.  For each crossing point of
the diagram we take a word $r$ from $W_S$ as follows: if the
intersection is like (i) below with the line crossing over going
from right to left take $r=cac^{-1}b^{-1}$, if as in (ii), then
take $r=c^{-1}acb^{-1}$. Let $R$ be the subset of $W_S$ obtained
by selecting an element for each crossing point of the knot
diagram.  Let $G$ be the group with $S$ as set of generators and
$R$ as set of relations.  It can be shown that two different
diagrams of the same knot give rise to isomorphic groups.
Therefore $G$ is called the group of the knot $K$.  Actually $G$
is isomorphic to the fundamental group of ${\bf R}-K$.

\medskip

\setlength{\unitlength}{.3in}

\begin{center}
\begin{picture}(10,4)
\put(0,4){(i)} \put(1,2){\line(1,0){3}}
\put(5,2){\vector(-1,0){1}} \put(4,1.6){$c$}
\put(3,0){\line(0,1){1.8}} \put(3,0){\vector(0,1){1}}
\put(3.2,0.8){$b$} \put(3,2.2){\line(0,1){1.8}}
\put(3,2.2){\vector(0,1){1}} \put(3.2,3){$a$}

\put(6,4){(ii)} \put(7,2){\line(1,0){4}}
\put(7,2){\vector(1,0){3}} \put(9.8,1.6){$c$}
\put(9,0){\line(0,1){1.8}} \put(9,0){\vector(0,1){1}}
\put(9.2,0.8){$b$} \put(9,2.2){\line(0,1){1.8}}
\put(9,2.2){\vector(0,1){1}} \put(9.2,3){$a$}

\end{picture}
\end{center}

\bigskip

EXAMPLES

If $K$ is the trivial group, then $G$ is the group generated by
$a$ with no relations.  Thus $G$ is the infinite cyclic group with
elements $1,\ a,\ a^{-1},\ a^2,\ a^{-2},\cdots.$

\begin{center}
\epsfig{file=MJD-5.eps, width=40mm}
\end{center}


If $K$ is the trefoil, then

\bigskip $G=$

\bigskip
Note that $c=aba^{-1}$. So substituting for $c$ and $c^{-1}$
wherever they occur: $G=(a,\ b,\ |aba^{-1}aab^{-1}a^{-1}b^{-1},
baba^{-1}b^{-1}a^{-1})$,

\begin{center}
\epsfig{file=MJD-6.eps, width=40mm}
\end{center}
\bigskip


\bigskip
i.e. \begin{eqnarray*} G &amp; = &amp; (a,\ b,\
|aba^{-1}aab^{-1}a^{-1}b^{-1}, baba^{-1}b^{-1}a^{-1})\\ &amp; = &amp; (a,\
b\ |aba=bab). \end{eqnarray*}

\bigskip
We will be able to show that the trefoil is not the same as the
trivial knot if we can show that this group is not isomorphic to
the infinite cyclic group.

\bigskip
Now in $S_3$ if $\alpha = (12)$ and $\beta = (13)$, then

$$\alpha\beta\alpha=\ \ \ \ \ \ =\beta\alpha\beta.$$

\bigskip
Hence by Theorem 13.1 there is a homomorphism $\theta:\ G
\longrightarrow S_3$  such that $a\theta=\alpha$, $b\theta=\beta$.
This homomorphism is surjective, and so $S_3 \cong G / {\rm
Ker}\theta$. But $S_3$ is not abelian and any factor of an abelian
group is abelian.  Hence $G$ is not abelian and therefore cannot
be infinite cyclic.


\begin{center}
\underline{Exercises}
\end{center}

\begin{description}
\item[13.1.]
Let $G=(a,\ b\ |a^2=b^3=(ab)^5=1)$.  Prove that if $H$ is abelian
and $\theta:\ G \longrightarrow H$ is a homomorphism, then
$a\theta=b\theta=1$.  Prove that there is a non-trivial
homomorphism $\phi:\ G \longrightarrow A_5$.
\bigskip

\item[13.2.]
Use Theorem 13.1 to show that

$$ (a,\ b\ |aba=bab) \cong (c,\ d\ |c^2=d^3).$$
\bigskip

\item[13.3.]
Write down the knot group of the figure eight knot.  Write down
the knot group of the granny knot.  Show that the granny knot is
not the same as the trivial knot.
\end{description}
\end{document}
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