Theorem 14.1.
Let $G$ be a group and $a,b,c\in G.$
-
If $ab=ac$ then $b=c.$ (left cancellation law)
-
If $ba=ca$ then $b=c.$ (right cancellation law)
-
$ax=b$ has a unique solution in $G,$ $x=a^{-1}b.$
$xa=b$ has unique solutions in $G,$ $x=ba^{-1}.$
-
$(a^{-1})^{-1}=a.$
-
$(ab)^{-1}=b^{-1}a^{-1}.$
Note.
In the Cayley table of $G,$
$ax$ will be in the row labeled by $a.$
Note.
If $a,x\in G,$ a finite group, Theorem 14.1
implies $ax$ appears in the row labeled by
$a$ in
$G$'s
Cayley table.
Note.
If $b\in G,$ a finite group, and $ax=b$,
Theorem 14.1(c) implies $b$ appears
exactly once in the row labeled by $a$ in
$G$'s
Cayley table.
Theorem.
Ignoring row labels, each element of a
finite group appears exactly once in
each row of the Cayley table for the
group.
Theorem.
Ignoring column labels, each element of a
finite group appears exactly once in each
column of the Cayley table for the group.
Definition.
Define integral powers of $a\in G:$
\begin{align*}
a^0 &=e\\
a^1 &=a\\
a^2 &=aa\\
&\;\;\vdots\\
a^{n+1} &=a^na
\end{align*}
so that $a^n$ equals the product
of $n$
$a$'s
for each positive integer $n.$
Also, for each positive integer $n,$
\[
a^{-n}=(a^{-1})^n
\]
Theorem.
Laws of exponents.
For all integers $m,n,$
\begin{align*}
a^ma^n=a^{m+n}\\
(a^m)^n=a^{mn}\\
\end{align*}
Note.
In additive notation, for all integers $m,n:$
\begin{align*}
na &=a+\cdots+a\quad(n\text{ terms})\\
(-n)a &=n(-a)\\
(ma)+(na) &=(m+n)a\\
n(ma) &=(mn)a
\end{align*}
Definition.
If $G$ is a group and $a\in G$
then $\gen{a}$
will denote the set of all integral
powers of $a.$
\[
\gen{a}=\{a^n\mid n\in\Z\}
\]
Example 14.1.
The set of all integral powers of
$(1\ 2\ 3)$ is
\[
\langle(1\ 2\ 3 )\rangle=
\{(1),(1\ 2\ 3),(1\ 3\ 2)\}
\]
Theorem 14.2.
If $G$ is a group and $a\in G$
then $\gen{a},$
the set of all integral powers of $a,$
is a subgroup of $G.$
The subgroup $\gen{a},$
is called
the subgroup generated by $a.$
Definition.
If $H$ is a subgroup and
$H=\gen{a},$
for some $a\in H,$
then $H$ is said to be a
cyclic subgroup.
Guevara Corollary.
If $a\in G,$ then the subgroup
$\gen{a}$ is cyclic.
Example.
The group of integers is cyclic:
\[
\Z=\gen{1}=\gen{-1}
\]
Theorem 14.3.
If $G$ is a group, $a\in G,$
and $r,s\in\Z$ such that $r\ne s$
and $a^r=a^s,$ then
-
There is a smallest positive integer $n$
such that $a^n=e.$
-
If $t$ is an integer, then $a^t=e$
iff $n$ is a divisor of $t.$
-
The elements
$e=a^0,$
$a,$
$a^2,$
$\ldots,$
$a^{n-1}$
are distinct, and
$\gen{a}=\{e,a,a^2,\ldots,a^{n-1}\}.$
Definition.
If $a$ is an element of a group,
then the smallest positive integer $n$
such that $a^n=e,$ if it exists,
is called
the order of $a.$
If there is no such integer, then $a$
is said to have
infinite order.
The order of an element $a$ will be
denoted by $o(a).$
Note.
If $a\in G$ then $a^n=e$
automatically if $n=0.$
By the preceding definition,
the order of $a$ is never $0$
That is, if $n$ is the order
of $a,$ then $n\gt0$ by definition,
and is the smallest such $n$
that $a^n=e.$
Example 14.2.
-
In $S_3,$ $o((1\ 2\ 3 ))=3.$
-
In the group of nonzero rational numbers
(operation multiplication),
$2$ has infinite order,
because $2^n\ne1$ for every
positive integer $n.$
Note.
In additive notation, the condition
$a^n=e$ becomes $na=0.$
In $\Z_n,$ the condition $a^n=e$
becomes $n[a]=[0].$
Example 14.3.
In $\Z_6,$ $o([2])=3,$
because $[2]\ne[0]$ and
\[
2[2]=[2]\oplus[2]=[4]\ne0
\]
but
\[
3[2]=[2]\oplus[2]\oplus[2]=[6]=[0].
\]
(We also see that $\langle[2]\rangle=\{[0],[2],[4]\}.$)
Corollary.
If $a$ is an element of a group,
then $o(a)=\abs{\gen{a}}.$
Alex Notes