Suppose $p$ is a positive prime number,we now define a binary operation $\bigotimes$:If $rs\equiv t (\hbox{mod}p)(0\leq t\leq p-1)$,then we say $r\bigotimes s=t$.
Theorem : $(\{1,2,...,p-1\},\bigotimes)$ is a group.
Proof: We want to prove $(\{1,2,...,p-1\},\bigotimes)$ is a group,which means that
1.$\forall a,b\in \{1,2,...,p-1\}$,we have $a\bigotimes b\in \{1,2,...,p-1\}$.
2.$\forall a,b,c\in \{1,2,...,p-1\}$,we have $(a\bigotimes b)\bigotimes c=a\bigotimes (b\bigotimes c)$.
3.There is an identity element $e\in \{1,2,...,p-1\}$ such that $\forall a\in \{1,2,...,p-1\}$,we have $e\bigotimes a=a\bigotimes e=a$.
4.There is an inverse for every member of $\{1,2,...,p-1\}$ ,which means that $\forall a\in \{1,2,...,p-1\}$,we have $a^{-1}\in \{1,2,...,p-1\}$ such that $a\bigotimes a^{-1}=a^{-1}\bigotimes a=e$.
Property 1 can be easily verified as follows : $a\bigotimes b$ must not be 0.(otherwise,$p|ab$.Because $p$ is a prime number,so either $p|a$ or $p|b$.But we know that both $a$ and $b$ are less than $p$,so it is impossible).So $a\bigotimes b\in \{1,2,...,p-1\}$.As for property 3, it is very easy to verify that $e=1$.In order to prove property 2,we need three lemmas.
lemma 1: $a\bigotimes b=b\bigotimes a$.
Proof:Trival.
lemma 2: $(a\bigotimes 1)\bigotimes c=(ac)\bigotimes 1=a\bigotimes c$.
Proof:Trival.
lemma 3: $(a\bigotimes b)\bigotimes c=(abc)\bigotimes 1$.
According to lemma 2,$(abc)\bigotimes 1=(ab\bigotimes 1)\bigotimes c=((a\bigotimes 1)\bigotimes b)\bigotimes c=(a\bigotimes b)\bigotimes c$.$\Box$
According to the above three lemmas,property 2 can be easily proved(How?).Now we prove property 4.In order to prove property 4,we need lemma 4.
lemma 4:If $1\leq i<j\leq p-1$,then $\forall n\in \{1,2,...,p-1\}$,we have $i\bigotimes n\neq j\bigotimes n$.
Proof:Otherwise,$i\bigotimes n=j\bigotimes n$.Let $in=k_1p+t_1,jn=k_2p+t_1$.So $(j-i)n=(k_2-k_1)p$.Because $p$ is prime,so $p|n$ or $p|(j-i)$.But $n<p$,$j-i<p$,so it is impossible.
Now it is the right time to prove property 4:We have the following p-1 sequences :
$$
\left(
\begin{array}{cccc}
1\bigotimes 1&2\bigotimes 1&\cdots&(p-1)\bigotimes 1\\
1\bigotimes 2&2\bigotimes 2&\cdots&(p-1)\bigotimes 2\\
\vdots&\vdots&\vdots&\vdots\\
1\bigotimes (p-1)&2\bigotimes (p-1)&\cdots&(p-1)\bigotimes (p-1)
\end{array}
\right)
$$
We can regard these $p-1$ sequences as a matrix which has $p-1$ columns and $p-1$ rows.The first column of the matrix ,according to lemma 4,is consisted by $p-1$ different numbers in $\{1,2,...,p-1\}$.Thus there must be only one number in this column which is equal to 1.The second column of the matrix is also consisted by $p-1$ different numbers in $\{1,2,...,p-1\}$.Thus there must be only one number in this column which is equal to 1.Apply similar to every column,it can be verified that there must be only one number in each column be 1.Thus the property 4 is proved. $\Box$.