For positive integers N>1, the positive integers less than N and coprime to N form a multiplicative group modulo N.
How can I prove that the multiplication table for this group is well defined?
How can I prove that this group satisfies the associative group axiom? It's easy to prove the closure axiom and identity/inverse axiom, but I have no idea how to prove associativity.
Best Answer
Let $\overline a$ denote the class of the integer $a\in \mathbb Z$ modulo $N$. We are working in $$G=\{\overline a\in \mathbb Z/N\mathbb Z\,:\,\textrm{gcd}\,(a,n)=1\}.$$ You said that you proved the "closure" axiom so I assume you know that the product in $G$ is just $\overline a\cdot\overline b:=\overline{ab}$, where $ab$ denotes the product in $\mathbb Z$. Hence $$(\overline{a}\cdot\overline b)\cdot \overline c=\overline{ab}\cdot \overline c=\overline{(ab)c}\overset{\ast}{=}\overline{a(bc)}=\overline{a}\cdot\overline{bc}=\overline a\cdot(\overline b\cdot \overline c).$$ We used associativity in $\mathbb Z$ in $\ast$.
Added. Let us show $\overline a\cdot \overline b=\overline{ab}$ is well-defined. Let us change the representatives: $$\overline a=\overline{a'},\,\,\,\,\,\overline b=\overline{b'}.$$ This means $$a-a'=hN,\,\,\,\,\,b-b'=kN\,\,\,\,\,\textrm{some }h,k\in\mathbb Z.$$ We have to show that changing representatives does not change the result, i.e. our claim is: $\overline{a'}\cdot \overline{b'}=\overline{ab}$. So: $$\overline{a'}\cdot \overline{b'}=\overline{a'b'}=\overline{(a+hN)(b+kN)}=\overline{ab+N(ak+bh+hkN)}=\overline{ab}.$$