# Evidence about the group $\left ( (\mathbb{Z}/p\mathbb{Z})^*,\underset{p}{ \odot} \right )$

Be $p$ an odd prime number. Show that the group $\left ( (\mathbb{Z}/p\mathbb{Z})^*,\underset{p}{ \odot} \right )$ has a unique element of order $2$, namely $\overline{p-1}$, and show that $(p-1)!\equiv p-1 \pmod{p}$ (Wilson's Theorem)

The first part, I managed to do.

$$\overline{(p-1)}\underset{p}{\odot}\overline{(p-1)}=\overline{p^2}-\overline{2p}+\overline{1}=\overline1$$Because $p\mid p^2-2p$, then $\overline{p-1}$ has order $2$.

To prove that it is unique. And Wilson's theorem, how do?

If $x$ has the order $2$ then $$x^2=1\iff (x-1)(x+1)=0$$ but since $x\ne1$ and $\Bbb Z/p\Bbb Z$ is a field since $p$ is prime then we have $$x+1=0\iff x=-1$$ hence $-1=p-1$ is the only element equal to his inverse (except also $1$).
Now we have $$(p-1)!=1\times \underbrace{2\times\cdots (p-2)}_{=1}\times(p-1)$$ since every element from $2$ to $p-2$ has his inverse in this set which's different to this element hence $$(p-1)!=p-1$$
• $2$ for $p-2$ we have that each element has its inverse, but this needs to be Abelian, right? This group is abelian? – marcelolpjunior Apr 23 '14 at 20:05
• Yes it's abelian. To be more scientist notice that I already said in my answer that $\Bbb Z/p\Bbb Z$ is a field and it's finite. Do you know a theorem that affirm that every finite field is commutative: wedderburn theorem? – user63181 Apr 23 '14 at 20:15