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?


I will use a lighter notation

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 $$(p-1)!=p-1$$

  • $\begingroup$ $2$ for $p-2$ we have that each element has its inverse, but this needs to be Abelian, right? This group is abelian? $\endgroup$ – marcelolpjunior Apr 23 '14 at 20:05
  • $\begingroup$ 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? $\endgroup$ – user63181 Apr 23 '14 at 20:15
  • $\begingroup$ No, unfortunately I do not know the theorem. $\endgroup$ – marcelolpjunior Apr 23 '14 at 20:19
  • 1
    $\begingroup$ There's no problem since fortunately the commutativity in our case is trivial:-) $\endgroup$ – user63181 Apr 23 '14 at 20:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.