Proof of Wilson's Theorem using concept of group.

I am studying group theory so I do it by using the concept of group.

What I am trying to prove is if p is prime then $(p-1)!\equiv-1\mod p$

Note that $\mathbb{Z_p}$ forms a multiplicative group. Hence $\forall a \in \{1,2,\dots,p-1\},\exists a^{-1}\in \{1,2,\dots,p-1\}$

This means that $aa^{-1}\equiv 1\mod p$

If $a=a^{-1}, 1 \equiv aa^{-1}= a^2 \mod p$, So this means that $a \equiv \pm 1 \mod p$. (I am not sure that why this is true, is it because the group is abelian?)

"We have seen that this necessitates $a ≡ ±1 (\mod p)$ and so a = 1 or a = p − 1." ( I totally can't understand this sentence.)

Then we pair up $(p-1)!=1(2)\dots(p-2)(p-1)$ such that every element is with their multiplicative inverse.

So all the other pairs get $-1$ except for the pair ((1)(p-1)). Why this will happens?

I am not familiar with number theory so I hope I can get some idea or explaination about those statements.

• @Pierre-GuyPlamondon: In fact only half of it, since the question is only about one direction of the "if and only if" in Wilson's theorem. Jul 21 '15 at 7:39
• The fact that the only idempotent (i.e. self-inverse) elements in $\Bbb{Z}_p^*$ are $\pm 1$ is due to the fact that $\Bbb{Z}_p^*$ has exactly one subgroup of order $2$. It isn't related to commutativity.
– A.P.
Jul 21 '15 at 7:46
• As @dREaM noted, you've got the wrong pairing -- you're pairing the numbers with their additive inverses, but you should be pairing them with their multiplicative inverses. Jul 21 '15 at 7:47
• I deleted the comment @joriki was mentioning. It said $2(p-2)$ is congruent to $-4\bmod p$ which is not always $-1\bmod p$. Jul 21 '15 at 7:49
• see wiki en.wikipedia.org/wiki/Wilson%27s_theorem for a proof of Wilson's theorem using sylow theorem... Jul 21 '15 at 7:52

Your first question: If $a^2=1$ then $a^2-1=(a-1)(a+1)=0$. Since $\mathbb{Z}/(p)$ is a field, this implies that either $a-1=0$ or $a+1=0$ or that $a=\pm 1$.

For the next one. The only numbers between $0$ and $p-1$, that satisfy $a=\pm 1$ are $1$ and $p-1$, this is clear since $a=-1=p-1$ mod $p$.

Lastly, if we think of $(p-1)!$, this is the product of all the elements of the multiplicative group together. Group theoretically, consider $\prod_{g\in G} g$. Then, for each $g$, we can find the $g^{-1}\in G$ in the product and cancel. This aurgument only breaks down if $g^{-1}=g$, so in the product we are left with $\prod_{g\in G, g^2=1}g$. In the case of $G=\mathbb{Z}/(p)^*$, we have just shown that the only such elements at $p-1$ and $1$, so that $(p-1)!=\prod_{g\in \mathbb{Z}/(p)^*}g=\prod_{g\in \mathbb{Z}/(p)^*, g^2=1}g=(p-1)1=-1$ mod $p$. This is probably what they mean by pairing.

• @joriki Wow, I might be up a bit too late. Thanks for catching that!
– pax
Jul 21 '15 at 7:49

Let's do it for odd primes:

The multiplicative group of a finite field is cyclic. In this case the cyclic group is $\mathbb Z_{p-1}$.

We know a lot of things about groups, we know each element has a unique inverse, we know even more about cyclic groups, we know there are $\varphi(d)$ elements of order $d$ when $d$ divides the order of the group. Of course $2|(p-1)$ so we deduce there are $\varphi(2)=1$ elements of order $2$, in other words there are exactly two self inverses, $1$ and $-1$.

So when you take the product of all numbers what happens is that all of the pairs of inverses cancel out and you are only left with $1\cdot (-1)=-1$