Why is the product of all units of a finite field equal to $-1$?

Suppose $F=\{0,a_1,\dots,a_{q-1}\}$ is a finite field with $q=p^n$ elements. I'm curious, why is the product of all elements of $F^\ast$ equal to $-1$? I know that $F^\ast$ is cyclic, say generated by $a$. Then the product in question can be written as $$ a_1\cdots a_{q-1}=1\cdot a\cdot a^2\cdots a^{q-2}=a^{(q-2)(q-1)/2}. $$ However, I'm having trouble jumping from that product to $-1$. What is the key observation to make here? Thanks!

Solution 1:

The only elements of a field so that $x^2=1$ are $1$ and $-1$. Therefore, when you multiply all the non-zero elements together, each element but $1$ and $-1$ will be paired with its inverse. Thus, the product is $-1$.

Note for Fields with Characteristic 2:

In a field of characteristic $2$, $-1=1$, so there is only one element so that $x^2=1$. Thus, the product is still $-1=1$.

Solution 2:

If the inverse of $x$ is not equal to $x$, pair $x$ with its inverse. Note that if $x$ gets paired with $y$, then $y$ gets paired with $x$. The only objects that do not get paired are the ones that are their own inverse.

The object $x$ is its own inverse iff $x^2=1$, or equivalently $(x-1)(x+1)=0$. If the characteristic of the field is $\ne 2$, there are two solutions, $1$ and $-1$. The product of any two paired elements is $1$, so the product of all paired elements is $1$. The product of $1$ and $-1$ is $-1$, so the full product is $-1$.

If the characteristic is $2$, then $-1=1$. The product of the paired elements is $1$, so the full product is $1$. But this is equal to $-1$.

Remark: While I prefer the pairing argument, yours will work. Take for example the case characteristic $\ne 2$. If $w=a^{(q-1)/2}$, then $w^2=1$. But $w\ne 1$, so $w=-1$. Since $q-2$ is odd, it follows that $\left(a^{(q-1)/2}\right)^{q-2}=-1$.

Solution 3:

This is a special case of Wilson's Theorem in a Finite Commutative Group. See, for instance, pp. 253-255 of these notes for the statement and proof.

Added: Here is a more direct answer to your question. Write

$a^{(q-2)(q-1)/2}$ as $\left( a^{\frac{q-1}{2}} \right)^{q-2}$. Now try to convince yourself that

(i) $a^{\frac{q-1}{2}} = -1$ and
(ii) $(-1)^{q-2} = -1$.

For (i), there is a general recipe for the order of a power of a generator in a cyclic group. If you don't know about this, see e.g. Proposition 250 on page 241 of the same notes. You should be able to do (ii) easily; for extra credit, you might want to check that $a^{q-2} = a^{-1}$ for all $a \in \mathbb{F}_q^{\times}$.

Solution 4:

The $q-1$ elements of $F^*$ are roots of the polynomial $x^{q-1}-1$. Hence, $$-1 = \prod_{\alpha \in F^{*}} (-\alpha) = (-1)^{q-1}\prod_{\alpha \in F^{*}} \alpha \Rightarrow \prod_{\alpha \in F^{*}} \alpha = (-1)^q = \begin{cases}-1,&q ~ \text{odd},\\ +1,&q ~ \text{even},\end{cases}$$ but, as André reminded you, $-1 = +1$ if the characteristic of the field is $2$, and so we need not make special cases but just write that $\displaystyle \prod_{\alpha \in F^{*}} \alpha = -1$.