This is problem 3.9 of Falko Algebra I, Fields and Galois Theory.

Let $G\subset K$ be a finite multiplicative abelian group and $-1\in G$ where $K$ is a field. Denote $\mu(G)$ as the product of all elements of $G$.

Show $\mu(G)=-1$.

It is clear that $\mu(G)^2=1$ as all elements have their own inverse included. So $\mu(G)=x$ where $x\in G$ such that $x^2=1$.

  1. Since $G$ is finite, it must come from a prime field of $K$? There is no reason to believe $G$ must come from a prime field of $K$. There might be other finite subfields of $K$. What are other subfields $F_{p^n}$ for $p$ prime?

So I conclude that $G$ has even order. Eliminate 1 and -1. Since every other element have an inverse, hence, whenever I have an element being its own inverse, I will have another element being its own inverse.

  1. How do I rule out that other elements cannot be its own inverse?

Note: At this point, the units of $K$ has not been discussed. So I presumed that I cannot use the conclusion that unit group is cyclic. I would hope there is a more basic way to deal with it.

  • 2
    $\begingroup$ If you are in a field, then the equation $x^2=1$ can have at most $2$ roots. Be careful with characteristic $2$ though. $\endgroup$ – lulu Aug 6 '17 at 20:55
  • $\begingroup$ @Severin That would in fact make the group trivial. But it is not torsion free here. $\endgroup$ – Tobias Kildetoft Aug 6 '17 at 21:13

Since $\mathbb{F}$ is a finite field, every element of $\mathbb{F}$ has finite order. This is clear in the case of $\mathbb{F}_2$, where $1=-1$. So let of $\text{char }\mathbb{F}\neq 2$. Let $x_0,x_1,x_2,\ldots,x_n$ be the nonzero elements of $\mathbb{F}$. Without loss of generality, we can order $x_0,x_1,x_2,\cdots,x_n$ so that $x_0=1$, $x_1=-1$ and $x_{2m}=x_{2m+1}^{-1}$ for $m \geq 1$. Inverses in a group or ring are unique. There are also no repeats in this list as no element besides $\pm 1$ can be its own inverse: if there were such an element, say $x$, then $x^{-1}=x$. But as $x \neq 0$, this implies $x^2=1$. But then $x^2-1=0$ which is to say $(x-1)(x+1)=0$ so that $x \in \{1,-1\}$. But then each element is paired with its inverse in the product so that

$$x_0x_1x_2\cdots x_n=x_0x_1 \cdot 1 \cdot 1 \cdot \cdots \cdot 1=x_0x_1=1 \cdot -1 = -1$$

  • $\begingroup$ Yes. I forgot to use $F[x]$ has UFD property which implies $x=\pm 1$. $\endgroup$ – user45765 Aug 6 '17 at 21:11
  • $\begingroup$ While it does not really change the argument, it was just the group, not the field that was assumed to be finite. $\endgroup$ – Tobias Kildetoft Aug 6 '17 at 21:15
  • $\begingroup$ Yes, after I posted I realized I answered not quite the question asked but a related question. However, making appropriate changes to the argument presented (for a whole finite field $\mathbb{F}$) the argument is essentially the same, mutatis mutandis. $\endgroup$ – mathematics2x2life Aug 6 '17 at 21:18

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.