# product of all elements of finite abelian group containing -1 of a field is always -1?

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.

• If you are in a field, then the equation $x^2=1$ can have at most $2$ roots. Be careful with characteristic $2$ though. – lulu Aug 6 '17 at 20:55
• @Severin That would in fact make the group trivial. But it is not torsion free here. – 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
$$x_0x_1x_2\cdots x_n=x_0x_1 \cdot 1 \cdot 1 \cdot \cdots \cdot 1=x_0x_1=1 \cdot -1 = -1$$
• Yes. I forgot to use $F[x]$ has UFD property which implies $x=\pm 1$. – user45765 Aug 6 '17 at 21:11
• 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. – mathematics2x2life Aug 6 '17 at 21:18