# Thread: Multiplicative group of a field.

1. ## Multiplicative group of a field.

Hi:
Let $G$ be an abelian group (written multiplicatively), $|G|=n$. Denote by $\mu (G)$ the product of all elements of $G$. Consider the subgroups $G_2= \{x \in G: x^{2}=1\}$ and $G^{2}=\{x^{2}: x \in G\}$. Prove:
(a) $\mu(G)= \mu(G_2)$.
(b) If $G$ is a subgr of the multiplicative group of a field $K$ and $-1 \in G$, then $\mu(G)= -1 = (-1)^{n/2} \mu(G^{2})$.

Up to here, the statement of the problem. And it is in the statement, point (b), that I find a counter-example: For the statement seems to assert n is an even number. However, let $K$ be finite, $|K|= 2^{m}$ and take $G$ to be the whole multiplicative group of $K$. Then $|K|$ even and $|G|$ is odd. Yet, $1$, and therefore $-1$ belong to G, as the author of the problem requires (that is, $-1 \in K$ and $-1$ not equal $0$. So, $-1 \in G$).

Am I right? Is the statement wrong? Any hint will be greatly appreciated. Good bye.

2. Originally Posted by ENRIQUESTEFANINI
Hi:
Let $G$ be an abelian group (written multiplicatively), $|G|=n$. Denote by $\mu (G)$ the product of all elements of $G$. Consider the subgroups $G_2= \{x \in G: x^{2}=1\}$ and $G^{2}=\{x^{2}: x \in G\}$. Prove:
(a) $\mu(G)= \mu(G_2)$.
(b) If $G$ is a subgr of the multiplicative group of a field $K$ and $-1 \in G$, then $\mu(G)= -1 = (-1)^{n/2} \mu(G^{2})$.

Up to here, the statement of the problem. And it is in the statement, point (b), that I find a counter-example: For the statement seems to assert n is an even number. However, let $K$ be finite, $|K|= 2^{m}$ and take $G$ to be the whole multiplicative group of $K$. Then $|K|$ even and $|G|$ is odd. Yet, $1$, and therefore $-1$ belong to G, as the author of the problem requires (that is, $-1 \in K$ and $-1$ not equal $0$. So, $-1 \in G$).

Am I right? Is the statement wrong? Any hint will be greatly appreciated. Good bye.

From where is this statement? If it is a from a book it may well a mistraslation and or a typo, and I think it should be $\mu(G)= -1 = (-1)^{n/2\cdot\mu(G^{2})}$ , otherwise the very statment of the result is basically non-sensical: how can -1 equal -1 to some power times the number of elements in $\mu(G^2)$?? In the best of the cases this would mean $\mu(G^2)=1$ , which of course is false.
Check this, I think it works fine with the correction.

Tonio

Pd. Si estás estudiando de un libro traducido ten cuidado: son comunes los errores de traducción y los tipográficos.

3. It is from a German book on algebra, translated into English. I don't think it is in error. If char $K$ not equal 2, then n is even. Not only this. In this case the statement is valid. I think I have the correct proof.

You say $\mu(G^{2})=1$ is false. But consider $K=\{0,1\}$. In K, $-1=1$. So $G=\{1\}$ and $-1 \in G$. As a result, $G^{2}=\{1^{2}\}$ and $\mu(G^{2})=1$. Bear in mind: if $G=\{a_1,a_2,...,a_k\}$ then, assuming the $a_i^{2}$ are all diferent, $\mu(G^{2})= a_1^{2}a_2^{2}...a_k^{2}$
Enrique.

P.D.: estaba ayer contestando tu correo, cuando el sistema me dejo afuera. Decidi dejarlo para el dia siguiente. Aunque ya es seguro que nadie conteste este.

4. Originally Posted by ENRIQUESTEFANINI
It is from a German book on algebra, translated into English. I don't think it is in error. If char $K$ not equal 2, then n is even. Not only this. In this case the statement is valid. I think I have the correct proof.

You say $\mu(G^{2})=1$ is false.

I meant it was false in a general context, but I misread: I though $\mu(G)$ was the number of elements in G and etc. My bad.

Tonio

But consider $K=\{0,1\}$. In K, $-1=1$. So $G=\{1\}$ and $-1 \in G$. As a result, $G^{2}=\{1^{2}\}$ and $\mu(G^{2})=1$. Bear in mind: if $G=\{a_1,a_2,...,a_k\}$ then, assuming the $a_i^{2}$ are all diferent, $\mu(G^{2})= a_1^{2}a_2^{2}...a_k^{2}$