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**ENRIQUESTEFANINI** Hi:

Let $\displaystyle G$ be an abelian group (written multiplicatively), $\displaystyle |G|=n$. Denote by $\displaystyle \mu (G)$ the product of all elements of $\displaystyle G$. Consider the subgroups $\displaystyle G_2= \{x \in G: x^{2}=1\}$ and $\displaystyle G^{2}=\{x^{2}: x \in G\}$. Prove:

(a) $\displaystyle \mu(G)= \mu(G_2)$.

(b) If $\displaystyle G$ is a subgr of the multiplicative group of a field $\displaystyle K$ and $\displaystyle -1 \in G$, then $\displaystyle \mu(G)= -1 = (-1)^{n/2} \mu(G^{2})$.

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 $\displaystyle K$ be finite, $\displaystyle |K|= 2^{m}$ and take $\displaystyle G$ to be the whole multiplicative group of $\displaystyle K$. Then $\displaystyle |K|$ even and $\displaystyle |G|$ is odd. Yet, $\displaystyle 1$, and therefore $\displaystyle -1$ belong to G, as the author of the problem requires (that is, $\displaystyle -1 \in K$ and $\displaystyle -1 $ not equal $\displaystyle 0$. So, $\displaystyle -1 \in G$).

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