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Regard $K=\mathbb{R}-\lbrace{0\rbrace}$ as a multiplication group. Let $f:K\to K$ be a multiplication homormorphism.

Question 1. Whether that $f$ is surjective implies that $f$ is injective?

Question 2. Whether that $f$ is injective implies that $f$ is surjective?

Question 3. $g: x\to x^b$ is a multiplication homormorphism of $K$ where $b=n/m, (n,m)=1$,$n\in\mathbb{Z}$ and $m$ is an odd integer. How to find any other multiplication homormorphism of $K$ than this form. Any example?

Edit. Emil Jeřábek gave other explicit examples: $h: x\to |x|^r$ or $x\to sgn(x)|x|^r.$ Of course, $hg$ is also ok.

Any other explicit ones?

Perhaps, these questions look like homework, but not easy to me to answer (my major is not in algebraic theory).

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  • $\begingroup$ What do you mean by "multiplication homomorphism", perhaps just a homomorphism? And why the tag "algebraic groups"? $\endgroup$
    – Martin Brandenburg
    Commented Jan 2, 2013 at 17:22
  • $\begingroup$ To add to Toink’s answer, there are also simple explicit examples for Q3: consider $f(x)=|x|^r$ or $f(x)=\mathrm{sgn}(x)|x|^r$ for any $r\in\mathbb R$. $\endgroup$
    – Emil Jeřábek
    Commented Jan 2, 2013 at 17:34
  • $\begingroup$ Yeah. I should not have omitted these possibility. $\endgroup$
    – woodbass
    Commented Jan 2, 2013 at 17:52
  • $\begingroup$ any other form of example except Tonik's? $\endgroup$
    – woodbass
    Commented Jan 2, 2013 at 17:54
  • $\begingroup$ The endomorphisms I mentioned are continuous. The other half of the story is that is every other endomorphism is neither Baire measurable nor Lebesgue measurable, and you need the axiom of choice to prove their existence, hence there is pretty much no way to describe them explicitly. Toink’s construction (with the proviso that elements of the basis can be mapped to arbitrary elements of $K$, and you have additionally a choice to map $−1$ to either $1$ or $−1$) describes all endomorphisms. $\endgroup$
    – Emil Jeřábek
    Commented Jan 2, 2013 at 18:21

1 Answer 1

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There is an isomorphism $\mathbb{Z}/2\mathbb{Z}\times (\mathbb{R}^+,\cdot )\to K$ given by $(x,y)\mapsto (-1)^xy$ (considering $\mathbb{Z}/2\mathbb{Z}$ to contain 0 and 1). There is also an isomorphism $(\mathbb R,+)\to(\mathbb R^+,\cdot)$ given by $x\mapsto exp(x)$.

So for all your questions it is enough to consider $K':= (\mathbb R,+)$, since $K \cong \mathbb{Z}/2\mathbb{Z} \times K'$ But $K'$ is a $\mathbb{Q}$-vector space. So you can pick any basis (which will be uncountable) and then do something on it giving you a lot of endomorphisms of $K'$.

Both questions 1 and 2 are false, since on the uncountable basis of $K'$ there are injective but not surjective set-maps (and vice versa), that extend to an endomorphism of $K'$.

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