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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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What do you mean by "multiplication homomorphism", perhaps just a homomorphism? And why the tag "algebraic groups"? –  Martin Brandenburg Jan 2 '13 at 17:22
    
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 $r\in\mathbb R$. –  Emil Jeřábek Jan 2 '13 at 17:34
    
Yeah. I should not have omitted these possibility. –  woodbass Jan 2 '13 at 17:52
    
any other form of example except Tonik's? –  woodbass Jan 2 '13 at 17:54
2  
You should accept the answer to avoid that the question prompts in the list. –  Benoît Kloeckner Jan 3 '13 at 8:47

1 Answer 1

up vote 5 down vote accepted

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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Thank you for your prompt answer. –  woodbass Jan 2 '13 at 17:25

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