Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I faced to the following problem and could to verify the first part of it:

Let $q\in\mathbb Q-\{0\}$ and let $v_q:(\mathbb Q,+)\to (\mathbb Q,+)$ is defined as $$v_q(t)=qt$$ Then proved that $v_q$ is an automorphism of $(\mathbb Q,+)$ and moreover, conclude that the characteristics subgroup of $(\mathbb Q,+)$ are only $(\mathbb Q,+)$ and $\{1\}$.

Indeed, $v_q$ is a homomorphism and $q\neq 0$ leads me to this point that it is injective. Also, after examining the map, I could see $v_q(q^{-1}t)=t$, so it is onto. Thanks for your hints. I really like this problem.

share|improve this question
    
I think you meant to define $\,v_q(t):=qt\,$ –  DonAntonio Feb 5 '13 at 4:39
1  
I don't understand: what exactly is your question? –  DonAntonio Feb 5 '13 at 4:40
    
babak jan, aya ketabi hastesh ke por az test dar zamine jabre khati bashe? –  aliakbar Feb 5 '13 at 14:40
    
@BabakSorouh mer30 azizam. Comment shoma ro dide boodam ama alan daghighan yadam nist kodom soal bood –  aliakbar Feb 6 '13 at 4:26
    
@aliakbar: math.stackexchange.com/a/285113/8581 wa ino math.stackexchange.com/a/287249/8581. Mamnunam age nazareto bedunam. Mamnunam az waghti ke mizari. Age umadi teh sari be man bezan. daneshgahe azad wahede jonub. –  Babak S. Feb 6 '13 at 5:21

1 Answer 1

up vote 4 down vote accepted

Hint: Let $G\subseteq\mathbb{Q}$ be characteristic. Suppose that $G\ne\mathbb{Q}$ and $G\ne \{0\}$. Choose $x\in \mathbb{Q}-G$ and $y\in G-\{0\}$. Then, the equation $v_q(y)=yq=x$ is solvable...so...

share|improve this answer
    
...is solvable for $\,q\,$, showing that there exists $\,v_q\,$ as defined which doesn't map $\,G\,$ to itself. +1 –  DonAntonio Feb 5 '13 at 4:48
    
@DonAntonio: I see that... –  Babak S. Feb 5 '13 at 4:51

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.