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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.

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I think you meant to define $\,v_q(t):=qt\,$ – DonAntonio Feb 5 '13 at 4:39
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: wa ino 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
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

share|cite|improve this answer 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

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