Finding the charctristics subgroup of $(\mathbb Q,+)$

Question

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.

babak jan, aya ketabi hastesh ke por az test dar zamine jabre khati bashe? — Aliakbar, Feb 05 '13 at 14:40
@BabakSorouh mer30 azizam. Comment shoma ro dide boodam ama alan daghighan yadam nist kodom soal bood — Aliakbar, Feb 06 '13 at 04:26
@aliakbar: http://math.stackexchange.com/a/285113/8581 wa ino http://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. — Mikasa, Feb 06 '13 at 05:21

score 4 · Accepted Answer · answered Feb 05 '13 at 04:37

4

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...

answered Feb 05 '13 at 04:37

Alex Youcis

54,059

...is solvable for $,q,$, showing that there exists $,v_q,$ as defined which doesn't map $,G,$ to itself. +1 – DonAntonio Feb 05 '13 at 04:48
@DonAntonio: I see that... – Mikasa Feb 05 '13 at 04:51

Finding the charctristics subgroup of $(\mathbb Q,+)$

1 Answers1