# Q\to Q by q\to kq

• Oct 7th 2011, 01:01 PM
dwsmith
Q\to Q by q\to kq
Prove that for all fixed $\displaystyle k\in\mathbb{Q}$ such that k isn't 0 the map $\displaystyle \varphi :\mathbb{Q}\to\mathbb{Q}$ defined by $\displaystyle \varphi (q)=kq$ is an automorphism of $\displaystyle \mathbb{Q}$.

So I need to show it is an isomorphism. I am having a problem with the homomorphism part though.

Let $\displaystyle q,r\in\mathbb{Q}$. $\displaystyle \varphi (qr)=kqr$ but $\displaystyle \varphi (q)\varphi(r)=kqkr=k^2qr$.

What do I need to do?
• Oct 7th 2011, 01:10 PM
Swlabr
Re: Q\to Q by q\to kq
Quote:

Originally Posted by dwsmith
Prove that for all fixed $\displaystyle k\in\mathbb{Q}$ such that k isn't 0 the map $\displaystyle \varphi :\mathbb{Q}\to\mathbb{Q}$ defined by $\displaystyle \varphi (q)=kq$ is an automorphism of $\displaystyle \mathbb{Q}$.

So I need to show it is an isomorphism. I am having a problem with the homomorphism part though.

Let $\displaystyle q,r\in\mathbb{Q}$. $\displaystyle \varphi (qr)=kqr$ but $\displaystyle \varphi (q)\varphi(r)=kqkr=k^2qr$.

What do I need to do?

You need to remember that you are looking at the rationals under addition! So,

$\displaystyle \varphi(p+q)=k(p+q)=kp+kq=\varphi(p)+\varphi(q)$.
• Oct 7th 2011, 01:12 PM
dwsmith
Re: Q\to Q by q\to kq
Quote:

Originally Posted by Swlabr
You need to remember that you are looking at the rationals under addition! So,

$\displaystyle \varphi(p+q)=k(p+q)=kp+kq=\varphi(p)+\varphi(q)$.

Why are you able to assume it is addition?
• Oct 7th 2011, 01:14 PM
Swlabr
Re: Q\to Q by q\to kq
Quote:

Originally Posted by dwsmith
Why are you able to assume it is addition?

Because it isn't an automorphism otherwise, as you have already shown...
• Oct 7th 2011, 10:48 PM
zoek
Re: Q\to Q by q\to kq
Quote:

Originally Posted by dwsmith
Why are you able to assume it is addition?

Because $\displaystyle (\Mathbb{Q}, +)$ is a group while $\displaystyle (\mathbb{Q}, \cdot)$ is not.
• Oct 7th 2011, 11:00 PM
Deveno
Re: Q\to Q by q\to kq
in dwsmith's defense, it is bad form to speak of "the group Q". a group is not merely a set, but a set and a binary operation. failure to specify the intended binary operation requires an unwarranted amount of mysticism on the student's part.