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# TIFR 2013 Problem 2 Solution - A characterization of automorphisms of $\mathbb{Q}$

TIFR 2013 Problem 2 Solutions is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India's premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Contemporary Abstract Algebra by Joseph A. Gallian. This book is very useful for the preparation of TIFR Entrance.

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## Problem:True/False?

Any automorphism of the group $\mathbb{Q}$ under addition is of the form x→qx for some q ∈ $\mathbb{Q}$.

## Solution:

Hint: Does there exist an element of $\mathbb{Q}$ whose image determines the whole map?

Let $\phi:\mathbb{Q}\rightarrow \mathbb{Q}$ be an automorphism. That is, $\phi$ is a one-to-one, onto, homomorphism (of groups).
Using 1 we can 'create' the whole set of integers. (In the language of groups, we say Z is a cyclic subgroup generated by 1) Using the property of homomorphism, we can now know the image of any integer.

For m$\in$$\mathbb{N}$ we have:

$\phi(m)=\phi\left( 1+1+...+1)=\phi(1)+\phi(1)+...+\phi(1\right)$

and consequently, $\phi(m)=\phi\left( 1\right) =m\phi\left( 1\right)$

Since $\phi(1+(-1))=\phi(1)+\phi(-1)$ and $\phi(0)=0$, we have $\phi(-1)=-\phi(1)$

Now, let m$\in$$\mathbb{Z}$. Already, we know the value of $\phi$ at m$\ge$0. Let m$\lt$0. Then $m=-n$, where n$\in$$\mathbb{N}$. Then we get, $\phi(m)=\phi(-n)=\phi(-1)+\phi(-1)+...+\phi(-1)$.

But since, $\phi(-1)=-\phi(1)$, we finally get, $\phi(m)=-nq$.

That is, we get, $\phi(m)=mq$ for all m$\in$$\mathbb{Z}$.

We have finally figured out $\phi$ for integers. It is completely determined by $\phi(1)$.

Now we move on to $\mathbb{Q}$.  As $\mathbb{Q}$ is the field of fractions of $\mathbb{Z}$ it is not much to expect that we can get values of $\phi$ at fractional points using values at integer points.

What is the value of $\phi$ at $\frac{1}{n}$, n$\in$$\mathbb{N}$?

We exploit the homomorphism property of the map $\phi$ one more time.

Since $$\frac{1}{n}+\frac{1}{n}+...+\frac{1}{n}=1 (n times)$$

We have $$\phi\left(\frac{1}{n}\right)+\phi\left(\frac{1}{n}\right)+...+\phi\left(\frac{1}{n}\right)=\phi(1)=q$$

Hence $$\phi\left(\frac{1}{n}\right)=\frac{q}{n}$$

Can you show that $\phi$ at $\frac{1}{n}$ is \frac{q}{n} for n negative integer also? (Hint: Use the fact that $\phi(-1)=-\phi(1)$ )

Finally, we are ready for attacking the most general form of rational numbers.

For $\frac{m}{n}$, where m and n $\in$$\mathbb{Z}$, n$\neq 0$ we get,

$$\phi\left(\frac{1}{n}\right) + \phi\left(\frac{1}{n}\right) + ... +\phi\left(\frac{1}{n}\right)= \phi\left(\frac{m}{n}\right)$$ for all m $\ge0$ (m times addition)

Notice that, since we can always have the numerator (m) part of the rational number nonnegative ( by adjusting sign of denominator ) and already, $\phi\left(\frac{1}{n}\right)=\frac{q}{n}$ for all $n\in\mathbb{Z}$, we have

$$\phi\left(\frac{m}{n}\right) = m\frac{q}{n} = q \frac{m}{n}$$ for all $m,n\in\mathbb{Z}$ with n$\neq 0$.

That is for all x$\in\mathbb{Q}$, we have $$\phi(x)=qx$$

## Helpdesk

• What is this topic:Abstract Algebra
• What are some of the associated concept: Group Homomorphism
• Book Suggestions: Contemporary Abstract Algebra by Joseph A. Gallian