# A characterization of automorphisms of $$\mathbb{Q}$$ (TIFR 2013 Problem 2)

Question:

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

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

Solution: 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 $$\phi(x)=qx$$