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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?

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 $$ \phi(x)=qx $$


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