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There exists an onto group homomorphism

A. from \(S_5\) to \(S_4\)

B. from \(S_4\) to \(S_2\)

C. from \(S_5\) to \(\mathbb{Z}_5\)

D. from \(S_4\) to \(\mathbb{Z}_4\)


\(S_2\) is the permutation group on 2 letters. It has order 2, so it is isomorphic to \(\mathbb{Z}_2\).

And we know a group homomorphism from \(S_n\) to \(\mathbb{Z}_2\), namely the signature map.

\(\sigma \to 0 \) if \( \sigma \) is even

\(\sigma \to 1 \) if \( \sigma \) is odd.

This map is onto. And it is a homomorphism. This is fairly well known fact so let us not prove it here.

So we know for sure that B is true.

How do we know that C is not true?

For that, suppose \(\phi:S_5 \to \mathbb{Z}_5 \) be an onto group homomorphism. Apply first isomorphism theorem for groups, then \(S_5/ker{\phi} \) is isomorphic to \(\mathbb{Z}_5\). That implies their orders are same. And \(|G/H|=|G|/|H| \) so we get \(|ker{\phi}| = 4! =24 \).

Now remember that \(ker{\phi} \) is a normal subgroup of \(S_5\). Also, notice that any 4-cycle in \(S_5\) must go to the identity of \(\mathbb{Z}_5\) since the order of image must be divisible by the order of domain which in the case of 4-cycles are 4. So the image of any 4-cycle must have order 1,2 or 4 in \(\mathbb{Z}_5\). But out of these, only order 1 is possible in \(\mathbb{Z}_5\). Hence the image of any 4-cycle is the identity. So, the kernel must contain all the 4-cycles in \(S_5\). Now we count the number of 5-cycles in \(S_5\).

There are \({{5}\choose{4}} \) ways to choose 4 elements out of 1,2,3,4,5. Next, given any 4 elements, we can always write them in increasing order. Let’s say we chose 2,4,5,3. Then we write this as 2,3,4,5. Now to count the number of possible 4-cycles (distinct) we fix 2 in the first position and permute the rest. Each of these permutations will give a different 4-cycle. There are \(3!\) such permutations. So in total, there are \({{5}\choose{4}}\times 3! =30 \) 4-cycles in \(S_5\). This is more than our cardinality of \(ker{\phi}\). This is a contradiction.

So option C is false.

The same kind of argument will apply to option D as well. Here one can consider the 3 cycles in \(S_4\) and get a contradiction as above.

For option A, consider the 5-cycles in \(S_5\). They must map to identity (because of order reasons as above). And there are \(4!=24\) 5-cycles (start the cycles with 1, permute rest). But first isomorphism theorem will give the cardinality of the kernel as \(\frac{5!}{4!}=5\), a contradiction. This disproves A.


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