 TIFR 2014 Problem 23 Solution 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.

Also Visit: College Mathematics Program

## Problem:True/False?

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 $$\mathbb{Z}_4$$

## Discussion:

$$S_2$$ 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.

## Helpdesk

• 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