INTRODUCING 5 - days-a-week problem solving session for Math Olympiad and ISI Entrance. Learn More 

November 18, 2017

TIFR 2014 Problem 25 Solution - Identifying Isomorphic Groups


TIFR 2014 Problem 25 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 Topics in Algebra by I.N.Herstein. This book is very useful for the preparation of TIFR Entrance.

Also Visit: College Mathematics Program


PROBLEM:TRUE/FALSE?


Which of the following groups are isomorphic?

A. (\mathbb{R}) and (\mathbb{C})

B. (\mathbb{R}^* ) and   (\mathbb{C}^*)

C. (S_3 \times \mathbb{Z}_4) and (S_4)

D. (\mathbb{Z}_2 \times \mathbb{Z}_2 ) and ( \mathbb{Z}_4 )


DISCUSSION:


(\mathbb{R}) and (\mathbb{C}) are isomorphic:

Both these spaces are vector spaces ove (\mathbb{Q}).

We state a result without proof.

Theorem: Let (V) be an infinite dimensional vector space over a countable field (F). Then the dimension of (V) over (F) is (|V|) (the cardinality of (V) ).

This immediately tells us that (\mathbb{R}) has dimension (|R| = 2^{\aleph _0 } ).

Now the dimension of (\mathbb{C}) over (\mathbb{Q}) is (dim(\mathbb{R}^2) = 2 \times 2^{\aleph _0 } =2^{\aleph _0 } ).

What did we just show? We showed that (dim(\mathbb{C})=dim(\mathbb{R}) ) over (\mathbb{Q}).

Therefore they are isomorphic as vector spaces. So they are isomorphic as groups with respect to addition.

(\mathbb{R}^* ) and   (\mathbb{C}^*) are not isomorphic:

(\mathbb{C}^) has an element of order 4, namely (i) has order 4. If there was an isomorphism then the corresponding element in (\mathbb{R}^) will also have order 4. But there is no element of order 4 in (\mathbb{R}^*). Hence we conclude that these two groups are not isomorphic.

(S_3 \times \mathbb{Z}_4) and (S_4) are not isomorphic:

The order of (((1 2 3), [1] ) \in S_3 \times \mathbb{Z}_4) is 12. (S_4) does not contain an element of order 12.

 (\mathbb{Z}_2 \times \mathbb{Z}_2 ) and ( \mathbb{Z}_4 ) are not isomorphic:

(\mathbb{Z}_2 \times \mathbb{Z}_2 ) is not cyclic and ( \mathbb{Z}_4 ) is cyclic. So they can not be isomorphic.


HELPDESK

  • What is this topic:Modern Algebra
  • What are some of the associated concept: Isomorphism
  • Book Suggestions: Topics in Algebra by I.N.Herstein

Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.
JOIN TRIAL
support@cheenta.com