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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

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 )

**(\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.

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

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

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 )

**(\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.

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

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