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Identifying isomorphic groups (TIFR 2014 problem 25)

Question:

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.

November 18, 2017

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