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## 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 $$((1 2 3),  ) \in S_3 \times \mathbb{Z}_4$$ 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