Which pair of groups are isomorphic?

Question

For a ring $R,$ let $R^*$ denote the set of units of $R.$ Are the pair of groups isomorphic?

(a) $\left (\Bbb C^*, \cdot \right ), \left (\left (\Bbb R^2 \right )^*,\cdot \right )$

(b) $\left (\Bbb R, + \right ),$ $\left (\Bbb C, + \right )$

I think (a) is false because $\Bbb C^*$ contains an element of order $3$ but $\left (\Bbb R^2 \right )^*$ doesn't contain any element of order $3.$ Now we know that two additive groups are isomorphic if and only if they are isomorphic as $\Bbb Q$-vector spaces. So (b) is true if and only if $\Bbb R$ and $\Bbb C$ are isomorphic as $\Bbb Q$-vector spaces. But I think I have read somewhere that two $\Bbb Q$-vector spaces are isomorphic if and only if both of them have the same cardinality (Though I am not quite sure about that). I believe that for any $\Bbb Q$ vector space $V$ we have $\dim_{\Bbb Q} V = \text {Card}\ (V)$ and two $\Bbb Q$-vector spaces are isomorphic if and only if they both have the same $\Bbb Q$-dimension. So if my argument were correct then (b) is true since $\text {Card}\ (\Bbb C) = \text {Card}\ (\Bbb R).$ Am I right?

Thanks in advance.

Answer 1

(a) is correct.

Now we know that two additive groups are isomorphic if and only if they are isomorphic as $\mathbb{Q}$-vector spaces.

Only if they actually are $\mathbb{Q}$-spaces. Which is the case for $\mathbb{R}$ and $\mathbb{C}$, but not in general (e.g. any finite abelian group or $\mathbb{Z}$ is not a $\mathbb{Q}$-space). Also be warned that for "$\Rightarrow$" you have to show that every group homomorphism is actually $\mathbb{Q}$-linear map (which indeed holds).

But I think I have read somewhere that two $\mathbb{Q}$-vector spaces are isomorphic if and only if both of them have the same cardinality (Though I am not quite sure about that)

This is only true for infinite dimensional spaces over $\mathbb{Q}$. That's because if $V$ is a vector space over $F$ and $B\subseteq V$ its basis then $V$ can be represented as

$$\{f:B\to F\ |\ f(b)=0\text{ for all but finitely many }b\}$$

which for infinite $F$ has cardinality $|V|=|F|\cdot |B|=\max\{|F|, |B|\}$. Now if $|F|=\aleph_0$ (which is our case) and $V$ is of infinite dimension then this simplifies to $|V|=|B|$, i.e. $|V|=\dim V$.

And so, yes, once we have this the rest is simple, since $|\mathbb{R}|=|\mathbb{C}|$ and both are of infinite dimension over $\mathbb{Q}$ and thus they have to be isomorphic.