Which of these groups are isomorphic to each other? $(\mathbb{Q}_{>0},\times,1)$, $(\mathbb{R}_{>0},\times,1)$, $(\mathbb{R},+,0)$, $(\mathbb{C}\setminus \{0\},\times,1)$
The answer seems to me to be none. Since all groups are either strict subsets of each other or strict subsets excepts for one element. Clearly they do not have the same order.
Does this make sense?
On
First of all, note that all of the groups have the same order, namely $\infty$. So you cannot argue using the order of the groups.
But if you think of the exponential function $x \mapsto e^x$ you see that this is an operation preserving map because $x + y = e^x e^y$ so it takes additive reals to multiplicative reals. Also, its image is $\mathbb R_{>0}$ and its domain is $\mathbb R$. So we have found an isomorphism from $(\mathbb R, +) \to (\mathbb R_{>0}, \times)$:
$$(\mathbb{R}_{>0},\times,1) \cong (\mathbb{R},+,0)$$
For $\mathbb{Q}_{>0}$ you can argue with cardinality: $|\mathbb Q| < |\mathbb R| = |\mathbb R_{>0}| = |\mathbb C|$ hence $(\mathbb{Q}_{>0},\times,1)$ cannot be isomorphic to any of the others (because none of the maps can be bijective).
Finally, we can think about whether $(\mathbb{C}\setminus \{0\},\times,1)$ is isomorphic to $(\mathbb{R}_{>0},\times,1)$ or not. To this end, note that isomorphic groups have the same number of elements of each order.
This means that if $G$ and $H$ are isomorphic groups and $G$ contains $5$ elements of order $2$ then $H$ also contains $5$ elements of order $2$.
How many elements of order $4$ does $\mathbb C - \{0\}$ contain?
The answer is two: $i$ and $-i$.
And how many elements of order $4$ does $\mathbb R$ contain?
The answer is: none.
It makes sense; unfortunately, it's wrong. The group depends not just on the set, but also on the operation; knowing that the underlying set of one group is a subset of the underlying set of the other tells you very little. In particular, think about the middle two examples and the map $f(x)=\ln(x)$ . . .