Find two arguments to justify $(\mathbb{C}-\{0\},.)\not\simeq(\mathbb{R-\{0\}},.)$.
An isomorphism implies that there exist $f:\mathbb{C}-\{0\}\to\mathbb{R-\{0\}}$. such that f is bijective and f is a homomorphism.
1) Since the cardinality of $\mathbb{R}\subset\mathbb{C}$ and the cardinality of $\mathbb{C}$ is higher than the cardinality of $\mathbb{R}$, the function cannot be surjectivce hence not bijective.
2) Let´s assume $f$ is a homomorphism. If I pick the unitary root $i=\sqrt{-1}$, the order of $i$ would be 4, and there is no element in $\mathbb{R}$ whose order is $4$.
Question:
I am not sure about the first point since the both groups are innumerable. Is it right? What other arguments could I come up with?
Thanks in advance!
Note that $$\mathbb{R}\subset\mathbb{C}$$ does not mean that the cardinality of $\mathbb{R}$ is less than the cardinality of $ \mathbb{C}$.
So the cardinality argument does not work.
In fact they have the same cardinality.
The argument using the order of $i$ is a valid argument. An isomorphism should keep orders unchanged.