2 arguments that prove$(\mathbb{C}-\{0\},.)\not\simeq(\mathbb{R-\{0\}},.)$. > 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.