Multiplicative Groups of Real Numbers and Complex Numbers are not Isomorphic

Group Theory Problems and Solutions in Mathematics

Problem 130

Let $\R^{\times}=\R\setminus \{0\}$ be the multiplicative group of real numbers.
Let $\C^{\times}=\C\setminus \{0\}$ be the multiplicative group of complex numbers.
Then show that $\R^{\times}$ and $\C^{\times}$ are not isomorphic as groups.

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Let $G$ and $K$ be groups.
Recall that a map $f: G \to K$ is a group homomorphism if
\[f(ab)=f(a)f(b)\] for all $a, b \in G$.
A group isomorphism is a bijective homomorphism.
If there is a group isomorphism from $G$ to $K$, we say that $G$ and $K$ are isomorphic (as groups).
We give two proofs.

Proof 1.

Seeking a contradiction, assume that there is a group isomorphism $\phi: \C^{\times} \to \R^{\times}$.
Since $\phi$ is a group homomorphism, $\phi(1)=1$. Thus we have
\[1=\phi(1)=\phi((-1)(-1))=\phi(-1)\phi(-1)=\phi(-1)^2.\] Hence $\phi(-1)=\pm 1$. But since $\phi$ is injective and $\phi(1)=1$, we must have $\phi(-1)=-1$.

Now we have

Since $\phi(i)\in \R^{\times}$, $\phi(i)^2$ must be a positive number. Thus we reached a contradiction.
Hence there is no isomorphism between $\R^{\times}$ and $\C^{\times}$.

Proof 2.

Suppose that there is a group isomorphism $\phi: \C^{\times} \to \R^{\times}$.
We want to find a contradiction.
Let $\zeta=e^{2\pi i /3}$ be a primitive third root of unity.

Since $\zeta^3=1$ and $\phi$ is a group homomorphism, we have
Since $\phi(\zeta)$ is a real number, this implies that $\phi(\zeta)=1$.

This is a contradiction since $\phi$ is injective, but we have $\phi(1)=1=\phi(\zeta)$.
Therefore, there cannot be a group isomorphism between $\R^{\times}$ and $\C^{\times}$.

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