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Group homomorphism with an infinite kernel

Let f:\mathbb{C}^* \rightarrow \mathbb{R}^* be a homomorphism. Prove that kernel of f is infinite.

Solution

If \mathcal{G} < \mathbb{C}^\ast is a finite subgroup then all z \in \mathcal{G} must have norm 1, i.e. z \in \mathbb{S}^1. ( Otherwise otherwise z, z^2, z^3, \dotsc is an infinite sequence of distinct elements in \mathcal{G}. )

Suppose , that \mathcal{G} is finite and is the kernel of f:\mathbb{C}^\ast \rightarrow \mathbb{R}^\ast. Then let z \in \mathcal{G} and

    \[f(2z) = f(2)f(z) = 2\cdot 0 = 0\]

So,  2z \in \ker f = G. But this is a contradiction since |2z| = 2 \neq 1.

Hence,  no finite subgroup \mathal{G} of \mathbb{C}^\ast can be a kernel.

Note: The homomorphisms that can easily be described are for the form f(z)=|z|^\alpha.

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