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Let $f: \mathbb{C}^* \rightarrow \mathbb{R}^*$ be a homomorphism of the multiplicative group of complex numbers to the multiplicative group of real numbers. I need to show that the kernel of $f$ must be infinite.

I do know that $\mathbb{C}^*$ and $\mathbb{R}^*$ are not isomorphic to each other from here. So does that mean $f$ is not onto? But how will I be able to show that the kernel is infinite?

Thanks in advance.

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what homomorphisms are you familiar with from $\mathbb{C}^*$ to $\mathbb{R}^*$ something like $z\rightarrow |z|$? – Praphulla Koushik Feb 14 '14 at 6:44
up vote 9 down vote accepted

Hint: There are infinitely many complex numbers $z$ that satisfy the equation $z^n=1$ for some odd integer $n$. What can you say about their homomorphic images?

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Sir, could you please extend this a bit.... are you saying any homomorphism is of the form $z\rightarrow z^n$ :O – Praphulla Koushik Feb 14 '14 at 6:53
No :-). I'm fairly sure that all the continuous homomorphisms are of the form $z\mapsto |z|^\alpha$ for some $\alpha$, but all?? Looks like an application of Zorn's lemon will give us plenty of others. But we are not required to show that the kernel is uncountable. Countably infinite will do ;-> And how many solutions are there to $x^n=1$, $n$ odd, in the reals anyway? – Jyrki Lahtonen Feb 14 '14 at 7:06
@JyrkiLahtonen Well, formal has two meanings and one of it - i.e. in the sense of logical stringence - is of course often required :) – Hagen von Eitzen Feb 14 '14 at 7:18
@PraphullaKoushik: If $z$ is a complex root of unity of an odd order, i.e. $z^n=1$ for some odd $n$, then, no matter what the homomorphism $f$ is, we must have $f(z)^n=f(z^n)=f(1)=1$. But $f(z)$ is a real number. And $z$ is in the kernel, iff $f(z)=1$. – Jyrki Lahtonen Feb 14 '14 at 8:32
Oh yes...I got it.. take any complex number of odd order then $f(z)=1$ i.e., $z$ is in the kernel....there would be at least one complex number of each odd order thus kernel is countably infinite... Excuse me for being a bit dumb... Thank you :) – Praphulla Koushik Feb 14 '14 at 8:37

I'm not terribly sure if this is the same thing Jyrki is hinting at, and I wouldn't like to give the game away, but at any rate I believe that if this is not a different solution, it at least approaches it from a slightly different angle:

We know that $\ker f$ is a subgroup of $\mathbb{C}^\ast$, so let's ask what are the finite subgroups of $\mathbb{C}^\ast$?

The rest I hide in a spoiler tag below to let you think about this a bit more.

First observation: if $G < \mathbb{C}^\ast$ is a finite subgroup then all $z \in G$ must have norm $1$, i.e. $z \in S^1$. (For otherwise $z, z^2, z^3, \dotsc$ is an infinite sequence of distinct elements in $G$.) Suppose then that $G$ is finite and is the kernel of $f:\mathbb{C}^\ast \rightarrow \mathbb{R}^\ast$. Then let $z \in G$, and then $$f(2z) = f(2)f(z) = 2\cdot 0 = 0,$$ so $2z \in \ker f = G$. But this is a contradiction since $|2z| = 2 \neq 1$.

In other words, no finite subgroup $G$ of $\mathbb{C}^\ast$ can be a kernel.

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$0$ isn't an element of $\mathbb{R}^*$: no $z$ can satisfy $f(z) = 0$. The kernel is the set of elements that map to $1$. However, this method can prove that any monoid homomorphism from $\mathbb{C}^*$ to the monoid of all real numbers under multiplication that maps any nonzero complex number to $0$ must actually map all complex numbers to zero. – Hurkyl Feb 14 '14 at 7:51
Argh whoops, multiplicative! Not additive! Much confuse! So lame! My bad! Wow. I'd downvote this answer if I could. – chiaroscuro Feb 14 '14 at 13:02
no problem josh, it was pretty appreciated for your help anyway! – PandaMan Feb 16 '14 at 6:54

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