Talk:Circle Group is Group

hey Linus44, what was wrong (apart from the fact that the proof of infinity was missing) with the original proof posted? --prime mover 15:33, 6 March 2011 (CST)

Interesting and nice would be to prove this from showing that it is the kernel of the group homomorphism $\phi: \left({\C^\times, \cdot}\right) \to \left({\R_{>0}, \cdot}\right), z \mapsto \left\vert{z}\right\vert$. --Lord_Farin 17:56, 24 March 2012 (EDT)

I think that another possible proof would be to show that the modulus is a norm (which is not done) and use it to show that $\left|{x\cdot y^{-1}}\right|=\dfrac {\left|{x}\right|}{\left|{y}\right|}=1$ when $x,\ y\in\mathbb S$; and the apply the One-Step Subgroup Test. Using this aproach it is not necessary to define $\exp:\C\to\C$.--Dan232 18:29, 24 March 2012 (EDT)


 * There's all sorts of approaches, feel free to add any that occur to you. The bulk of the contents of Proof 2 may move soon, to another page I have coming up. --prime mover 19:20, 24 March 2012 (EDT)


 * UPDATE: This has been moved back to Circle Group is Group after refactoring. --prime mover (talk) 14:30, 5 June 2021 (UTC)