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I want to show that $(\mathbb{C}^{*}/\mathbb{R}^{*},\cdot)\cong (U/U_2,\cdot)$, where $U$ is the unit circle and $U_2=\{-1,1\}$.

I considered the function $f:\mathbb{C}^{*}\to (U/U_2,\cdot)$, $f(z)=\hat{\frac{z}{|z|}}$.

Then $f$ is well defined and it is a surjective homomorphism.

Let $z\in \ker(f)$. Then $f(z)=\hat{1}\iff \hat{\frac{z}{|z|}}=\hat{1}\iff\frac{z}{|z|}\in U_2\iff z\in \mathbb{R}^{*}$. So $\ker(f)=\mathbb{R}^{*}$ and now we are done by the fundamental isomorphism theorem.

I think that this proof should work, but I want to ask a further question. If I considered $g:\mathbb{C}^{*}\to U$, $g(z)=\frac{z}{|z|}$, I think that I would have got in the same manner that $(\mathbb{C}^{*}/\mathbb{R}^{*},\cdot)\cong (U,\cdot)$. This together with the other isomorphism implies that $(U/U_2,\cdot) \cong (U,\cdot)$.

So, could I say that factoring by $U_2$ is basically pointless?

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Yes, modding out by $U_2$ is "trivial".

Note that if $G \leq S^1$is a finite subgroup, then it consists of roots of unity. One can actually show that $G = G_n$ consists of all the $n$-th roots of unity for some $n \geq 1$.

But $z \mapsto z^n$ is a surjective endomorphism of the sphere, and thus $S^1/G_n \simeq S^1$.

Edit: your second map has kernel $z \in \Bbb C^\times$ such that $z = |z|$. That's not $\Bbb R^\times$ but rather $\Bbb R_{> 0}$.

Here's another way to see your first construction: you first map gives an iso

$$ \tau \colon \Bbb C^\times/\Bbb R_{>0} \to S^1. $$

The preimage of $G_2$ are precisely the classes $[z]$ for which $z/|z| = \pm 1$, so $z = \pm|z|$ i.e. $z \in \mathbb{R}^\times$. Inside the quotient $\Bbb C^\times /\mathbb{R}_{>0}$ that's just $[-1]$ and $[1]$, but we can also think of it as $\Bbb R^\times /\Bbb R_{>0}$, hence since $\tau(\Bbb R^\times / \Bbb R_{>0}) = G_2$ we get

$$ \Bbb C^\times / \Bbb R^\times \simeq (\Bbb C^\times/ \Bbb R_{>0}) / (\Bbb R^\times / \Bbb R_{>0}) \simeq S^1/G_2. $$

This is still $S^1$, though. An explicit map $\Bbb C^\times / \Bbb R^\times \to S^1$ would be induced by $z \mapsto z^2/|z|^2$.

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  • $\begingroup$ thank you! So, basically, the original problem where I was asked for the isomorphism with $U/U_2$ was kind of misleading, because there is nothing special about this group (that's why I wanted to make sure that both my isomorphisms are right, the conclusion looked a bit odd, but now I see that it is truly right). $\endgroup$
    – TheZone
    Jan 24, 2021 at 0:59
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    $\begingroup$ I don't think your last map is the iso you want, though. For example, it should send $-1$ to $1 \in S^1$ but it sends it to $-1/|-1| = -1$; it's important to note that the isomorphism I mention is not the projection $S^1 \to S^1/G_n$. That is never injective unless $n = 1$! If the second map were an iso, and its composition with the projection too, we would have that $S^1 \to S^1/G_2$ is an iso and that's not true $\endgroup$
    – qualcuno
    Jan 24, 2021 at 1:08
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    $\begingroup$ @TheZone I've added some remarks on the lines of my comment. $\endgroup$
    – qualcuno
    Jan 24, 2021 at 1:19
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    $\begingroup$ thank you very much! Now I see why my kernel was not right. It also seemed kind of strange just to modify the range of my first map and to obtain the second isomorphism $\endgroup$
    – TheZone
    Jan 24, 2021 at 12:42
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    $\begingroup$ Yes, the first map is alright $\endgroup$
    – qualcuno
    Jan 24, 2021 at 12:48

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