# Showing that $(\mathbb{C}^{*}/\mathbb{R}^{*},\cdot)\cong (U/U_2,\cdot)$

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?

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 $$z \mapsto z^2/|z|^2$$.

• 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). Jan 24, 2021 at 0:59
• 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 Jan 24, 2021 at 1:08
• @TheZone I've added some remarks on the lines of my comment. Jan 24, 2021 at 1:19
• 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 Jan 24, 2021 at 12:42
• Yes, the first map is alright Jan 24, 2021 at 12:48