Is $\mathop{Hom}(O(1),\mathbb{C}^*)$ isomorphic to $\mathop{Hom}(\mathbb{C}^*,O(1))$?

I have an elementary question on homomorphisms.

Let $O(1)=\{A: A^t A=1 \}$ and let $\mathbb{C}^*= \{ z\in\mathbb{C}:z\not=0\}$.

Then what is the character group $\mathop{Hom}(O(1),\mathbb{C}^*)$ isomorphic to, and what is $\mathop{Hom}(\mathbb{C}^*,O(1))$ isomorphic to?

Is $\mathop{Hom}(O(1),\mathbb{C}^*)$ isomorphic to $\mathbb{Z}/2\mathbb{Z}$ since $f^2 = 1$ for any $f\in \mathop{Hom}(O(1),\mathbb{C}^*)$?

At the moment, I am not sure about $\mathop{Hom}(\mathbb{C}^*,O(1))$.

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Are you sure you want $O(1)$ rather than $O(n)$? Because $O(1)$ is simply the group $\{\pm 1\}$. – Alex Becker Jul 11 '12 at 8:30
How would the answer change if I change $O(1)$ to $U(1)$-- is this what you are suggesting? – math-visitor Jul 11 '12 at 8:33
Yeah, if you go to the answer in this link: math.stackexchange.com/questions/169316/… $O(1)$ is mentioned there. So I wanted to figure out what are $Hom(O(1),\mathbb{C}^*)$ and $Hom(\mathbb{C}^*,O(1))$. – math-visitor Jul 11 '12 at 8:34
Here's how $O(1)$ arose in that link: what is $G$ which will satisfy $Hom(G,\mathbb{C}^*)=\mathbb{Z}/2\mathbb{Z}$? Now, I'm trying to go backwards to make sure that the abelian group is $\mathbb{Z}/2\mathbb{Z}$. – math-visitor Jul 11 '12 at 8:44

Note that $O(1)$ is the group $\{\pm 1\}$. Any homomorphism into $\mathbb C^*$ sends $1$ to $1$ and $-1$ to some square root of $1$, that is to either $1$ or $-1$. Thus there are two maps in $Hom(O(1),\mathbb C^*)$, the constant map $1$ and the identity, so $Hom(O(1),\mathbb C^*)\cong \mathbb Z/2\mathbb Z\cong O(1)$.

On the other hand, any homomorphism from $\mathbb C^*$ to $O(1)$ has kernel of index $1$ or $2$. No subgroup of $\mathbb C^*$ has index $2$, or any proper subgroup of finite index. To see this, note that $\mathbb C^*$ can be factored into the circle group and the real numbers under addition (by mapping $re^{i\theta}$ to $(e^{i\theta},\log r)$). A finite index subgroup of $\mathbb R$ would give us a decomposition $\mathbb R=H\cup (a_1+H)\cup \cdots \cup (a_n+H)$. Similarly, restricted to the rational numbers it gives us the rationals as a union of disjoint cosets. We can restrict our attention to those $a_i$ that are rational. We have some subfield $K$ which contains none of the $a_i$, and the denominators in $K$ are coprime to all $a_i$, thus $K\subseteq H$ as it can contain no element of the other cosets. We can then take realize $H$ as a vector subspace of $\mathbb R$ over $K$, which must correspond to a proper subset of some basis for $\mathbb R$. Thus the quotient of $\mathbb R$ by $H$ as a vector space contains at least one copy of $K$, which is infinite, contradicting the fact that $H$ has finite index. Thus any finite index subgroup would have to correspond to a finite index subgroup of the circle group, but this is isomorphic to $\mathbb R/\mathbb Z$ and the same approach works here (I think). Thus the kernel has index $1$ so is all of $\mathbb C^*$, hence the only homomorphism from $\mathbb C^*$ to $\mathbb C^*$ is the trivial one, and so $Hom(\mathbb C^*,O(1))$ is the trivial group.

As a corollary, the two groups are not isomorphic.

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Wow, thank you Alex! That was fast! – math-visitor Jul 11 '12 at 9:19
Why does being dense at 0 imply that it is the whole group? Thanks for the explanation. – awllower Jul 11 '12 at 9:21
@awllower Actually, that isn't quite enough. There's a bit I forgot to add. Editing. – Alex Becker Jul 11 '12 at 9:23
@awllower Actually, I'm wrong about the countable index thing. Assuming AoC, such subgroups do exist. – Alex Becker Jul 11 '12 at 9:26
Thanks for the clarification! It saves in a wonderful manner the argument, albeit it is a different proof this way... – awllower Jul 11 '12 at 11:31

Allow me to write $\{ \pm 1 \}$ instead of $\mathrm{O}(1)$ and $\mathbb{C}^\times$ instead of $\mathbb{C}^*$.

It's clear that there are only two possible homomorphisms $\{ \pm 1 \} \to \mathbb{C}^\times$: the constant $1$ homomorphism and the inclusion $\{ \pm 1 \} \hookrightarrow \mathbb{C}^\times$. Thus, $\textrm{Hom}(\{ \pm 1 \}, \mathbb{C}^\times) \cong \{ \pm 1 \}$.

Now, consider any homomorphism $\phi : \mathbb{C}^\times \to \{ \pm 1 \}$. It is either trivial or has kernel a normal subgroup of index $2$. I claim the latter is impossible, so $\textrm{Hom}(\mathbb{C}^\times, \{ \pm 1 \}) = \{ 1 \}$. Indeed, observe that $\mathbb{C}^\times$ is a divisible group: this means, for any element $z$ and any positive integer $n$, there is an element $w$ such that $w^n = z$.

We will now show that any normal subgroup $K$ of a divisible group $G$ with finite index must be the whole group $G$ itself. Indeed, consider the quotient group $H = G / K$. Let $h \in H$ and let $n$ be a positive integer. Then there is an element $g$ in $G$ such that $h = g K$, and since $g$ has a $n$-th root, so does $h$. Thus, $H$ is also a divisible group. But it is clear that a finite group is divisible if and only if it is the trivial group, so $G = K$, as claimed.

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