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 $\mathop{Hom}(\mathbb{C}^*,O(1))$. 
 A: 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.
A: 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.
