Calculating $\operatorname{Hom}(\mathbb C^*, \mathbb C^*)$? I want to calculate this group.
So far I noticed that if $h:\mathbb C^* \to \mathbb C^*$ is a homomorphism, then $h(z)=h(1\cdot z)=h(1)\cdot h(z)$ for any $z\in \mathbb C^*$. Thus $h(1)=1$.
Further I know that for any $t\in \mathbb C$ $h_t: z\mapsto z^t$ satisfies $h(xy)=(xy)^t=x^ty^t=h(x)h(y)$. So this is an homomorphism, hence  $\operatorname{Hom} (\mathbb C^*, \mathbb C^*)\subset \mathbb C.$
But is this an equality, or are there more homomorphisms?
 A: One can classify simply the continuous homomorphisms from $\Bbb C^*$
to itself. They are the maps
$$z\mapsto z^n\exp(a\ln|z|)$$
where $n\in\Bbb Z$ and $a\in\Bbb C$.
But assuming the axiom of choice, one proves that $\Bbb C^*$
is isomorphic to the additive $(\Bbb Q/\Bbb Z)\times\bigoplus_I\Bbb Q$
where $I$ is an uncountable index set. This group has $2^{2^{\aleph_0}}$
automorphisms, an awful lot.
A: Multiplicative notation is not common in abelian group theory.
We can use the fact that $\mathbb{C}^*\cong\mathbb{R}\oplus\mathbb{T}$, where $\mathbb{T}=\mathbb{R}/\mathbb{Z}$, via
$$
(r,u+\mathbb{Z})\mapsto e^r(\cos(2\pi u)+i\sin(2\pi u))
$$
Thus the group of endomorphisms of $\mathbb{C}^*$ is the direct sum of
$$\DeclareMathOperator{\Hom}{Hom}
\Hom(\mathbb{R},\mathbb{R})\oplus\Hom(\mathbb{R},\mathbb{T})\oplus
\Hom(\mathbb{T},\mathbb{T})\oplus\Hom(\mathbb{T},\mathbb{R})
$$
The first two groups are big $\mathbb{Q}$-vector spaces, the third one is a wild beast. If we consider the exact sequence $0\to\mathbb{Z}\to\mathbb{R}\to\mathbb{T}\to0$, then we have
$$
0\to\Hom(\mathbb{T},\mathbb{T})\to\Hom(\mathbb{R},\mathbb{T})\to\Hom(\mathbb{Z},\mathbb{T})\to0
$$
and $\Hom(\mathbb{Z},\mathbb{T})\cong\mathbb{T}$. So $\Hom(\mathbb{T},\mathbb{T})$ sits in a big $\mathbb{Q}$-vector space so that the cokernel is $\mathbb{T}$.
What's $\Hom(\mathbb{T},\mathbb{R})$? A similar sequence arises:
$$
0\to\Hom(\mathbb{T},\mathbb{R})\to\Hom(\mathbb{R},\mathbb{R})\to\Hom(\mathbb{Z},\mathbb{R})\to0
$$
As you see, the group $\Hom(\mathbb{C}^*,\mathbb{C}^*)$ is much bigger than $\mathbb{C}$.
