If $G$ is a finite group of odd order, show there is only one homomorphism between $G$ and $\mathbb{R}^{\times}$ which is the trivial one.
I know there is a similar case if the order of G and H is coprime. then the order of image of $G$ is 1. But in this case, order of $\mathbb{R}^{\times}$ is infinite. Any hints would be helpful. Thanks in advance.
A homomorphism from $G$ to $\mathbb R^\times$ satisfies $\phi(a^n) = (\phi(a))^n$ for all $a \in G$ and $n \in \mathbb N$.
With this in mind, note that $a \in G$ always satisfies $a^{|G|} = e_G$. Applying $\phi$ on both sides yields $(\phi(a))^{|G|} = 1$.
That is, $\phi(a)$ is a real number, which satisfies $\phi(a)^{|G|} = 1$, where $|G|$ is odd. This forces $\phi(a) = 1$ for all $a$. So $\phi$ is trivial.
Note that non-trivial homomorphisms in the reverse direction i.e. from $\mathbb R^\times$ to $G$ may very much exist. Can you think of an obvious map from $\mathbb R^\times$ to the multiplicative group $\pm 1$ of two elements, for example?
Note that if $|G|$ is even, then the statement is not true. Can you think of a non-trivial homomorphism from $\pm 1$ to $\mathbb R^\times$? This is even more obvious than the previous question.