Cfrgtkky

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.