Homomorphisms between a finite group and $(\mathbb R,+)$

Question

I'm trying to prove that any homomorphism from a finite group to $(\mathbb R,+)$ and from $(\mathbb R,+)$ to a finite group must be trivial.

What I have so far: let $G$ be a finite group and $f: G\to \mathbb R$ a homomorphism. Since $G$ is finite, for every element $x\in G$ there is a positive $n$ such that $x^n=e$. Then $0=f(e)=f(x^n)=f(x)\cdot n\implies f(x)=0$. So $f$ is trivial. Is that correct?

For the other statement, that's what I have (not sure if it's the right direction). Let $f: \mathbb R\to G$ be a homomorphism. Let $a=f(1)$. Since $G$ is finite, $a^n=e$ for $n\in \mathbb N$. Then $e=a^n=[f(1)]^n=f(n)$. This implies $f(n\mathbb Z)=e$. is this helpful at all?

Answer 1

Your first proof is correct.

To prove the second statement, notice indeed that by setting $n=|G|$, we find out that for any $x\in \mathbb{R}$, $f(nx)=f(x)^n=e$. Then, taking any $y\in \mathbb{R}$ and setting $x=\frac{y}{n}$, we obtain $$f(y)=f(nx)=e$$ so that $f$ is the trivial morphism.

(As Lord Shark the Unknown states it, this trick is due to the fact that $(\mathbb{R},+)$ is divisible).

Answer 2

The first way round is fine.

For the converse, the group $\Bbb R^+$ has a property called divisibility. An additive (resp. multiplicative) group $G$ is divisible if whenever $a\in G$ and $n\in\Bbb N$, then there is $b\in G$ with $nb=a$ (resp. $b^n=a$).

Prove (i) $\Bbb R^+$ is divisible, (ii) the image of a divisible group under a homomorphism is divisible.