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Let $A$ be a divisible group, let $B$ be a finite group, and let $f: A \rightarrow B$ be a homomorphism. Show that $f$ is trivial.

(A group $A$ is divisible if for each $a \in A$ and $n \ge 1$ there exists some $b \in A$ such that $b^n = a$)

I wanted to know if my solution is correct -

Let $a \in A$. And assume that $|B| = n$ for some $n \in \mathbb{N}$. So we can see that -

$f(a)=f(b^n)=(f(b))^n=e_B$ and therefore $f$ is trivial.

The first $=$ is because of $A$ being a divisible group, and the second is because of $f$ being an homomorphism.

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    $\begingroup$ Looks very good. $\endgroup$ – Lee Mosher Jul 20 '18 at 21:51
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The proof is correct, but I'd add something for clarity.

Suppose $A$ is divisible and that $B$ is finite with $|B|=n$. If $f\colon A\to B$ is a homomorphism and $a\in A$, then there exists $x\in A$ such that $x^n=a$. Therefore $$ f(a)=f(x^n)=f(x)^n=e_B $$ Since $a$ was arbitrary, we conclude that $f$ is trivial.

A similar technique shows that the homomorphic image of a divisible group is divisible. You may want to show that a nontrivial divisible group is infinite.

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