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.

  • 1
    $\begingroup$ Looks very good. $\endgroup$ – Lee Mosher Jul 20 '18 at 21:51

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.