1
$\begingroup$

This is a homework question for an abstract algebra class. As such, I don't expect a complete answer but a hint in the right direction would be great. The question is as follows:

Suppose that $f: (G, ∗) \rightarrow (G', ∗')$ is a surjective homomorphism from a cyclic group $G$ to a group $G'$.

  1. Show that $G'$ is also cyclic.
  2. Show that if $f$ is not injective, then $G'$ must be finite.

The proof for part 1 seems quite simple: because $f$ is surjective, $G' = f(G)$. Since $G$ is cyclic, $G = \langle x \rangle$ so $G' = f(G) = f(\langle x \rangle) = \langle f(x) \rangle$. Since $G$ is generated by $\langle f(x) \rangle$, it is cyclic.

I am unclear on how to show the link between $f$ not being one-to-one and the size of $G'$. I have considered dividing it into two cases: $G$ is either finite or infinite. If $G$ is finite, $G'$ must be as well. For the other case, I tried to assume by way of contradiction that $f$ is a non-injective, surjective homomorphism and $G'$ is infinite, but was unable to produce the contradiction.

Can someone provide an intuition about how to tackle this problem?

$\endgroup$
  • $\begingroup$ Hint: If $G$ is finite then $G'$ is finite. There is only one infinite cyclic group up to isomorphism. If $f(a)=f(b)$ you can use what you know of its structure to conclude your proof. $\endgroup$ – Mark Bennet Feb 23 '15 at 23:13
  • $\begingroup$ Okay - if $G$ is infinite, $G\simeq\mathbb Z$. However if $f(a) = f(b)$ (that is, there are two elements in $G$ that map to one element in $G'$), $G'$ cannot be isomorphic to $G$ and therefore must be finite. Correct or too simple? $\endgroup$ – bkaiser Feb 23 '15 at 23:25
  • $\begingroup$ You have to use that $f$ is a homomorphism - I've put that bit in an answer. $\endgroup$ – Mark Bennet Feb 24 '15 at 7:46
1
$\begingroup$

Consider the kernel of $f$. If $f$ is not injective then this kernel is nonzero and $G/\ker f \simeq G'$. So show that $G/\ker f$ is finite when $\ker f \neq 1$.

$\endgroup$
0
$\begingroup$

To fill out the hint in comments, suppose $G=\mathbb Z$ and $f(a)=f(b)$ with $b\gt a$. Now call $c=b-a\neq 0$ and note that $f(c)=f(b-a)=f(b)-f(a)=0$ because $f$ is a homomorphism.

Now use the division algorithm to write an arbitrary number $n=cq+r$ with $0\leq r \lt c$, then show that $f(n)=f(r)$, and the options for $r$ are finite.

To complete the analysis of the image of $f$, rather than just the finiteness argument, work instead with the least integer $d$ for which $f(d)=0$ (it is not necessary to do this to prove finiteness, we just need some positive integer $c$ with $f(c)=0$).

$\endgroup$

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.