5
$\begingroup$

For $2$ i get that $C_2 \times C_2$ is not cyclic and I understand that if the homomorphism is surjective it must cover the entirety of $C_2 \times C_2$, but i don't follow why the image must be cyclic.


$8. )$ Does there exist a surjective homomorphism

  1. from $C_{12}$ onto $C_{4}$ ?
  2. from $C_{12}$ onto $C_{2} \times C_{2}$ ?
  3. from $D_{8}$ onto $C_{4}$ ?
  4. from $D_{8}$ onto $C_{2} \times C_{2}$ ?

Give reasons for your answers.


(2) No: the image of any homomorphism $C_{12} \rightarrow C_{2} \times C_{2}$ must be cyclic ( as it will be generated by the image of a generator of $C_{12}$ ) So it can't be surjective .

$\endgroup$

3 Answers 3

9
$\begingroup$

Let's say $G$ is a cyclic group. So $G = \langle g \rangle$, where $g$ is a generator. That means that any element of $G$ must be of the form $g^n$.

What does its image look like under a homomorphism $\phi$?

Take $y$ in the image of $\phi$. $y$ must be of the form $y = \phi(x)$, where $x$ is an element of $G$. But $x$ must be of the form $g^n$ (because it is cyclic).

Therefore, $ y= \phi(x) = \phi(g^n) = \phi(g)^n$.

We conclude that any element of the image is of the form $\phi(g)^n$.

From here (with a bit more thought) you should be able to conclude why $\phi(g)$ is a generator for the image.

$\endgroup$
2
  • $\begingroup$ I get that \phi(g)^12 = identity of the image. Does this imply \phi(g) is the generator? $\endgroup$ Apr 12, 2014 at 14:39
  • $\begingroup$ Not quite. The fact that $\phi(g)^{12} = id$ says just that. It doesn't even tell us what the order of $\phi(g)$ is. The point in that every element of the image comes from taking $\phi(g)$, and iterating it. $\endgroup$
    – Braindead
    Apr 12, 2014 at 15:11
4
$\begingroup$

Any homomorphism $\phi : C_{12} \rightarrow C_2 \times C_2$ is entirely determined by where it maps the generator of $C_{12}$.

If we suppose that $C_{12} = \langle g \rangle$, then $\phi(g^k) = \phi(g)^k$ for all $k\in \mathbb{Z}$ (using the homomorphism property of $\phi$). Hence it should be easy to see that $\mathrm{im}\, \phi = \{\phi(g)^k : k \in \mathbb Z\}$, a cyclic subgroup of $C_2 \times C_2$.

$\endgroup$
3
$\begingroup$

The key is to carefully parse the parenthetical remark in the solution. It is telling you more than just that the image is cyclic, it is giving you a purported generator.

That is, suppose you have a homomorphism $\phi : C_{12} \rightarrow C_2 \times C_2$. Let $x$ be a generator for $C_{12}$ and let $g = \phi(x)$. The claim is that $g$ must generate the image of $\phi$. How do you prove such a claim? Well, take any element $h$ in the image of $\phi$. You need to prove that $h = g^m$ for some integer $m$. You will need to use that $C_{12}$ is cyclic and that $\phi$ is a homomorphism to prove this.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .