1
$\begingroup$

I was given the projection homomorphism $\mathbb{Z}_4 \times \mathbb{Z}_3 \to \mathbb{Z}_3$ and asked to find it and come up with the kernel. I came up with $\phi(x,y)= x$ such that $x \in \mathbb{Z}_4$ and $y \in \mathbb{Z}_3$

$$\ker(\phi)=\{(n,0\}|n\in\{0,1,2,3\}\}$$

What I am trying to do from here is find the general case: given $\mathbb{Z}_n \times \mathbb{Z}_m \to \mathbb{Z}_l$ define the projection homomorphism and come up with the kernel.

Even if you could just give me some different examples of this with real numbers that would be incredibly helpful.

Example: How would you define $$\mathbb{Z}_4 \times \mathbb{Z}_5 \to \mathbb{Z}_3$$ or is this not possible?

Or if someone could just give me an easier to understand definition of a projection homomorphism that would be great as well!

$\endgroup$
  • $\begingroup$ In general, there is no nontrivial homomorphism ${\bf Z}_n\times {\bf Z}_m\to {\bf Z}_l$. $\endgroup$ – tomasz Dec 19 '14 at 9:34
  • $\begingroup$ I am basically just trying to get a better understanding of this because I do not understand it at all... for example, how would you define Z4 x Z5 -> Z3 or is this not possible? $\endgroup$ – Lydia Dec 19 '14 at 9:56
  • $\begingroup$ let there exist nontrivial $\Bbb{Z}_4\times \Bbb{Z}_5\to \Bbb{Z}_3$, consider $(a,b)$ in $\Bbb{Z}_4\times \Bbb{Z}_5$. its order is divide $20$. $\endgroup$ – Mirin Dec 19 '14 at 10:01
  • $\begingroup$ How would you define phi(x,y)= though? $\endgroup$ – Lydia Dec 19 '14 at 10:14
  • 1
    $\begingroup$ Lydia, there's a difference between a homomorphism and a projection homomorphism. While the latter may not be formally defined, it is used to indicate one of the homomorphisms from a product of groups to one of its factors, i.e. $\phi_i : G_1\times\dots\times G_n \to G_i$ given by $(g_1, \dots, g_n) \mapsto g_i$. If you have a homomorphism $G_1\times\dots\times G_n \to H$ where $H \neq G_i$ for $i = 1, \dots, n$, then that would not be called a projection homomorphism, but rather just a homomorphism. $\endgroup$ – Michael Albanese Dec 22 '14 at 5:35
3
$\begingroup$

In your particular case, every element $x\in\mathbb{Z}_4\times\mathbb{Z}_5$ satisfies $x^{20} = e$. So if $\varphi:\mathbb{Z}_4\times\mathbb{Z}_5\to\mathbb{Z}_3$, then $$\varphi(x)^{20} = \varphi(x^{20}) = \varphi(e) = e,$$ so that the image of any element of $\mathbb{Z}_4\times\mathbb{Z}_5$ must be such that its $20^{\mathrm{th}}$ power is the identity. But the order of any element of $\mathbb{Z}_3$ is either $1$ or $3$, so if $\varphi(x)$ is an element of order $3$, its $20^{\mathrm{th}}$ power will not be the identity. Therefore $\varphi(x) = e$ for every $x\in \mathbb{Z}_4\times\mathbb{Z}_5$, so the only such homomorphism is the trivial one.

In general, if you have a homomorphism $\varphi:G\to H$, and $x\in G$ with $|x|=n$, then if $n$ is prime to the order of $H$, it must be the case that $\varphi(x) = e$, by an argument very similar to the one above.

$\endgroup$

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.