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!
Best Answer
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.