I was exercising for group theory trying to find the indicated quantities for the given homomorphisms. What I am struggling to understand is to find the quantities when it comes to homomorphism from a group product to a group or another group product. Here's the example
Find $Ker(\alpha)$ and $\alpha(-3,2) $ for the homomorphism $\mathbb{Z}$ x $\mathbb{Z} \to \mathbb{Z}$ where $\alpha(1,0) = 3$ and $\alpha(0,1) = -5$
To find $\alpha(-3,2)$ I tried doing $$ \alpha(-3,2) = -3\alpha(1,0)+2\alpha(0,1) = -3*3 + 2*(-5) = -19$$ and for $Ker(\alpha)$ I tried $$ \alpha(a,b) = 0 \implies a\alpha(1,0)+b\alpha(0,1) = 3*a - 5*b = 0$$
So what is Ker now? Maybe some help or suggestions to tackle these type of problems differently would be huge for me.
Thank you and be safe.
You can write what you currently have as $$\ker(\alpha) = \{(a, b) \in \mathbb Z \times \mathbb Z: 3a - 5b = 0\}.$$
As it turns out, all solutions to this equation are of the form $(5n, 3n)$ for $n \in \mathbb Z$. You can look up the phrase homogeneous linear Diophantine equation like here to get more information on why (or just convince yourself that all of these tuples are solutions). However, these number theory details are disjoint from the group-theoretic questions about homomorphisms that you're asking.
As for this kind of problem in general: the key is that you knew the value of $\alpha$ on a $\mathbb Z$-basis $\{(1, 0), (0, 1)\}$ for $\mathbb Z \times \mathbb Z$. You can then use the homomorphism property to determine $\alpha(a, b)$ for all other $(a, b)$. It's analogous to how in linear algebra we encode homomorphisms (there called linear transformations) by matrices, and matrices are keeping track of where the basis goes. In this case, we'd write $$\alpha = \begin{pmatrix} 3 & -5 \end{pmatrix}. $$