Let a, b, c be integers such that a divides b and b divides c. Then $c\mathbb{Z} \subset b\mathbb{Z} \subset a\mathbb{Z}$. Let $\phi :a\mathbb{Z} / c\mathbb{Z} \to a\mathbb{Z} / b\mathbb{Z}$ can be defined by $\phi \left(at + c\mathbb{Z}\right) = at + b\mathbb{Z}$.
a) Show that $\phi$ is a well-defined epimorphism
b) What is $ker(\phi)$?
I know that to be well-defined $g=h$ must imply that $\phi (g) = \phi (h)$.
I've worked mostly on the homomorphism part, where $\phi(ab)$ must equal $\phi(a)\phi(b)$. So I let $a_1 t_1 + c\mathbb{Z}, a_2 t_2 + c\mathbb{Z} \in a\mathbb{Z} / c\mathbb{Z}$. Then $\phi [(a_1 t_1 + c\mathbb{Z})(a_2 t_2 + c\mathbb{Z})] = \phi(a_1a_2t_1t_2 + a_1b_1c\mathbb{Z} + a_2b_2c\mathbb{Z} + c\mathbb{Z})$.
I'm not convinced that algebraic manipulation is the best way to prove this, but I didn't know how else to start.
For part b) I have a theorem that states if $N$ is normal in $G$, then there is a homomorphism $\psi$ of $G$ on $G/N$ such that $Ker \psi = N$. That seems like it will be useful if I can prove normality.
$\phi$ is well-defined if the outcome of $\phi$ does not depend on the presentation of the argument of $\phi$.
So it must be checked whether we have: $$x+c\mathbb Z=y+c\mathbb Z\implies x+b\mathbb Z=y+b\mathbb Z$$or equivalently
$$c\mid x-y\implies b\mid x-y$$ This can happily be recognized as a direct consequence of $b\mid c$.
In this context $\phi$ is a homomorphism if: $$\phi((x+c\mathbb Z)+(y+c\mathbb Z))=\phi(x+c\mathbb Z)+\phi(y+c\mathbb Z)$$
This follows immediately if we work out both sides.
$\phi$ will moreover be and epimorphism if it is surjective.
This follows directly from its definition:$$\phi(x+c\mathbb Z)=x+b\mathbb Z$$
$x+c\mathbb Z$ is an element of the kernel of $\phi$ if it is sent to the zero element of $a\mathbb Z/b\mathbb Z$, i.e. if we have:$$\phi(x+c\mathbb Z)=x+b\mathbb Z=b\mathbb Z$$ or equivalently if: $$b\mid x$$ or equivalently if:$$x+c\mathbb Z\in b\mathbb Z/c\mathbb Z$$
So $b\mathbb Z/c\mathbb Z$ can be recognized as the kernel of $\phi$.