Show that a function from $\Bbb Z / 3600\mathbb{Z}$ to $\mathbb{Z} / 1200\mathbb{Z}$ is a well defined group homomorphism of additive groups So, my question is: 
"Show that
\begin{align}
f : \mathbb{Z}/3600\mathbb{Z} &\longrightarrow \mathbb{Z}/1200\mathbb{Z}\\
[x]_{3600} &\longmapsto [88x]_{1200}
\end{align}
is a well defined group homomorphism of additive groups, then determine kernel and image of $f$ and their cardinality."
Whaaat? I really don't know where to start!
Thanks for helping me! I know what is a kernel, a homomorphism and whole group theory part, but the first part is really hard for me to see.
 A: Hints:  Well-definedness and that it's a homomorphism I leave to you.
Use the first isomorphism theorem, to get $\mid\operatorname{Im}\varphi \mid=150$ and $\mid\operatorname{ker}\varphi\mid=24$.  Also,  $\operatorname{ker}\varphi =[150]$.  This follows since $\varphi (x)=0\implies 88x\cong0\pmod {1200}\implies x=150$.
Finally  $\operatorname{Im}\varphi\cong \Bbb Z_{150}$.
Now, what is the kernel?
A: If the function is well defined it is certainly a group homomorphism (with respect to addition). So it makes sense to use the homomorphism theorems.
The map $g\colon\mathbb{Z}\to\mathbb{Z}/1200\mathbb{Z}$, $g(x)=[88x]_{1200}$ is certainly a group homomorphism. Its kernel consists of the integers $x$ such that $88x\in1200\mathbb{Z}$, which is equivalent to $11x\in150\mathbb{Z}$; since $\gcd(11,150)=1$, this is also equivalent to $x\in 150\mathbb{Z}$.
Since $3600\mathbb{Z}\subseteq 150\mathbb{Z}$, the homomorphism theorems guarantee that the map
$$
[x]_{3600}\mapsto g(x)=[88x]_{1200}
$$
is well defined.
The homomorphism theorem I'm referring to is: if $\varphi\colon G\to G'$ is a group homomorphism and $H$ is a normal subgroup of $G$ such that $H\subseteq\ker\varphi$, then there exists a unique group homomorphism $\psi\colon G/H\to G'$ such that $\psi(xH)=\varphi(x)$, for every $x\in G$.
A: Hint:
To show $f$ is well defined, you have to show that if $x'$ is another representative of the congruence class of $x\bmod 3600$, then $88x'$ belongs to the same congruence class $\bmod 1200$ as $88x$.
Then you have to show it is compatible with additions, $\bmod 3600$ and $\bmod 1200$ respectively.
