Understanding homomorphism and kernels Let $G$ be a group and $\phi$ a Homomorphism
$$
\phi:G\to G'
$$
Now I know that the size of the kernel tells you how many elements in $G$ map to the same element in $G'$
I couldn't find this in my book, but I have concluded the following. 
$$
\frac{|G|}{| \:\text{ker} \: \phi \:|} = |G'|
$$
Is that true?

It's true only if $\phi$ is surjective. Notice that we could let $\phi$ be the trivial homomorphism, sending every element of $G$ to the identity of $G'$. Then $\ker \phi = G$ and we are claiming that
$$\frac{|G|}{|\ker \phi|}=1=|G'|$$
for every group $G'$. This is obviously false.
However, if $\phi$ is surjective, then, clearly, the set $S_y$ of points $x\in G$ such that $\phi(x)=y$ clearly has that $|S_y|=|\ker \phi|$, since, for any $z\in S_y$, it clearly holds that $S_y \subseteq z\ker \phi$ and that $\ker \phi \subseteq z^{-1} S_y$ where the multiplication is taken to be a coset - this holds since, if $k\in \ker\phi$ then $zk\in z\ker\phi$ and $\phi(zk)=\phi(z)\phi(k)$, which is just $y$, since $\phi(z)=y$ and $\phi(k)$ is the identity. A similar argument holds for the other expression.
Thus, each element $y$ in $G'$ is associated to the set $S_y$, which has size equal to $|\ker \phi|$. Furthermore the $S_y$ partition $G$, implying the sum of their sizes is the size of $G$. Since there are $|G'|$ distinct $S_y$ and each has the same size as the kernel, $$|G'|\cdot |\ker \phi| = |G|.$$
