First Isomorphism Theorem Theorem: Let $G$ and $G'$ be groups  and let $f:G\to G'$ be a group homomorphism. Then
$G/\textrm{ker}\, f  \cong\textrm{im}\, f$. 
My question is how to understand this theorem  intuitively.
 A: The intuition is that the group looks similar to its image under the homomorphism when it is divided by a certain subgroup, since distinct elements may get mapped to the same element. By taking the quotient, elements which are mapped to the same are grouped together.
A: Maybe it gets more intuitive if you look at the situation with groups replaced by sets. 
If $f : M \rightarrow N$ is a map between sets $M$ and $N$, then you get an equivalenve relation $R_f := \{(x,y) \in M \times M \::\: f(x) = f(y)\}$.
The equivalence class of $x \in M$ is $f^{-1}(\{f(x)\})$, so we have a well-defined (and injective) map from $M/R_f$ (the set of equivalence classe) to $N$, sending $f^{-1}(\{f(x)\})$ to $f(x)$.
Thus, first collecting all elements, which map to the same element via $f$, and then mapping these collections to their respective values gives you a bijection onto the image.
Note that in the case of groups $G$ and $G'$ and group homomorphism $f: G \rightarrow G'$ you have $f^{-1}(\{f(x)\}) = x\ker(f)$ for all $x \in G$ and the induced map is a group homomorphism, yielding an isomorphism to the image.
