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.
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.
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.
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.