TITLE: What is the intuition behind the definition of the kernel of a homomorphism
QUESTION [7 upvotes]: I was starting to study some algebra (groups and homomorphisms in particular) and came across the definition of the kernel (for a group-homomorphism $f:G \rightarrow G'$):
$$\ker(f) = \{ x \in G \mid f(x) = e' \}$$
where $e'$ is the identity. I understand the definition but fail to appreciate its importance or what it means intuitively or the motivation to define such a set/object. So far I understand that it's the set of all input elements that once processed by our homomorphism, yield the identity under the other group $(G', e')$, however, I fail to really appreciate why mathematicians would bother to define (and give it a name) to such a set. 

What is the motivation to define such as set" If it has a name, it must have some importance. 
Why is it so special to map to the identity element of a set? What structure does it reveal about the relationship of both sets? What is the intuition? Why would mapping to the identity be useful?
What is the conceptual understanding of it beyond regurgitating/repeating the definition?

REPLY [9 votes]: There are many many reasons why kernels are important to group theory, but here's just one way of appreciating the kernel in a fairly isolated context.
If we zoom out a bit, any set-function $f: A \to B$ (here $A$ and $B$ are simply sets) naturally partitions $A$ into equivalence classes, and for $a \in A$, the equivalence class of $a$ is given by
$$[a] = \{a' \in A : f(a') = f(a)\};$$
the set of all elements of $A$ that get mapped to the same thing as $a$. The same logic applies if $f : G \to H$ isn't just a set function, but a homomorphism of groups.
With the equivalence class notation, the kernel of $f$ is simply the equivalence class of the identity $1_G$ of $G$,
$$\ker f = [e_G],$$
since any homomorphism $f: G \to H$ always sends the identity $e_G$ of $G$ to the identity $e_H$ of $H$. What can we say about arbitrary $g, g' \in G$ such that $f(g) = f(g')$? That is, what can we say about the equivalence class $[g]$ for any $g \in G$?

Claim: For a homomorphism $f: G \to H$ and $g \in G$, we have $f(g) = f(g')$ if and only if there exists some $k \in \ker f$ such that $gk = g'$; that is, $g$ and $g'$ differ by a multiple of something in the kernel of $f$. In particular, $[g] = \{gk: k \in \ker f\}$, and has size $|\ker f|$.
($\Longrightarrow$) Supposing $f(g) = f(g')$, note that there exists a unique $g^* = g^{-1}g' \in G$ such that $gg^* = g'$. Then
$$f(g) = f(g') = f(gg^*) = f(g)f(g^*),$$
and left-multiplication by $f(g)^{-1}$ shows that $e_H = f(g^*)$, hence $g^* \in \ker f$.
($\Longleftarrow$) Homework.

If you've ever heard homomorphisms described as functions that "respect" the group operation(s), the size of the kernel is a measure of just how "respectful" a given homomorphism is! A large kernel means that more of the structure of the group $G$ is "ignored" when transported to the group $H$.
Edit:
For "respectful", imagine two situations, considering $S_3$, the symmetric group of degree $3$. There's a sign homomorphism $\operatorname{sgn} : S_3 \to \{-1, 1\} = C_2$ to the multiplicative group $C_2$ sending each permutation to its sign. Its kernel is the alternating group $A_3 = \{1, (123), (132)\}$ of "even" permutations, and in the image $C_2$, almost all of the structure of $S_3$ is ignored; we forget everything but whether a permutation is even or odd.
On the other hand, we have a "copy" of $S_3$ as a subgroup of $S_4$ if we consider all permutations of $S_4$ that leave $4$ fixed. This leads to an "inclusion" homomorphism $\iota: S_3 \to S_4$, sending each permutation to its "copy" in $S_4$. This inclusion homomorphism has only the identity of $S_3$ in its kernel, and is considerably more "respectful" than the sign homomorphism; every bit of information about $S_3$ shows up in the image $\iota(S_3)$.