What is the intuition behind the definition of the kernel of a homomorphism

Question

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?

Answer 1

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 $\iota(S_3)$.

Answer 2

Here's one of the biggest reasons: $f(G)\cong G/\ker(f)$ for any group homomorphism.

Answer 3

There are so many beautiful results especially in Sylow theory that make use of the kernel of a homomorphism, but since you are just starting it can be a way of showing that two groups are isomorphic. You will soon encounter the First Isomorphism Theorem.

$G_1 \xrightarrow{\phi} G_2$ where $\phi$ is homomorphism

$G/\ker{\phi} \cong \textrm{Im}(\phi)$

Your next big adventure will be when you use this to show that every finite group is isomorphic to a subgroup of $S_n$ by considering the left multiplication of $G$ on itself. Such an action gives rise to a permutation representation $\theta$ which maps $G$ to $S_n$.

The quotient of $G$ with the kernel of this map is an imbedding into $S_n$. The action is faithful which means the kernel is trival which means $G/\ker(\theta) = G \cong \textrm{Im}(\theta) \leq S_n$. Please, this is already too much just wait for the build up, algebra is wonderful!