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?
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    $\begingroup$ Related: Intuition on group homomorphisms $\endgroup$ Commented May 26, 2015 at 23:00
  • $\begingroup$ Do you know about normal subgroups and quotient groups? Because its importance really starts to be clear when you consider those things. $\endgroup$ Commented May 26, 2015 at 23:00
  • $\begingroup$ @Meelo I don't know about them in detail. If you have any recommended sources to learn them, it would be awesome! $\endgroup$ Commented May 26, 2015 at 23:09
  • $\begingroup$ @ZevChonoles I will definitively take a look at that! Thanks for the link. $\endgroup$ Commented May 26, 2015 at 23:09
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    $\begingroup$ I found this video really helpful, since it explains the kernel in a visual and intuitive way youtube.com/watch?v=TngePpJ_x-I "The Kernel of a Group Homomorphism – Abstract Algebra" (Socratica) Matthew Salomone youtube.com/watch?v=rvhrngjKWzw is also an amazing teacher who is worth checking out! $\endgroup$
    – smokeink
    Commented Apr 17, 2018 at 3:29

3 Answers 3


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 $H$.


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 $\iota(S_3)$.

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    $\begingroup$ I see that you left the easy direction of the proof for me to fill. How nice of you :p thanks for the response. For future readers: HINT: consider f(gk) and f(g') for the other direction. $\endgroup$ Commented May 27, 2015 at 3:19
  • $\begingroup$ when you say "how respectful" the given homomorphism is, you mean that, the more elements in the kernel, the more multiplying by those elements will disappear in the image space of f, hence, not respecting the original space? $\endgroup$ Commented May 27, 2015 at 3:25
  • $\begingroup$ I see, so that is what $f(G)\cong G/\ker(f)$ means, right? $\endgroup$ Commented May 27, 2015 at 3:27
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    $\begingroup$ @Pinocchio Good question! (I've also made an edit to explain the "respectful" bit a little more) This isn't quite all there is to $f(G) \cong G/\ker f$, but it is enough to show that the size of $f(G)$ is $\frac{|G|}{|\ker f|}$. We would still need to show that there's a well-defined operation on $G/\ker(f)$, and describe the map from $f(G)$ to the quotient, but it should get us pretty close. $\endgroup$
    – pjs36
    Commented May 27, 2015 at 4:04

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

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    $\begingroup$ Or in words: "the image of the group under the homomorphism is essentially the same as the quotient group obtained by quotienting out the kernel." So using a homomorphism and forming a quotient group are kind of the same thing. $\endgroup$ Commented May 26, 2015 at 23:04

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!


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