11
$\begingroup$

I've done a lot of problems before but I am trying to get a really basic definition of kernel so that I may apply to any possible given question that I may be presented with.

Would I be correct in saying that the kernel (of a homomorphism) is basically what I can multiply any given function by to get the identity?

$\endgroup$
  • 2
    $\begingroup$ First thing is to get your statement grammatically correct. It makes no sense just to talk of "a kernel" in isolation, it must be "the kernel of a group homomorphism". $\endgroup$ – David Feb 20 '15 at 0:54
  • $\begingroup$ @David yes, I get in a lot of trouble with notation issues but what you both said is what I meant. I am trying to a clearer understanding of things so that I can be correct more often. Thank you $\endgroup$ – cele Feb 20 '15 at 0:59
  • $\begingroup$ Just to clarify: you would like the definition of "the kernel of a group homomorphism" correct? $\endgroup$ – Brian Fitzpatrick Feb 20 '15 at 1:14
19
$\begingroup$

Yes, sort of. The kernel of a group homomorphism $\phi:G\to H$ is defined as $$ \ker\phi=\{g\in G:\phi(g)=e_H\} $$ That is, $g\in\ker\phi$ if and only if $\phi(g)=e_H$ where $e_H$ is the identity of $H$.

It's somewhat misleading to refer to $\phi(g)$ as "multiplying $\phi$ by $g$". Rather, we use the language "applying $\phi$ to $g$" to emphasize that $\phi$ is a function between groups not an element of one of the groups in question.

Example. Note that $\Bbb Z$ and $\Bbb Z^2$ are groups under addition. Moreover, the identities are $e_{\Bbb Z}=0$ and $e_{\Bbb Z^2}=(0,0)$.

Let $\phi:\Bbb Z^2\to\Bbb Z$ be the group homomorphism defined by $\phi(a,b)=a+b$. Then $(a,b)\in \ker\phi$ if and only if $\phi(a,b)=0$. That is, $(a,b)\in\ker\phi$ if and only if $a+b=0$. Hence $(a,b)\in\ker\phi$ if and only if $b=-a$.

This proves that $\ker\phi=\{(a,-a):a\in\Bbb Z\}$.

As noted in the comments, kernels arise in lots of other contexts. If you're interested, see the "mathematics" section of the wikipedia entry for kernel.

If you're feeling extra ambitious, you could learn category theory and see how the kernel of a group homomorphism is a special case of an equalizer.

$\endgroup$
  • 1
    $\begingroup$ I think you should write the definition of the kernel of a ring homomorphism, too. It can sometimes be confusing to some people whether the kernel is the preimage of the multiplicative identity or additive identity. $\endgroup$ – layman Feb 20 '15 at 1:09
  • 2
    $\begingroup$ @user46944 How about I change $1_G$ and $1_H$ to $e_G$ and $e_H$. Seems useless to change the definition to ring homomorphism since OP tagged the question with "group theory". $\endgroup$ – Brian Fitzpatrick Feb 20 '15 at 1:10
  • 2
    $\begingroup$ Regardless of the tag, the question is asking what a kernel is. In this case, it wouldn't take much more work to explain that ring homomorphisms also have kernels, which are the things mapped to $0$, the additive identity. As the person answering the question, you typically have more knowledge than the person asking the question, and in my opinion should fill in such relevant details. $\endgroup$ – layman Feb 20 '15 at 1:11
  • 4
    $\begingroup$ @user46944 Ignoring the tag is tantamount to ignoring the question. Suppose OP is in a course that hasn't covered rings. Why should OP have to learn the definition of a ring, which is a different algebraic structure, when he or she could directly use his or her knowledge to understand the current definition? Also, why stop with rings? We could also include kernels of linear maps. Even more generally we could provide a definition of the kernel of module homomorphisms. I don't think further development is appropriate here. $\endgroup$ – Brian Fitzpatrick Feb 20 '15 at 1:13
  • $\begingroup$ If the OP doesn't understand what rings are, then they could easily ignore that part of your answer (it might help others, though). Is it really that much of a bother to add a sentence saying "if you are familiar with rings, then the kernel of a ring homomorphism is defined as the pre-image of the $0$ element, i.e., the additive identity"? $\endgroup$ – layman Feb 20 '15 at 1:14
5
$\begingroup$

To the OP: the other answers have given you the definition of the kernel of a group homomorphism only, probably because you tagged this question with group theory.

If you are familiar with ring theory (if not, you will be soon), we can have a homomorphism from a ring $R$ to a ring $S$. But a ring has both an additive identity, usually denoted $0$, and a multiplicative identity, usually denoted $1$. In this special case, the kernel of the homomorphism is defined as the stuff in $R$ that is mapped to $0$.

Also, if you have ever taken a linear algebra course, and know about linear transformations between vector spaces, the kernel of a linear transformation is the stuff in the domain that is mapped to the $0$ vector in the co-domain.

$\endgroup$
  • $\begingroup$ Since a ring is a group to begin with (as is a vector space and a module) homomorphisms of rings (and vector spaces and modules) are first homomorphisms of the underlying groups and their kernels are the kernels of those homomorphisms. So IMO, it is counterproductive and potentially confusing to talk about homomorphisms of these structures as some special case. (And of course some people like to work with rings that don't have a multiplicative identity.) $\endgroup$ – Stephen Meskin Dec 13 '17 at 7:41
4
$\begingroup$

If you're avoiding symbols, you could call it the preimage of the identity (of the target group), and think of it as "everything that gets sent to the identity (of the target group)".

$\endgroup$
  • $\begingroup$ Yes, I am looking for a very basic understanding of it. I've read too many definitions and I get very confused by the notation. I figure if I can first understand what it is, then I can work on noting it properly. $\endgroup$ – cele Feb 20 '15 at 1:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.