3
$\begingroup$

My book says that each subgroup $H\subset G$ can be the image of a homomorphism; consider the inclusion mapping $f\colon H\to G\colon x\mapsto x.$ However, not each subgroup can be the kernel of a homomorphism.

I don't see why this should be a problem. Consider $H\subset G$ a subgroup. We know that $H$ is not empty. Now consider the mapping $f\colon H\to G:x\mapsto e$. This way, $H$ is the kernel of $f$, and $f$ is a homomorphism, because $f(a+b)=e=e+e=f(a)+f(b)$.

So why did they say that not each subgroup can be the kernel of a homomorphism?

$\endgroup$
  • 5
    $\begingroup$ They meant a kernel of a homomorphism with domain $G$. $\endgroup$ – MatheinBoulomenos Mar 18 '17 at 12:35
  • $\begingroup$ Then we do they consider the inclusion mapping, that also doesn't have domain $G$? @MatheiBoulomenos $\endgroup$ – Sha Vuklia Mar 18 '17 at 12:36
  • 1
    $\begingroup$ The assertion means not each subgroup of a group $\color{red}G$ is the kernel of a homomorphism from $\color{red}G$ to another group. $\endgroup$ – Bernard Mar 18 '17 at 12:36
  • 1
    $\begingroup$ The inclusion mapping has domain $H$. $\endgroup$ – Bernard Mar 18 '17 at 12:36
  • 1
    $\begingroup$ It's only that your phrasing is not complete: it should begin with Let $G$ be group. […] . So we know the assertion is about a group (named $G$ in the assertion, but that is unimportant), which is the domain of a homomorphism. When you speak of the inclusion morphism $H\subset G$, the domain is $H$ by definition, that's all. And indeed, you observe $H$ is the kernel of a homomorphism with domain $H$. $\endgroup$ – Bernard Mar 18 '17 at 12:46
2
$\begingroup$

What is meant is that there does not exist a homomorphism $f:G\rightarrow K$ such that $\text{ker } f = H$. In fact, such an $f$ exists iff $H$ is normal. To see this, note that the kernel of a homomorphism is a normal subgroup. Conversely, if $H$ is normal, one can take the canonical projection $G\rightarrow G/H$.

$\endgroup$

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.