# Not each subgroup $H\subset G$ can be the kernel of a homomorphism

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?

• They meant a kernel of a homomorphism with domain $G$. – MatheinBoulomenos Mar 18 '17 at 12:35
• Then we do they consider the inclusion mapping, that also doesn't have domain $G$? @MatheiBoulomenos – Sha Vuklia Mar 18 '17 at 12:36
• 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. – Bernard Mar 18 '17 at 12:36
• The inclusion mapping has domain $H$. – Bernard Mar 18 '17 at 12:36
• 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 $H$. – Bernard Mar 18 '17 at 12:46

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 $G\rightarrow G/H$.