# Is every normal subgroup the kernel of some self-homomorphism? [duplicate]

Let $$G$$ be a group. If there is a homomorphism $$f:G\to G$$ (special case of the codomain being arbitrary group), then the kernel $$f^{-1}(id)$$ is a normal subgroup of $$G$$.

But now the other way around: Start out with the existence of a normal subgroup $$H$$ of $$G$$. Is there necessarily a homomorphism $$f:G\to G$$ such that the kernel of $$f$$ is $$H$$?

• I edited the title to clarify the question. Let me know if you agree with it. Mar 1, 2019 at 12:09
• @DietrichBurde that’s fine thanks! Mar 1, 2019 at 12:25
• Mar 1, 2019 at 14:09
• A more interesting question would be: for which groups is this true? Trivially true for simple groups and also true for finite cyclic groups. Are there any others? Mar 1, 2019 at 14:55

If $$G=\mathbb{Z}$$ then any homomorphism $$f:\mathbb{Z}\rightarrow\mathbb{Z}$$ takes the form $$f(a)=ma$$ for some $$m\in\mathbb{Z}$$. Clearly the kernel is trivial (unless $$m=0$$ then the kernel is everything). However for all $$n\in\mathbb{N}$$ we have that $$n\mathbb{Z}$$ is a normal subgroup of $$\mathbb{Z}$$. In particular $$2\mathbb{Z}$$ is not a kernel of any homomorphim from $$\mathbb{Z}$$ to itself.
However, it is possible to "correct" this statement. If you only request an homomorphism from $$G$$ to some other group. Given any normal subgroup $$H$$, the quotient homomorphism $$f:G\rightarrow G/H$$ which sends $$g\mapsto g+H$$ has $$H$$ as it's kernel. In other words, every normal subgroup is a kernel of some homomorphism, not necessarily from $$G$$ to itself.