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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 $H$?

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  • $\begingroup$ I edited the title to clarify the question. Let me know if you agree with it. $\endgroup$ Mar 1, 2019 at 12:09
  • $\begingroup$ @DietrichBurde that’s fine thanks! $\endgroup$
    – user56834
    Mar 1, 2019 at 12:25
  • $\begingroup$ See also math.stackexchange.com/questions/1826655/… $\endgroup$
    – Arnaud D.
    Mar 1, 2019 at 14:09
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    $\begingroup$ 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? $\endgroup$ Mar 1, 2019 at 14:55

1 Answer 1

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That's false.

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.

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