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I am looking for a necessary and sufficient condition for a subgroup $K$ of a group $G$ to be kernel of a homomorphism $\phi$ from $G$ to $G$. The tools that come into my mind is first isomorphism theorem (that $\frac{G}{\operatorname{kernel} \phi}$ is isomorphic to $\operatorname{Im}(\phi)$) and also that $\operatorname{kernel} \phi$ is normal subgroup of $G$. I don't know how to combine these two to get a necessary and sufficient condition?

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  • $\begingroup$ $K$ is the kernel of a group homomorphism iff $K$ is normal $\endgroup$
    – eepperly16
    Jun 15, 2016 at 2:49
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    $\begingroup$ @eepperly16: note that the question asks for a condition on $K$ to be a kernel of a homomorphism from $G$ to $G$, so the answer is a bit different. It seems to me that it is necessary and sufficient for $K$ to be a normal subgroup such that $G/K$ is isomorphic to a subgroup of $G$. $\endgroup$ Jun 15, 2016 at 2:52

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Let $G$ be a group, and let $K \subset G$ be a subgroup. I claim that $K$ is the kernel of some homomorphism $\varphi \colon G \to G$ if and only if $K$ is normal, and $G/K$ is isomorphic to a subgroup $K'$ of $G$. Indeed, if $K = \ker(\varphi)$ for some homomorphism $\varphi \colon G \to G$, then $K$ is normal and by the first isomorphism theorem, $G/K \cong \mathrm{im}(\varphi)$, which is a subgroup of $G$. On the other hand, if $G/K$ is isomorphic to the subgroup $K'$ of $G$, then the composition of the projection $G \to G/K$ with the isomorphism from $G/K$ to $K'$ and the inclusion $K' \hookrightarrow G$ gives a homomorphism from $G$ to $G$ with kernel $K$.

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  • $\begingroup$ @eepperly16: yes, thanks. $\endgroup$ Jun 15, 2016 at 3:06
  • $\begingroup$ Can you present an example when there does not exist a subgroup $K'$ isomorphic to the quotient group? $\endgroup$
    – eepperly16
    Jun 15, 2016 at 3:06
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    $\begingroup$ @eepperly16: Sure. Take $G = \mathbb{Z}$, and let $K$ be any proper subgroup of $G$. Then $G/K$ is finite and nonzero (isomorphic to $\mathbb{Z}/n\mathbb{Z}$ for some $n$), but the only finite subgroup of $\mathbb{Z}$ is the trivial subgroup. $\endgroup$ Jun 15, 2016 at 3:08
  • $\begingroup$ In the last projection, you have concluded that $K'\hookrightarrow G$ while the composition of the given projections is $G\hookrightarrow\frac{G}{K}\rightarrow K'$ which gives $G\rightarrow K'$ . The inclusion you have constructed needs the projection $\frac{G}{K}\rightarrow G$. Do we have such an inclusion? $\endgroup$
    – user160492
    Jun 15, 2016 at 4:04
  • $\begingroup$ @user160492: I'm confused by what you're asking. The point of having a subgroup isomorphic to $G/K$ is being able to include the isomorphic copy of $G/K$ into $G$. Explicitly, we have a sequence of maps $G \to G/K \to K' \to G$. The first map is the canonical quotient map, the second map is the isomorphism between $G/K$ and $K'$, and the third map is the inclusion of $K'$ into $G$. The latter two maps are injective, so the composition has kernel $K$. $\endgroup$ Jun 15, 2016 at 4:15

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