Given two groups $G,H$ and a group homomorphism $\varphi:G\to H$ its well-known that

$\varphi$ is injective if and only if the kernel $\ker\varphi$ is trivial.

In order to derive a similiar criterion for $\varphi$ being surjective (I think) I was able to prove the following proposition

$\varphi$ is surjective if and only if the quotient group $H/{\rm im}~\varphi$ exists.


If $\varphi$ is surjective, then ${\rm im}~\varphi=H$. Therefore the quotient group of our interest is given by $H/{\rm im}~\varphi=H/H\cong\{e\}$, in particular the quotient exists.
Now assume that $H/{\rm im}~\varphi$ exists. Then ${\rm im}~\varphi$ is a normal subgroup and as such kernel of a homomorphism with domain $H$. Let $G'$ be group and $\psi:H\to G'$ such that $\ker\psi={\rm im}~\varphi$. Composition yields the homomorphism $\psi\circ\varphi:G\to G'$ and by definition $\ker(\psi\circ\varphi)=G$. By the First Isomorphism Theorem we have $G/\ker(\psi\circ\varphi)\cong\{e\}\cong G'$. But then $\ker\psi=H$, hence by construction $\ker\psi=H={\rm im}~\varphi$. The result follows.

Is my argumentation sound; if so: why do I fail to locate a source actually stating this (occasionally) useful proposition? If not, where did I went wrong?

Thanks in advance!


From the comments I realised that I have overlooked a crucial part: if $H$ is abelian, then $H/{\rm im}~\varphi$ always admits a group structure; regardless of $\varphi$ being surjective as in an abelian group every subgroup is normal. So I would like to rephrase the claimed proposition.

Let $G,H$ be groups and consider $H$ to be non-abelian. A group homomorphism $\varphi:G\to H$ is surjective if and only if the coset $H/{\rm im}~\varphi$ is a group.

  • 3
    $\begingroup$ The quotient $H/\operatorname{im} \varphi$ exists independently of $\varphi$ being surjective. It's called the cokernel of $\varphi$. And the criterion you might be looking for is: $\varphi$ is surjective if and only if the cokernel of $\varphi$ is trivial. $\endgroup$
    – user291350
    Commented Dec 29, 2019 at 18:13
  • 1
    $\begingroup$ @Edu W. But $H/{\rm im}~\varphi$ hasn't to be group if the image isn't a normal subgroup of $H$; and the image of a $G$ under a group homomorphism is in general not a normal subgroup in $H$. Maybe I expressed myself badly: of course, the coset existence of the coset $H/{\rm im}~\varphi$ is independent of $\varphi$, but not this coset being a group. Anyway, thanks for the input! $\endgroup$
    – mrtaurho
    Commented Dec 29, 2019 at 18:17
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    $\begingroup$ What you wrote is false: take $H$ to be abelian but $\phi$ not onto. Your mistake is where you invoke the first iso theorem: we do not have that $G\prime$ is trivial. $\endgroup$ Commented Dec 29, 2019 at 18:29
  • $\begingroup$ @AndrewHubery I see. Didn't thought about the crucial difference between abelian and non-abelian groups when talking about the image. Beside that, how does the application of the First Isomorphism Theorem fails here? $\endgroup$
    – mrtaurho
    Commented Dec 29, 2019 at 18:32
  • $\begingroup$ Would the downvoter please elaborate on his issue with this question? $\endgroup$
    – mrtaurho
    Commented Mar 17, 2020 at 16:06

1 Answer 1


Your claim is false. Take $G$ to be arbitrary and take $H$ to be nontrivial. let $\psi\colon G\to H$ be the trivial map, $\phi(g)=e_H$ for all $g\in G$. Then $\mathrm{Im}(\phi)=\{e\}\triangleleft H$, so the quotient exists. But the map is very nonsurjective.

More generally, if $H$ is any nontrivial group and $N\triangleleft H$ is any proper normal subgroup (they always exist, since you can take $N=\{e\}$) then the embedding $i\colon N\hookrightarrow H$ is non-surjective, but $H/\mathrm{Im}(i) = H/N$ exists.

The real reason your argument fails is that while it is true that your composition has $G$ as the kernel, the isomorphism theorem only guarantees that the image of $\psi\circ\phi$ is isomorphic to $G/\mathrm{ker}(\psi\circ\phi)$. And by construction, that image is trivial, so there is no surprise that $\mathrm{ker}(\psi\circ\phi) = G$. You have no warrant for claiming that the image is all of $G'$. You can require (but did not do so) that $G'$ be equal to the image of $\psi$, but $\mathrm{Im}(\psi\circ\phi)\subseteq \mathrm{Im}(\psi)$, and equality need not hold. We know the equality always holds when $\phi$ is surjective... but that is what you are trying to prove, so you cannot assume it holds.

For abelian group, and more generally for modules, there is the dual concept to the kernel called the cokernel; given $f\colon M\to N$, the cokernel of $f$ is $\mathrm{coker}(f)=N/\mathrm{Im}(f)$. It is in fact the case that $f$ is surjective if and only if the cokernel is trivial, just like $f$ is injective if and only if the kernel is trivial. This does not work for arbitrary groups, though, because the image need not be normal. If you quotient out by the normal closure of the image you get the equivalent construction but it no longer "detects" surjectivity (because surjectivity is not a categorical notion).

  • $\begingroup$ At least this explains why I wasn't able to find my "proposed" proposition as it's just meaningless... Anyway, thank you kindly! $\endgroup$
    – mrtaurho
    Commented Dec 29, 2019 at 19:02
  • $\begingroup$ Thank you for the additional information! $\endgroup$
    – mrtaurho
    Commented Dec 29, 2019 at 19:20

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