Suppose $f\colon G\to G^{\prime}$ is a group homomorphism. Let us denote the groups additively.

It is well know that such a homomorphism always can be 'factored through' the quotient $G/\operatorname{ker}(f)$ by setting $\overline{f}(g+\operatorname{ker}(f))=f(g)$ and this is well defined because $f(g)=f(\overline{g})$ if and only if $g-\overline{g}\in\operatorname{ker}(f)$, which means $g+\operatorname{ker}(f)=\overline{g}+\operatorname{ker}(f)$.

Now, some book I am following had a homomorphism of groups $f$ defined and had a normal subgroup $H$ of $G$, which was inside $\operatorname{ker}(f)$. They are saying that we can factor $f$ through the quotient $G/H$. Now, I'm not sure how they can say that since for this to be well defined we would need that $f(g)=f(\overline{g})$ if and only if $g-\overline{g}\in H$, but we only know that $f(g)=f(\overline{g})$ if and only if $g-\overline{g}\in\operatorname{ker}(f)$, right? How can they do such a thing?

What is going on here? Thanks a lot.

I suspect that this could be something particular about the homomorphism in question, but I prefer to ask about all groups to see if this is some general phenomena I am missing.


1 Answer 1


If $g+H=g'+H\implies f(g)=f(g')$ or equivalently if $g-g'\in H\implies g-g'\in\ker f$ then the map prescribed by: $$g+H\mapsto f(g)$$ is well defined.

It is evident that this is the case under the condition $H\leq\ker f$.

  • 1
    $\begingroup$ Oh, I see. I realize now that we don't need if and only if. The failure on the other side only gives that it will not be injective in general. This was obvious. Thanks! $\endgroup$
    – Shoutre
    May 5, 2016 at 13:24
  • $\begingroup$ You are welcome. $\endgroup$
    – drhab
    May 5, 2016 at 13:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.