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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.

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1 Answer 1

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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 $H\leq\ker f$.

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    $\begingroup$ 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

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