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Theorem: Let $\phi: G \rightarrow H$ be a group homomorphism. Then $\operatorname{Ker}(\phi)$ is a normal subgroup of G. $\phi(G)$ is a subgroup of H and the map $\phi'$: $G/\operatorname{Ker}(\phi) \rightarrow \phi(G)$, $a\operatorname{Ker}(\phi)\mapsto \phi(a)$ is a group isomorphism.

My question relates to the part where it is shown that $\phi'$ is surjective. From the proof: Because of $\phi'(G/N)=\phi(G)$, $\phi'$ is obviously surjective. So why do we know that $\phi'(G/N)=\phi(G)$ holds?

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  • $\begingroup$ I'm assuming $N = \ker(\phi)$? $\endgroup$
    – Randall
    Apr 22 '21 at 16:11
  • $\begingroup$ @Randall yes, my bad $\endgroup$
    – WanyM
    Apr 22 '21 at 16:13
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    $\begingroup$ The definition of $\phi'$ should be $a\ker(\phi)\mapsto \phi(a)$, not "$\mapsto \phi(G)$". Using that, the claim is indeed obvious since it means that $\phi'(a\ker(\phi)) = \phi(a)$ for all $a\in G$. $\endgroup$
    – Christoph
    Apr 22 '21 at 16:17
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Anything in $\phi(G)$ takes the form $\phi(a)$ for $a \in G$, and the coset $aN$ will map to $\phi(a)$. Check the definition of $\phi'$.

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  • $\begingroup$ But isn't G/N:={aN:a in G}, and if there exists a,b in G such that aN=bN, then the set G/N should just contain one of them? $\endgroup$
    – WanyM
    Apr 22 '21 at 16:16
  • $\begingroup$ That isn't a question of it being onto, though. $\endgroup$
    – Randall
    Apr 22 '21 at 16:17
  • $\begingroup$ If $aN = bN$ then $G/N$ contains both since they're the same thing. Also, $G/N$ contains ALL cosets of the form $aN$. $\endgroup$
    – Randall
    Apr 22 '21 at 16:17
  • $\begingroup$ @WanyM Talking about "just one of them" is very misleading when there is just one thing. Note that $aN$ and $bN$ are just two names for the same thing when $aN=bN$. But this is not related to surjectivity of $\phi'$ at all. $\endgroup$
    – Christoph
    Apr 22 '21 at 16:29
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    $\begingroup$ You seem to be arguing something different, like possibly the well-defined-ness of $\phi'$. $\endgroup$
    – Randall
    Apr 22 '21 at 16:32

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