# Question about the proof of the Fundamental Theorem of Homomorphisms

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?

• I'm assuming $N = \ker(\phi)$? Apr 22 '21 at 16:11
• @Randall yes, my bad Apr 22 '21 at 16:13
• 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 $a\in G$. Apr 22 '21 at 16:17

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 $$\phi'$$.
• If $aN = bN$ then $G/N$ contains both since they're the same thing. Also, $G/N$ contains ALL cosets of the form $aN$. Apr 22 '21 at 16:17
• @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. Apr 22 '21 at 16:29
• You seem to be arguing something different, like possibly the well-defined-ness of $\phi'$. Apr 22 '21 at 16:32