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The book on abstract algebra that I'm reading uses the fact that, given the groups $G$ and $G'$ and a homomorphism $\phi:G \rightarrow G'$, then

$$ \operatorname{im}( \phi ) \cong G/\ker( \phi )$$

However the author doesn't provide a proof and simply states that it follows from "standard group theory". How could one prove the theorem above?


I would appreciate any help/thoughts!

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    $\begingroup$ This is the first isomorphism theorem and its proof is contained in virtually every elementary abstract algebra textbook. $\endgroup$
    – user296602
    Oct 1, 2018 at 23:46

1 Answer 1

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Here is a roadmap for the proof, which appears in all books:

Define $\pi : G/\ker( \phi ) \to \operatorname{im}( \phi )$ by $\pi(x \bmod \ker \phi) = \phi(x)$. Then prove:

  • $\pi$ is well defined

  • $\pi$ is a group homomorphism

  • $\pi$ is injective

  • $\pi$ is surjective

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