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

Question

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?

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I would appreciate any help/thoughts!

Answer 1

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