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

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!

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

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