I have a question regarding the application of the First Isomorphism Theorem for Groups in proofs; why are the proofs not concerned with whether the respective map is injective?
To clarify my question, let me state the First Isomorphism Theorem:
Let $G$ and $H$ be two groups and let $\phi: G \mapsto H$ be a homomorphism. Then, the First Isomorphism Theorem states that $ G/ Ker(\phi) \cong Im(\phi) $.
Then, to prove that $ Im(\phi) = H $ we have to show that $ \phi $ is surjective. This implies that $ G/ Ker(\phi) \cong H $.
My question then is: why are we not worried about whether $\phi$ is injective? Isn't an isomorphism a bijective homomorphism? Or is it that the quotient group $ G/ Ker(\phi) $ eliminates the need for the injectivity of $ \phi $?.
I wonder whether the fact that $\phi$ is injective $\implies G/Ker(\phi) = G$ has something to do with my question. Does the First Isomorphism Theorem allow us to say more things about homomorphisms that are not injective i.e whose kernel is not trivial?
Thus, I believe that the First Ismorphism Theorem essentially has three iterations:
- $\phi: G \mapsto H$ is a homomorphism $ \implies G/ Ker(\phi) \cong Im(\phi) $.
- $ \phi$ is a surjective homorphism $\implies G/ Ker(\phi) \cong H $.
- $\phi$ is an injective and surjective homomorphism i.e an isomorphism $\implies G/ Ker(\phi) = G \cong H $
Does my reasoning make sense? Thank you!
Best Answer
We are worried about injectivity of $\phi$! That's why we take the quotient by the kernel! $\phi$ need not inject on $G$, but the induced map from the quotient $G/\ker\phi$ - where we have killed off all the elements that contradict injectivity of $\phi$ ! - is always injective. This is true if we took $G/N$ for some normal $N$ containing $\ker\phi$ as well; however, if we took a strictly larger quotient then the image of $G/N\to H$ would be strictly smaller than $\mathrm{im}(\phi)$. So, killing off $\ker\phi$ in the quotient is exactly the correct thing to do.
Points in the image of $\phi$ are identifiable with points in $G$... up to a factor of $\ker\phi$, which is worrying (and we are worried) if $\phi$ isn't injective. "The" solution is to set this factor to be zero by passing to $G/\ker\phi$, which now gives a genuine $1$-$1$ identification.