What is the quotient group o mod the identity

Question

I am trying to prove something using the first isomorpism theorem, which basically says if $G,H$ are groups, and $f:G\to H$ is a group homomorphism, then $G/ker(f)\cong f(H)$. In this case, suppose $f$ is an epimorphism meaning $f(G)=H$.
I want to show that two groups are isomorphic, so one way I tried to do this was to find a homormorphism with trivial kernel. Which would then give $G/ \{e_G\} \cong H$ Does this even make sense? I think that $G/\{e_G\}$ is just $G$ but I have a feeling this doesn't make sense.

Any help is greatly appreciated.

Answer 1

What is an isomorphism? Is is an injective epimorphism, surjective monomorphism or an injective and surjective homomorphism. Let us take a very simple one for the theorem, the identity homomorphism, $id$. By the isomorphism theorem we have $$id(G)/\ker(id)\cong G$$ however we have $id(G)=G$ and $\ker(id)=e$ and as such we get $$G/e\cong G$$

Answer 2

You are correct in that if $f:G\to H$ is a surjective homomorphism (epimorphism) then $G/\text{Ker} f \cong H$. It is perfectly valid to find a map $f$ with $\text{Ker}f = \{e_G\}$. The then valid conclusion is that $G /\{e_G\} \cong G \cong H$.

It is worth noting however that in your search for such a map, showing that its kernel is trivial is the same as showing that its injective, and showing that that its an epimorphism guarantees surjectivity. So really the isomorphism theorem here is overkill, as the process to verify the hypotheses will prove your claim without the actual result of the first isomorphism theorem.