If f and g are two epimorphisms of a group G onto a group H such that ker f = ker g, then is f = g necessarily?
(I can think of disproving it by taking the specific case of kernel being null so that homomorphism becomes one-one, is the approach correct? Can this be disproven without a particular example? P.S.: I don't know the answer)
Let $G=H$ and take $f$ and $g$ automorphisms of $G$. Then clearly $\ker(f)=\ker(g)=1$, the trivial group, however not $f=g$ in general. For example, consider the automorphism group of $S_3$. It has $6$ elements.
Proving that Aut($S_3$) is isomorphic to $S_3$
Finding the automorphisms of $S_3$ by looking at the orders of the elements
Your reasoning is right, and there's no way to disprove this without a specific example.
To use your idea, you can compare the identity map to a nontrivial automorphism on $G$, which are obviously not equal but have the same kernel. You still have to exhibit this automorphism though, because you need to produce a specific counterexample to disprove this.
