This is a problem which I am trying to solve. Let $H$ be a normal subgroup of order $6$. If $f:G\to G_1$ ba an epimorphism of groups such that $H\subseteq\ker f$.Then show that $G_1$ is also a homomorphic image of $G/H$.
My try: By using 1st Isomorphism Theorem $G/\ker f$ is Isomorphic to $G_1$. Now I define a mapping $g:G/H\to G_1$ by $g(aH)=f(a)$ for all $a\in G$. Clearly it's easy to verify that $g$ is well defined and homomorphism. Now we take any $x\in G_1$, since $f$ is an epimorphism we must have $b\in G$ such that $f(b)=x$. Hence from the construction of $g$ we get $bH\in G/H$.So $g$ is on to.So my problem is solved.But i think somehow I have done something wrong as I didn't use the order of $H$ and the fact $H\subseteq\ker f$, so my question is where did I wrong in my try, I will be happy if someone give me another approch to solve the above problem.
You did use $H \subset \text{ker}(f)$, you used it in the step where you proved that $g$ is well-defined.
And no, you did not use that $H$ has order $6$. So ask yourself, does your proof remain valid when $H$ has order $42$? Or does it remain the proof valid without any assumptions at all on the order of $H$? If so, then you are free to keep that hypothesis or drop it, the theorem will still be true. And every theorem gets better if you can remove unnecessary hypotheses.