Let $H$ be a normal subgroup of order $6$. If $f :G\longrightarrow G_1$ be an epimorphism of groups such that $H\subset \ker(f)$, then show that $G_1$ is also a homomorphic image of $G/H$.
My solution goes as follows:
We consider, $\phi:G/H\longrightarrow G_1$, such that $\phi(Hg)=f(g)$. Then, we see that, $\forall Hg_1,Hg_2\in G/H$, $\phi(Hg_1Hg_2)=f(g_1g_2)=f(g_1)f(g_2)=\phi(Hg_1)\phi(Hg_2)$. So, $\phi$ is a homomorphism. Also, $\forall p\in G_1$,$\exists g\in G$, such that $f(g)=p$ and hence $\phi(Hg)=f(g)=p$. Hence, $f$ is onto. Therefore, $G_1$ is also a homomorphic image of $G/H$.
Is the above solution, correct? If not, where is it going wrong?