Is it possible to make a commutative homomorphism image non-commutative? I have worked out a surjective group homomorphism $\varphi: G\to K$, where $G$ is non-Abelian but $K$ is Abelian. Now, my question is: Is it possible to define two non-Abelian groups $\bar{G}$ and $\bar{K}$ as well as three group homomorphisms $\bar{\varphi}:\bar{G}\to\bar{K}$, $\rho:G\to\bar{G}$, and $\tau:K\to\bar{K}$ such that $\varphi,\rho,\bar{\varphi},\tau$ forms a commutative diagram?  
In short, for given $\varphi: G\to K$, how to obtain the following commutative diagram? 
$$\begin{array}
GG & \stackrel{\varphi}{\longrightarrow} & K \\
\downarrow{\rho} & & \downarrow{\tau} \\
\bar{G} & \stackrel{\bar{\varphi}}{\longrightarrow} & \bar{K}
\end{array}
$$
 A: This can be done in a kind of boring way. Since you did not require that $\bar\varphi$ remains surjective this simplest set-up is to keep $\bar G=G$ but to embed $K$ via $\tau$ into a into a non-Abelian group $\bar K$; $\bar\varphi=\tau\circ\varphi$.
If you want $\bar\varphi$ surjective, let $N$ be any non-Abelian group, put $\bar G=G\times N$, $\bar K=K\times N$, let the maps $\rho,\tau$ be the embeddings into the first component, and $\bar\varphi=\varphi\times1_N$.
This shows that there is no obstruction whatsoever to commutative diagrams as in the question. For a more concrete example, you could take $G=A_4$, and $K$ its quotient by the Klein subgroup $V_4$; since $|K|=3$, it is Abelian (and cyclic). Embedding $G$ in $\bar G=S_4$, the Klein subgroup is still normal, but now the quotient $\bar K$ by it is isomorphic to$~S_3$, not Abelian. You can realise this geometrically with $G$ the rotation of a regular tetrhedron, $K$ the induced permutations of the "axes" joining the midpoints of opposite edges, and $\bar G,\bar K$ the corresponding groups when allowing reflections.
It can of course not be done with a surjective morphism $\tau$, as this forces $\bar K$ to be Abelian.
