I'm trying to generaliz this proposition: Let $G$ an abelian group and $D$ a divisible grop, for every $H\leq G$ and every homomorphism $f:H\to D$ exists $\overline{f}:G\to D$ an extension of $f$.
I want to prove in the case when $G$ is non-abelian and $H$ is a normal subgroup of $G$.
I was trying to imitate the proof of the clasic result (I put this proof below as a picture) but I have a problem in the unique expressions as sumands of the elements.
Any help or hint is welcome.
It is no longer true for non-abelian groups.
Take any nilpotent group $G$ with $G_2 = \mathbb{Q}$. For instance take $$G:=\begin{bmatrix} 1& \mathbb{Q} &\mathbb{Q} \\ 0 &1 &\mathbb{Q} \\ 0& 0& 1\end{bmatrix}$$ then $$G_2=\begin{bmatrix} 1& 0 &\mathbb{Q} \\ 0 &1 &0 \\ 0& 0& 1\end{bmatrix}$$
Now let $H=G_2$ (it is classical that this group is normal, see e.g. How to show that the commutator subgroup is a normal subgroup) and define $\phi:G_2\rightarrow \mathbb{Q}$ be the projection to the coordinate in the top right corner. That is clearly a homomorphism.
However, any homomorphism $G\rightarrow \mathbb{Q}$ must be trivial on $G_2$ (because the image is abelian). Therefore one can not extend $\phi$ to $G$.