I was reading this post of stack exchange. So in the question if the circle group is replaced by $\mu_{p-1}$ which is the group of $(p-1)^{th}$ root of unity and if the group $G/H$ is assumed to a cyclic group, then the answer given by Alexander gruber shows that we can extend the morphism to $G$. This is because
Let $<x>=G/H $. Let the order of $x$ in $G$ be $d$. That is $x^d=1$. Then $d$ be the smallest positive integer such that $x^d \in H$. Let $f$ be a homomorphism from $H$ to $\mu_{p-1}$. Then $f(x^d)=1$. So we can define $g:G \rightarrow \mu_{p-1}$ by $g(x)=1$ and $g(h)=f(h)$ which will extend $f$. Hence we can extend a morphism $f$ to $G$ where the target group is not a divisble group.
Is my argument based on the above post right? Thanks in advance.