Extending of isomorphism $f: SO(2, \mathbb{R}) \rightarrow \mathbb{T}$ to a homomorphism $GL(2, \mathbb{R}) \rightarrow \mathbb{T}$. In Arhangel'skii-Tkachenko, Topological groups and related structures on page 6 there is a theorem 1.1.6:

Let $H$ be a subgroup of an Abelian group $G$. Then every homomorphism $f$ of $H$ to any divisible group $F$ can be extended to a homomorphism of $G$ to $F$. 

I am thinking about an example of non-commutative group $G$ to demonstrate that if we remove assumption that $G$ is Abelian, then the above theorem is no longer valid. 
I have an idea to start with well-known isomorphism $f: SO(2, \mathbb{R}) \rightarrow \mathbb{T}$ where $\mathbb{T}=\{z \in \mathbb{C} : |z|=1\}$ (multiplicative group).
My hypothesis is that there is no extension of isomorphism $f$ to some homomorphism of $GL(2, \mathbb{R})$ to $\mathbb{T}$, but I am not able to prove or reject it.
 A: The derived subgroup of $GL(2,\mathbb R)$ equals $SL(2,\mathbb R)$.   Thus any homomorphism from $GL(2,\mathbb R)$ to $T$
contains $SL(2,\mathbb R)$ in its kernel.  In particular, it contains
$SO(2,\mathbb R)$.  Thus no non-notrivial homomorphism from $SO(2,\mathbb R)$ to $T$ can be extended to $GL(2,\mathbb R)$.
A: It seems the following. 
We can construct a required counterexample as follows. Let $H$ be an arbitrary divisible group such that the group $Hom(H)$ of homomorphisms of the group $H$ contains a non-inner homorphism $\alpha$ (that is, there is no element $g\in H$ such that $\alpha(h)=g^{-1}hg$ for each element $x\in H$) (for instance, we may take $H=\mathbb R$ and $\alpha(x)=-x$ for each $x\in H$). Let $G$ be a semidirect product of the groups $H$ and $Hom(H)$, that is a direct product $H\times Hom(H)$ endowed with a multiplication $(h_1,\alpha_1) (h_2,\alpha_2)=
(h_1\alpha_1(h_2), \alpha_1\alpha_2)$ for every elements $h_1, h_2\in H$ and 
$\alpha_1,\alpha_2\in Hom(H)$. Put $g_0=(e,\alpha^{-1})$. Then $g_0^{-1}hg_0=\alpha(h)$ for each element $h\in H$. Let $f:H\to H$ be the indentity isomorphism. Suppose that there exists a homomorphic extension $f’:G\to H$ of the homomorphism $f$. Put $g=f'(g_0)$. Then 
$g^{-1}hg=\alpha(h)$ for each element $h\in H$, a contradiction.
