Extending a homomorphism $f:\left\to\Bbb T$ to $g:G\to \Bbb T$, where $G$ is abelian and $\mathbb{T}$ is the circle group. Suppose $G$ is an abelian group and $a\in G$ and
$$f:\left<a \right>\to\Bbb T$$
is a homomorphism. Can $f$ be extended to a homomorphism on $G$:
$$g:G\to \Bbb T$$
?
$\Bbb T$ is the circle group.
 A: Sure.  First consider the following construction, which works for general $H\leqslant G$.

Let $x\in G\setminus H$ and $d$ be the smallest positive integer so that $x^d\in H$.  Then $x^n\in H \iff d\mid n$.  Let $y\in\mathbb{C}$ be a $d^\text{th}$ root of $f(x^d)$, and note that $y\in\mathbb{T}$ since $f$ goes into $\mathbb{T}$.  Let $K=\langle x,H\rangle$, so every $k\in K$ may be written as $k=x^n h$ for some $h\in H$.  Define $g:K\rightarrow \mathbb{T}$ by $g(x^nh)=y^nf(h)$.
If our definition of $g$ makes sense, it is clearly a homomorphism, so it remains to be shown that $g$ is well defined. Suppose $x^nh=x^m\hat{h}$.  Then $\hat{h}h^{-1}=x^{n-m}\in H$ so $d\mid n-m$.  Write $n-m=qd$.  Then $$f(\hat{h})f(h)^{-1}=f(x^{n-m})=f(x^d)^q=y^{dq}=y^{n-m},$$ from which we ascertain that $y^nf(h)=y^mf(\hat{h})$.  Thus $g$ is well-defined.

So, first apply the above process to $H=\langle a \rangle$ with a random $x\in G\setminus H$.  The resulting homomorphism goes from $\langle x,H\rangle \rightarrow\mathbb{T}$, so we may apply the process using $\langle x,H\rangle$ and another random $x\in G\setminus \langle x,H\rangle$.  We repeat this process until we obtain a homomorphism $G\rightarrow \mathbb{T}$.
A: As mentioned in the other answer and comments, this is true and works even more generally. 
Suppose that $A$ is a divisible abelian group. If $G$ is abelian and $H \leq G$, then any homomorphism $f: H \rightarrow A$ can be extended to a homomorphism $g: G \rightarrow A$. As seen in the other answer, this can be proven with Zorn's lemma.
In fact, we can prove that for abelian groups, this type of extension is possible only if the target group is divisible. Suppose that $A$ is an abelian group such that for all $G$ abelian, $H \leq G$, any homomorphism $f: H \rightarrow A$ can be extended to a homomorphism $g: G \rightarrow A$. Consider the identity map $f: A \rightarrow A$. Now $A$ can be embedded to a divisible abelian group $D$ (see below), so we can extend $f$ to a surjective homomorphism $g: D \rightarrow A$. Thus $A$ is divisible as a quotient of a divisible group.

Proof that any abelian group can be embedded in a divisible group: If $A$ is abelian, then $A \cong H/K$ where $H$ is a direct sum of a certain amount of copies of $\mathbb{Z}$. Now $H$ can be embedded in a divisible group $D$ (take $D$ to be a direct sum of certain amount of copies of $\mathbb{Q}$). Then $H/K$ embeds into $D/K$ which is a divisible abelian group since $D$ is.
A: 
Let $G'$ be an abelian divisible group, $G$ be an abelian group, $H_0\le G$ and $f_0:H_0\to G'$ be a homomorphism. Then $f_0$ can be extended to a homomorphism $g_0:G\to G'$.

Sketch of a proof: Let $S$ be the set of all homomorphisms of the form $g:H\to G'$ which are extensions of $f_0$. $S$ can be partially ordered by
$$g_1\le g_2\leftrightarrow (g_2\text{ is an extension of }g_1)$$
part 1
One will prove that the domain of a maximal element of $S$ is $G$. To show this let $f:H\to G'$ be such a maximal element and $H\ne G$. Then there's some $x\in G\setminus H$. Let
$$K=\left<\{x\}\cup H\right>$$
one will show that $f$ can be extended to a homomorphism $g:K\to G'$. To prove this, let
$$A=\{ n\in  \Bbb N \mid x^n\in H\}$$
If $A$ is nonempty try to construct $g$ as shown in another anwser in this topic. Else use a definition similar to that answer.
part 2
Use Zorn's Lemma to prove that $S$ has a maximal element. Let $T$ be a chain in $S$. Define
$$g=\bigcup_{f\in T}f$$
Show that $g$ is a function. Extend $g$ to a homomorphism $h$ on $\left<  \text{Dom}(g) \right>$. $h$ will be an upper bound for $T$.
So $S$ has an maximal element $g_0$. According to part 1, it is of the form
$$g_0:G\to G'$$
and according to how $S$ is defined it is an extension of $f_0$.
