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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.

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    $\begingroup$ I think this is true, and even more generally when $\langle a \rangle$ is replaced by any subgroup of $G$ and $\mathbb{T}$ is replaced by any divisible abelian group. $\endgroup$ – Mikko Korhonen May 4 '13 at 17:00
  • $\begingroup$ @m.k. That must be true, it seems my proof below works under only those assumptions. I wonder if there is a sufficient condition. $\endgroup$ – Alexander Gruber May 4 '13 at 19:36
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    $\begingroup$ @AlexanderGruber: How about this? Let $A$ be an abelian group such that for any abelian group $G$ and subgroup $H \leq G$, any homomorphism $f: H \rightarrow A$ can be extended to a homomorphism $g: G \rightarrow A$. Now consider the identity map $f: A \rightarrow A$. We can embed $A$ (any abelian group, in fact) into a divisible group $D$, so we can extend $f$ to a surjective homomorphism $g: D \rightarrow A$. Quotients of divisible groups are divisible, so $A$ is divisible. Thus for abelian groups, this type of homomorphism extension is possible if and only if the target group is divisible. $\endgroup$ – Mikko Korhonen May 4 '13 at 20:26
  • $\begingroup$ you can add your comment as an answer if you are willing..you can edit and extend my question if necessary for forum rules. $\endgroup$ – user59671 May 4 '13 at 20:33
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    $\begingroup$ As speculated in other comments, and as carried out in answers, the point is that divisible abelian groups are injective in the category of abelian groups. This was maybe first noted by Reinhold Baer about 1940, and is called Baer's criterion. $\endgroup$ – paul garrett Sep 27 '13 at 22:15
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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 $G\rightarrow \mathbb{T}$.

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    $\begingroup$ We repeat?! It is needed to be at most a countable process?! is it? $\endgroup$ – user59671 May 4 '13 at 17:17
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    $\begingroup$ @CutieKrait In general this requires an application of Zorn's lemma. With it you can surely get a "maximal extension" of $f$ to some subgroup of $G$; then Alexander's reasoning shows this subgroup must be the whole of $G$, otherwise maximality would fail. $\endgroup$ – egreg May 4 '13 at 17:58
  • $\begingroup$ @egreg ahh thanks. $\endgroup$ – user59671 May 4 '13 at 18:47
  • $\begingroup$ Yu the Great $\endgroup$ – user59671 May 6 '13 at 8:59
  • $\begingroup$ Why such $d$ must exist? $\endgroup$ – Andre Gomes Feb 19 at 18:05
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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.

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  • $\begingroup$ The last paragraph assumes that $A$ is finitely generated. $\endgroup$ – Alex B. May 15 '13 at 12:37
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    $\begingroup$ @AlexB.: Let $H = \oplus_{a \in A} \mathbb{Z}$, ie. the free abelian group with basis $A$. Then there exists a surjective homomorphism $H \rightarrow A$. Also, we can choose $D = \oplus_{a \in A} \mathbb{Q}$. $\endgroup$ – Mikko Korhonen May 15 '13 at 18:48
  • $\begingroup$ Perhaps I am being dense. Let $A=\prod_{i\in \mathbb{N}}\mathbb{Z}$, i.e. the direct product of countably infinitely many copies of $\mathbb{Z}$. What is $H$? $\endgroup$ – Alex B. May 16 '13 at 23:24
  • $\begingroup$ Just like before, $H = \oplus_{a \in A} \mathbb{Z}$, the direct sum of $|A|$ copies of $\mathbb{Z}$, ie. the free abelian group with basis $A$. $\endgroup$ – Mikko Korhonen May 19 '13 at 16:56
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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 $f_0$.

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    $\begingroup$ I want this answer be deleted. $\endgroup$ – user79193 Sep 30 '13 at 14:38

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