In algebra we have the following problem to solve:

Describe the groups of homomorphisms of abelian groups.

(a) $\textrm{Hom}(\mathbb{Q} / \mathbb{Z}, \mathbb{Q})$

(b) $\textrm{Hom}(\mathbb{Q}, \mathbb{Q} / \mathbb{Z})$

So first of all, I'm not sure, if I really understand what to do. I assume that I just have to describe the group $\textrm{Hom}(\mathbb{Q/Z,Q})$, This means that I have to define a group operation and then prove that it satisfies the group axioms, like closedness, associativity, existence of inverses element, existence of an identity element, etc. Is this idea correct?

If so, I would define the operation like this: let $f,g \in \textrm{Hom}(\mathbb{Q/Z,Q})$ then: $$(f\odot g)(x):=f(x)+g(x), \qquad \forall x\in \mathbb{Q}.$$

to check that this operation is closed, we have to show that $$f\odot g \in \textrm{Hom}(\mathbb{Q/Z,Q}), \qquad \forall f,g \in \textrm{Hom}(\mathbb{Q/Z,Q})$$

Proof: Let $f,g \in \forall f,g \in \textrm{Hom}(\mathbb{Q/Z,Q})$. Then,

$$\begin{array}{rclcl} (f\odot g)(x+y) &=& f(x+y)+g(x+y) & \qquad & \text{(by def of $\odot$)} \\ &=& f(x)+f(y)+g(x)+g(y) & & \\ &=& f(x)+g(x)+f(y)+g(y) & & \\ &=& (f\odot g)(x)+(f\odot g)(y). & & \text{($\forall x,y \in \mathbb{Q/Z}$)} \end{array}$$

This follows only if $\mathbb{Q}$ is abelian, which is obviously true. Also from this fact follows that: $$(f\odot g)(x)=f(x)+g(x)=g(x)+f(x)=(g\odot f)(x), \qquad \forall x \in \mathbb{Q/Z},$$ and hence $\textrm{Hom}(\mathbb{Q/Z,Q})$ is abelian.

It remains only to show that there exist inverses and an identity element. As identity element we could take the embedding $e: \mathbb{Q/Z} \hookrightarrow \mathbb{Q}.$

For the inverse I did this: $(f\odot g)(x)=e(x)=x$, $\forall x \in \mathbb{Q/Z}$ , so $$(f\odot g)(x)=f(x)+g(x)=x$$ and so $f(x)^{-1}:=g(x)=x-f(x)$ should be the inverse. Is this correct, or am I completely on the wrong trail?

  • $\begingroup$ I think that the exercise assumes as a know fact that for abelian groups $A,B$, $\hom(A,B)$ is naturally embedded with an abelian group structure. The goal of the exercise is to give an expression of the groups. $\endgroup$ – Pece Oct 9 '14 at 6:57
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    $\begingroup$ I don't know exactly what you mean by "expression" of the group, could you explain it a bit $\endgroup$ – Bobby Oct 9 '14 at 7:03
  • $\begingroup$ What is the embedding $\mathbb{Q} / \mathbb{Z} \hookrightarrow \mathbb{Q}$? $\endgroup$ – Travis Oct 9 '14 at 7:13
  • $\begingroup$ the embedding $e:\mathbb{Q/Z} \hookrightarrow \mathbb{Q}$ just takes values $x \in \mathbb{Q/Z}$ and maps them to $x$, because $\mathbb{Q/Z} \subset \mathbb{Q}$ $\Rightarrow e(x)=x, \forall x \in \mathbb{Q/Z}$ $\endgroup$ – Bobby Oct 9 '14 at 7:17
  • $\begingroup$ No, it doesn't. Let that $x$ be represented by $a/b$. Then $x+x+\cdots+x$ $b$ times equals zero in $\Bbb Q/\Bbb Z$. $\endgroup$ – Arthur Oct 9 '14 at 7:23

If $A$ and $B$ are abelian groups, then $\operatorname{Hom}(A,B)$ is an abelian group under the obvious operations and I don't think you have to discuss them, because it's general knowledge.

If $A$ is a torsion abelian group and $B$ is a torsion free abelian group, then $\operatorname{Hom}(A,B)=\{0\}$ and this settles the first part. Why is that? If $f\colon A\to B$ is a homomorphism and $x\in A$, choose $n>0$ such that $nx=0$; then $0=f(0)=f(nx)=nf(x)$; since $B$ is torsion free, this implies $f(x)=0$.

A good starting point for the second part is to consider the exact sequence $$ \def\Z{\mathbb{Z}}\def\Q{\mathbb{Q}} \def\Hom{\operatorname{Hom}}\def\Ext{\operatorname{Ext}} 0\to\Z\to\Q\to\Q/\Z\to0 $$ and apply the $\Hom(\Q,-)$ functor, which gives the exact sequence $$ 0=\Hom(\Q,\Z)\to\Hom(\Q,\Q)\to\Hom(\Q,\Q/\Z)\to\Ext(\Q,\Z)\to\Ext(\Q,\Q)=0 $$ Now, what's $\Ext(\Q,\Z)$? It's isomorphic to the additive group of the real numbers, as shown by Wiegold (Bull. Austral. Math. Soc. 1 (1969), 341–343).

Since $\Hom(\Q,\Q)\cong\Q$ is divisible, the sequence splits, so we get that $$ \Hom(\Q,\Q/\Z)\cong \Q\oplus\mathbb{R}\cong\mathbb{R}. $$

  • $\begingroup$ why $\mathbb{Q} \oplus \mathbb{R} \cong \mathbb{R} $ ? $\endgroup$ – WLOG Oct 9 '14 at 13:10
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    $\begingroup$ @WLOG Because $\mathbb{R}$ is an infinite dimensional vector space over $\mathbb{Q}$, so adding a copy of $\mathbb{Q}$ gives an isomorphic vector space. There's no “canonical” isomorphism, though. $\endgroup$ – egreg Oct 9 '14 at 13:12
  • $\begingroup$ thanks a lot for your answer $\endgroup$ – Bobby Oct 11 '14 at 9:21

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