7
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
3
$\begingroup$

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 $$ \Hom(\Q,\Q/\Z)\cong \Q\oplus\mathbb{R}\cong\mathbb{R}. $$

$\endgroup$
  • $\begingroup$ why $\mathbb{Q} \oplus \mathbb{R} \cong \mathbb{R} $ ? $\endgroup$ – WLOG Oct 9 '14 at 13:10
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.