Let $(G,+)$ be an additive abelian group.

Let us suppose $G$ divisible (i.e. $G=nG\;\;\;\forall n\ge1$). Let then $x,y\in G$. Then there exists $z\in G$ and $n,m\ge1$ such that $x=nz$ and $y=mz$.

Being $x\in G=nG$, we have that for every $n\ge1,\;\;\exists h_n\in G$ such that $x=nh_n$. Similarly we have that $y=mk_m\;\;\;\forall m\ge1$.

It easily seen that, being $x,n$ fixed, we have that $h_n$ is unique (idem for $k_m$).

Thus the claim is true iff $\{h_n\}_{n\ge1}\cap\{k_m\}_{m\ge 1}\neq\emptyset$.

From this I can't go on. Can someone help me please?

EDIT: the problem above comes from what follows.

We must show that

$(G,+)$ abelian group is divisible $\Longleftrightarrow$ it's an homomorphic image of $\Bbb Q^{(X)}$ for some set $X$.

Thus I began to show "$\Rightarrow$". I built $\varphi:\Bbb Q^{(X)}\to G$ as follows (observe that $\Bbb Q^{(X)}=\{f:X\to\Bbb Q\;:\;f(x)=0\; \mbox{for all except a finite number of}\;x\in X\}$): fix an arbitrary $g\in G$; let then $f\in \Bbb Q^{(X)}$; write $q_f:=\sum_{x\in X}f(x)=\frac ab\in\Bbb Q$. So take the unique $h\in G$ s.t. $g=bh$; finally define $\varphi(f):=ah$.

I proved that such a $\varphi$ is well defined, and that it is a group homomorphism ($\Bbb Q^{(X)}$ is a group wrt the pointwise sum). My problem is to show that $\varphi$ is onto. And, if you write down some detail, it should follows that surjectivity of $\varphi$ is equivalent my initial claim.

  • 2
    $\begingroup$ It seems that $G = \mathbb{Q}$, $x = 1$, $y = -1$ is a counter-example? Am I missing something? $\endgroup$ – Marc Paul Jan 14 '16 at 21:37
  • $\begingroup$ $G$ is divisible iff $G=nG$ for every $n\ge1$. But it is clearly equivalent to consider $n\in\Bbb Z,\;n\neq0$, thus, the $n$'s and $m$'s above are nonzero integers. Thus, your counterexample doesn't work. Do you agree? $\endgroup$ – Joe Jan 14 '16 at 22:18
  • $\begingroup$ I understand what you mean. Perhaps $G = \mathbb{Q}\times \mathbb{Q}$, $x = (1, 0)$, $y = (0,1)$ would be a more satisfying counterexample? $\endgroup$ – Marc Paul Jan 14 '16 at 22:32
  • $\begingroup$ See my edit Marc! However thanks! $\endgroup$ – Joe Jan 14 '16 at 22:45
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    $\begingroup$ You're welcome! Considering your edit: it could be useful to run your argument for the specific case of $G = \mathbb{Q} \times \mathbb{Q}$ mentioned before, to see why the map $\varphi$ you constructed is indeed not always onto. By the way, it is not always the case that the $h \in G$ you pick above is unique! An example of a group in which this fails is the group $\mathbb{Q}/\mathbb{Z}$ (i.e. addition of fractions modulo $1$). $\endgroup$ – Marc Paul Jan 14 '16 at 22:58

We will prove that every abelian divisible group $(G, +)$ is the homomorphic image of $\mathbb{Q}^{(X)}$ for some set $X$. The first thing you will want to do is select a set $X$. After all, not every $X$ will work. We define $X = G$, and we will show that there exists a surjective homomorphism $\mathbb{Q}^{(G)} \to G$.

We will begin by defining a map $\psi: \mathbb{Z}^{(G)} \to G$ by setting $\psi(f) = \sum_{g \in G} f(g)\cdot g$ for every $f \in \mathbb{Z}^{(G)}$. This is a well-defined map, since every such $f$ is equal to $0$ almost always (so the sum in the definition is finite) and $G$ is abelian (so the order of the summation doesn't matter). It is routine to show that $\psi$ is a homomorphism, and we easily see that $\psi$ is surjective.

Now we have a homomorphism $\mathbb{Z}^{(G)} \to G$. But we also have an inclusion $\mathbb{Z}^{(G)} \hookrightarrow \mathbb{Q}^{(G)}$. Since $G$ is divisible, it is injective in the category of abelian groups, so $\psi$ extends to a group homomorphism $\phi: \mathbb{Q}^{(G)} \to G$ (assuming the axiom of choice). Since $\psi$ was already surjective, $\phi$ will be as well.


Note that your attempt is doomed to failure: division in $G$ is not necessarily unique. It is only if $G$ is torsion-free.

Consider all homomorphisms $\mathbb{Q}\to G$ and index them; you get the family $(f_i)_{i\in I}$, for some set $I$. This defines a homomorphism $$ \varphi\colon \mathbb{Q}^{(I)}\to G $$ by $$ \varphi\colon (x_i)_{i\in I}\mapsto \sum_{i\in I}f_i(x_i) $$ The image of $\varphi$ is a divisible subgroup $H$ of $G$, hence it is a direct summand: $G=H\oplus K$, for some $K$. Now it's a question of showing $K=0$.

Thus we are reduced to prove that, if $K$ is a nonzero divisible group, there exists a nonzero homomorphism $f\colon\mathbb{Q}\to K$. Indeed, once proved this, if $K\ne0$ we'd get a homomorphism $\mathbb{Q}\to G$ whose image is not contained in $H$, which is a contradiction.

Now we can use the fact that the torsion part $t(K)$ of $K$ is divisible, hence a direct summand. If $K/t(K)$ is nonzero, it is a divisible torsion-free group, so we find a nonzero homomorphism $\mathbb{Q}\to K/t(K)$; using the fact that $t(K)$ is a summand, we are done.

So we can assume $K$ is torsion. It is easy to see that $K$ can be written as the direct sum of its $p$-components: for $p$ a prime, the $p$-component $t_p(K)$ is the set of $x\in K$ such that $p^nx=0$, for some $n\ge0$. As $K$ is nonzero, we have that $t_p(K)\ne0$, for some prime $p$. Hence we are reduced to the case $K$ is a $p$-group and we can make a further reduction to the case $K$ is indecomposable.

Now, up to isomorphisms, the only indecomposable divisible $p$-group is the Prüfer group $\mathbb{Z}(p^\infty)$ which is the $p$-component of $\mathbb{Q}/\mathbb{Z}$. Therefore we have our nonzero homomorphism from $\mathbb{Q}$.

  • $\begingroup$ You can also use divisible abelian groups are injective. $\endgroup$ – Pedro Tamaroff Jan 15 '16 at 0:06
  • $\begingroup$ @PedroTamaroff I used it implicitly: a divisible group is a summand of every group it's embedded in. $\endgroup$ – egreg Jan 15 '16 at 0:17
  • $\begingroup$ That's true. I was thinking more of the other answer, however. $\endgroup$ – Pedro Tamaroff Jan 15 '16 at 0:22
  • $\begingroup$ @PedroTamaroff This one is, basically, the structure theorem for divisible abelian groups. The final part is not straightforward, but I think this is more informative, particularly for better showing what really goes wrong in the OP's attempt. $\endgroup$ – egreg Jan 15 '16 at 0:27
  • $\begingroup$ I agree. =) ${}{}{}{}{}{}{}$ $\endgroup$ – Pedro Tamaroff Jan 15 '16 at 0:34

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