I am looking for a reference for the fact that an Abelian group is divisible if and only if its isomorphic to a direct sum of $\mathbb Q$ and quasicyclic groups so I can study the proof. Can someone suggest me a reference?
[Math] Abelian group is divisible if and only if its isomorphic to a direct sum of $\mathbb Q$ and quasicyclic groups
abstract-algebragroup-theoryreference-request
Related Solutions
A more natural problem would be to decide when the canonical morphism $\bigoplus_i A_i \to \prod_i A_i$ is an isomorphism. Clearly, this is the case if almost all $A_i$ are trivial. Notice that often when people say that $X,Y$ are not isomorphic, they really mean that a certain canonical morphism $X \to Y$ is not an isomorphism.
The question if there is any "random" isomorphism between $\bigoplus_i A_i$ and $\prod_i A_i$ is not that natural, because this isomorphism may just ignore the inclusions and the projections of the direct sum resp. direct product and therefore can be "wild".
Let us replace for the moment the category of abelian groups with the category of vector spaces over some fixed field $k$. Let $V_i$ be an infinite-dimensional vector space. Then it is known (MO/49551) that $\dim(\prod_i V_i)=\prod_i |V_i|$, where $|V_i|$ is the cardinality of (the underlying set of) $V_i$. Clearly, we also have $\dim(\oplus_i V_i) = \sum_i \dim(V_i)$. It is also well-known (math.SE/194281) that $|V_i|=\max(\dim(V_i),|k|)$. So the question if $\bigoplus_i V_i$ and $\prod_i V_i$ are isomorphic really only depends on the cardinal number equation $\sum_i d_i = \prod_i \max(d_i,q)$, where $q=|k|$ and $d_i=\dim(V_i)$. (The whole algebraic structure has disappeared!) Let us assume for the moment that $k$ is countable, i.e. $q \leq \aleph_0$. Then the equation becomes $\sum_i d_i = \prod_i d_i$. If $d:=d_i$ is constant and $|I|=\aleph_0$, the equation becomes $d = d^{\aleph_0}$. And this is perfectly possible, for example when $d=2^{\aleph_\alpha}$. Of course there are also examples when $d_i$ is not constant and where $I,k$ are arbitrary. (On the other hand, for some cardinals $d$, the equation $d=d^{\aleph_0}$ might as well be independent from ZFC!)
So you see, there are many families of infinite-dimensional vector spaces $V_i$ with $\bigoplus_i V_i \cong \prod_i V_i$, for example $V_i=\mathbb{R}$ over the base field $\mathbb{Q}$. In particular, their underlying abelian groups are isomorphic, contradicting the claim. (Notice that you cannot really write down this isomorphism, and therefore it is of no practical importance.)
So the fundamental theorem of abelian groups says that if $G$ is a finite abelian group then $G\cong \Pi \mathbb{Z}_{p_i^{n_i}}$ with $\vert G \vert = \Pi p_i^{n_i} $ where the product is taken over the index $i$. Overall it is clear that we need to have $p$ show up as many times as it does in the prime factorization of the size of $G$. As a more concrete example all of the abelian groups of order 4 are $\mathbb{Z}_{4}$ and $\mathbb{Z}_2 \times \mathbb{Z}_2$. So in your case you are looking for the different ways to split 2 and 5 up so that the total number of them appearing is 3. The easiest 2 ways would be $\mathbb{Z}_2 \times\mathbb{Z}_2 \times\mathbb{Z}_2 \times\mathbb{Z}_5 \times\mathbb{Z}_5 $ and $\mathbb{Z}_8 \times \mathbb{Z}_{25}$ thought this isn't an exhaustive list.
Best Answer
You should find it in Fuchs' “Infinite Abelian Groups” (volume 1).
However, the proof is not really so difficult. Here's a sketch.
Let $G$ be a divisible abelian group. Then the torsion part $t(G)$ is divisible as well and so it splits: $G\cong t(G)\oplus (G/t(G))$. Since $G/t(G)$ is torsionfree, it has unique division by integers and so it becomes a vector space over $\mathbb{Q}$, hence a direct sum of copies of $\mathbb{Q}$.
It remains to analyze $t(G)$, which can be written as the direct sum of the $p$-components, like every torsion group. Hence we are reduced to a torsion divisible $p$-group ($p$ a prime).
This is perhaps the most difficult part. But some module theory can help: since $\mathbb{Z}$ is noetherian, every injective module is a direct sum of indecomposable modules. Thus we just need to characterize the indecomposable divisible $p$-groups (injectivity on $\mathbb{Z}$-modules is the same as divisibility).
In an indecomposable divisible $p$-group the socle must be simple and so the group is the injective envelope of $\mathbb{Z}/p\mathbb{Z}$ which is easily seen to be the Prüfer $p$-group.