# Direct sum and direct product of infinitely many abelian groups are not isomorphic

Let $I$ be an infinite set, and for each $i$ let $A_i$ be an abelian group with order $o(A_i) \ge 2$. Prove that the direct product $\prod A_i$ and the direct sum (coproduct) $\bigoplus A_i$ are not isomorphic.

Here the product and coproduct are taken in the category of abelian groups. So an element of the direct product is a "list" of one element from each group, and an element of the coproduct is such a list where all but finitely many terms are zero.

This problem appeared on an algebra problem set a couple weeks ago and I haven't been able to solve it then or since then. Any ideas?

An initial observation is that the direct sum is of course a subgroup of the direct product. But the direct product could be a subgroup of the direct sum as well! For example let $I = \mathbb{N}$, and let $A_1 = B \times B \times B \times \cdots$ where $B = A_2 \times A_3 \times A_4 \times \cdots$. Then $$\prod A_i = A_1 \times B = (B \times B \times B \times \cdots) \times B = A_1 \subset \bigoplus A_i$$

This causes many of my ideas to fail; for instance I cannot rely on there existing a generating set of a certain cardinality, nor can I argue there are a certainly cardinality of subgroups, nor in general can I make any sort of argument regarding the size of the two groups.

It also seems promising to appeal directly to the universal mapping properties satisfied by the product and coproduct. The problem I ran into in this case was that the projections from the direct product to the individual $A_i$s and the inverse projections from the $A_i$s to the direct sum need not have anything to do with each other. Also, this approach would somehow have to make use of the fact that $I$ is infinite...

• The claim is not correct (see my answer). – Martin Brandenburg Oct 8 '14 at 7:38

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.)