In "A Basic Course in Algebraic Topology" by Massey, the weak product (or direct sum) of a family $\{G_i\}_{i \in I}$ of groups is defined to be the subgroup of their product direct $\prod_{i \in I} G_i$ consisting of all elements $g \in G$ such that $g_i$ is the identity element of $G_i$ for all except a finite number of indices $i$.
Massey also mentions that for any finite indexing set $I$, the direct product and the direct sum are the same. (Not sure if he means "same" as in the underlying sets are equal or if the groups are isomorphic)
If we let $(\mathbb{Z}, +)$ denote the additive group of integers, then let $(\mathbb{Z} \times \mathbb{Z}, +_{\times})$ denote the direct product, and if we pick $(a, b), (c, d) \in (\mathbb{Z} \times \mathbb{Z}, +_{\times})$, then $(a, b) +_{\times} (c, d) = (a + c, b +d) \in (\mathbb{Z} \times \mathbb{Z}, +_{\times})$.
But by the definition above $\mathbb{Z} \oplus \mathbb{Z} = \{ (0, n) | n \in \mathbb{Z}\}$, since $0$ is the identity of $(\mathbb{Z}, +)$, clearly $\mathbb{Z} \oplus \mathbb{Z} \neq \mathbb{Z} \times \mathbb{Z}$, and neither are they isomorphic.
I must have misinterpreted the definition, or made a mistake in the example, where have I gone wrong?
Best Answer
It is not true that $\mathbb Z \oplus \mathbb Z = \{(0,n) \mid n \in \mathbb Z\}$; in fact $\mathbb Z \oplus \mathbb Z = \mathbb Z \times \mathbb Z$ in this case, because an arbitrarily element $(n,m) \in \mathbb Z \times \mathbb Z$ is non-zero on a finite number of indices: namely, at most the index set $\{1,2\}$.
The definition is just not interesting when you are "summing" a finite number of groups, since then there are a finite number of indices, and thus any element of the (finite) product also belongs to the (finite) sum. However, if you consider for example $$ \prod_{n=1}^\infty \mathbb Z/n\mathbb Z $$ then this contains, say, the element $(1, 1, 1, 1, 1, \ldots)$, while the sum $$ \bigoplus_{n=1}^\infty \mathbb Z/n\mathbb Z $$ does not contain it, since it is non-zero on an infinite index set: namely, all of $\mathbb N$.