For example. In Lang's Algebra there's a theorem:
Theorem 8.1 Let $A$ be a torsion abelian group. Then $A$ is the direct sum of
its subgroups $A(p)$ for all primes $p$ such that $A(p) \neq 0$.Proof. There is a homomorphism
$\bigoplus_p A(P)\rightarrow A, \ $ which to each element $(x_p)$ in the direct sum associates the element $\sum_p x_p$ in $A$. We prove that this homomorphism is both surjective and injective….
So, reading the theorem it sounds to me like they're talking about the internal direct product. But then they talk about the tuple $(x_p)$ which means they're talking about the external direct product. Is there an isomorphism?
Thanks.
By internal direct sum, I mean the subgroup formed (additively) by $\sum_i A_i = A$, where $A_i$ are subgroups of $A$ and $A_i + A_j = \{a_i + a_j : a_i\in A_i, a_j \in A_j\}$ in the sum. Additionally $A_{i+1} \cap (A_1 + \dots + A_i) = \{0\}$. So I guess the only definition I know if is the countable internal direct sum.
Best Answer
There's an isomorphism between an internal direct some and it's naturally corresponding external direct sum. Let $A = \oplus_i A_i$ internally and write $\bigoplus_i A_i$ to mean external direct sum. Define $A \rightarrow \bigoplus_i A_i$ by $x_1 + x_2 + \dots \rightarrow (x_1, x_2, \dots )$. The mapping is a homomorphism. The kernel is clearly trivial, and it's onto. So you can look at the direct sum either way.
This should work for countable index set for $i$. But the notation is wrong for an arbitrary direct sum.